Question 6

   Question: Let $\vec{a}$ and $\vec{b}$ be two unit vectors such that the angle between them is $\frac{\pi}{3}$. If $\lambda \vec{a} + 2 \vec{b}$ and $3 \vec{a} - \lambda \vec{b}$ are perpendicular to each other, then the number of values of $\lambda$ in $[-1, 3]$ is :

   Options: 

    A. 1 

    B. 3 

    C. 2 

    D. 0

   Correct Answer: D

   Year: JEE Main 2025 (Online) 22nd January Evening Shift

   Solution: We are given two unit vectors $\vec{a}$ and $\vec{b}$ with an angle of $\pi/3$ between them. This means $\vec{a} \cdot \vec{a} = \vec{b} \cdot \vec{b} = 1$ and $\vec{a} \cdot \vec{b} = \cos(\pi/3) = 1/2$. We need to find the number of values of $\lambda$ in the interval $[-1, 3]$ for which $(\lambda \vec{a} + 2 \vec{b}) \cdot (3 \vec{a} - \lambda \vec{b}) = 0$. Expanding: $3\lambda(\vec{a} \cdot \vec{a}) - \lambda^2(\vec{a} \cdot \vec{b}) + 6(\vec{b} \cdot \vec{a}) - 2\lambda(\vec{b} \cdot \vec{b}) = 0$. Substituting values: $3\lambda - \lambda^2/2 + 3 - 2\lambda = 0$. This simplifies to $\lambda - \lambda^2/2 + 3 = 0$, or $\lambda^2 - 2\lambda - 6 = 0$. Solving for $\lambda$ gives $1 \pm \sqrt{7}$. $\lambda \approx 3.6458$ and $\lambda \approx -1.6458$. Neither value lies in $[-1, 3]$.

   Step Solution:

    1.  Determine dot product of unit vectors: $\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}| \cos(\pi/3) = (1)(1)(1/2) = 1/2$.

    2.  Set up perpendicularity condition: $(\lambda \vec{a} + 2 \vec{b}) \cdot (3 \vec{a} - \lambda \vec{b}) = 0$.

    3.  Expand and substitute: $3\lambda(1) - \lambda^2(1/2) + 6(1/2) - 2\lambda(1) = 0 \Rightarrow \lambda - \frac{\lambda^2}{2} + 3 = 0$.

    4.  Solve the quadratic $\lambda^2 - 2\lambda - 6 = 0$: $\lambda = \frac{2 \pm \sqrt{4 - 4(1)(-6)}}{2} = \frac{2 \pm \sqrt{28}}{2} = 1 \pm \sqrt{7}$.

    5.  Check the interval: $1 + 2.64 = 3.64$ (out) and $1 - 2.64 = -1.64$ (out). Count is 0.

   Difficulty level: Medium

   Concept Name: Scalar Product (Dot Product)

   Short cut solution: Rapidly expand the dot product into a quadratic in $\lambda$ and use the discriminant/roots to check if they fall within the specified range.

 Question 14

   Question: If the components of $\vec{a} = \alpha \hat{i} + \beta \hat{j} + \gamma \hat{k}$ along and perpendicular to $\hat{b} = 3 \hat{i} + \hat{j} - \hat{k}$ respectively, are $\frac{16}{11} (3 \hat{i} + \hat{j} - \hat{k})$ and $\frac{1}{11} (-4 \hat{i} - 5 \hat{j} - 17 \hat{k})$, then $\alpha^2 + \beta^2 + \gamma^2$ is equal to :

   Options: 

    A. 16 

    B. 23 

    C. 26 

    D. 18

   Correct Answer: C

   Year: JEE Main 2025 (Online) 28th January Evening Shift

   Solution: Any vector can be expressed as the sum of its components along and perpendicular to another vector. Therefore, $\vec{a} = \vec{a}_{||} + \vec{a}_{\perp}$. Adding the given vectors: $\vec{a} = \frac{16}{11}(3\hat{i} + \hat{j} - \hat{k}) + \frac{1}{11}(-4\hat{i} - 5\hat{j} - 17\hat{k})$. This results in $\vec{a} = 4\hat{i} + \hat{j} - 3\hat{k}$. Thus $\alpha=4, \beta=1, \gamma=-3$. The value of $\alpha^2 + \beta^2 + \gamma^2 = 16 + 1 + 9 = 26$.

   Step Solution:

    1.  Combine the components to find $\vec{a}$: $\vec{a} = \text{Component}_{||} + \text{Component}_{\perp}$.

    2.  Calculate the sum: $\vec{a} = \frac{1}{11} [ (48-4)\hat{i} + (16-5)\hat{j} + (-16-17)\hat{k} ]$.

    3.  Simplify to find vector components: $\vec{a} = \frac{44\hat{i} + 11\hat{j} - 33\hat{k}}{11} = 4\hat{i} + \hat{j} - 3\hat{k}$.

    4.  Identify coefficients: $\alpha = 4, \beta = 1, \gamma = -3$.

    5.  Calculate magnitude squared: $\alpha^2 + \beta^2 + \gamma^2 = 4^2 + 1^2 + (-3)^2 = 16 + 1 + 9 = 26$.

   Difficulty level: Easy

   Concept Name: Resolution of a Vector into Components

   Short cut solution: Use the property $|\vec{a}|^2 = |\vec{a}_{||}|^2 + |\vec{a}_{\perp}|^2$. Find the squares of the magnitudes of the two given component vectors and add them directly.

 Question 20

   Question: Consider two vectors $\vec{u} = 3 \hat{i} - \hat{j}$ and $\vec{v} = 2 \hat{i} + \hat{j} - \lambda \hat{k}, \lambda > 0$. The angle between them is given by $\cos^{-1} \left( \frac{\sqrt{5}}{2 \sqrt{7}} \right)$. Let $\vec{v} = \vec{v}_1 + \vec{v}_2$, where $\vec{v}_1$ is parallel to $\vec{u}$ and $\vec{v}_2$ is perpendicular to $\vec{u}$. Then the value $|\vec{v}_1|^2 + |\vec{v}_2|^2$ is equal to

   Options: 

    A. 23/2 

    B. 25/2 

    C. 10 

    D. 14

   Correct Answer: D

   Year: JEE Main 2025 (Online) 4th April Morning Shift

   Solution: First find $\lambda$ using the dot product: $\vec{u} \cdot \vec{v} = |\vec{u}||\vec{v}| \cos \theta \Rightarrow 5 = \sqrt{10} \sqrt{5+\lambda^2} \frac{\sqrt{5}}{2\sqrt{7}}$. Solving this gives $\lambda = 3$. Given $\vec{v} = \vec{v}_1 + \vec{v}_2$ with $\vec{v}_1 \parallel \vec{u}$ and $\vec{v}_2 \perp \vec{u}$, these are orthogonal components. By Pythagoras, $|\vec{v}_1|^2 + |\vec{v}_2|^2 = |\vec{v}|^2$. Calculating $|\vec{v}|^2 = 2^2 + 1^2 + 3^2 = 14$.

   Step Solution:

    1.  Apply dot product formula to find $\lambda$: $5 = \sqrt{10}\sqrt{5+\lambda^2} \cos\theta$.

    2.  Substitute $\cos\theta = \frac{\sqrt{5}}{2\sqrt{7}}$: $5 = \frac{5\sqrt{2}\sqrt{5+\lambda^2}}{2\sqrt{7}} \Rightarrow \sqrt{14} = \sqrt{5+\lambda^2} \Rightarrow \lambda^2 = 9 \Rightarrow \lambda = 3$.

    3.  Identify required sum as the magnitude of $\vec{v}$ squared: $|\vec{v}_1|^2 + |\vec{v}_2|^2 = |\vec{v}|^2$.

    4.  Calculate $|\vec{v}|^2$ using $\lambda = 3$: $|\vec{v}|^2 = (2)^2 + (1)^2 + (-3)^2$.

    5.  Result: $4 + 1 + 9 = 14$.

   Difficulty level: Medium

   Concept Name: Scalar Product and Orthogonal Vector Decomposition

   Short cut solution: Realize that $|\vec{v}_1|^2 + |\vec{v}_2|^2$ is always equal to $|\vec{v}|^2$ when the components are perpendicular. Thus, you only need to find $\lambda$ to find the total magnitude of $\vec{v}$.

Question 21

   Question: Let the angle $\theta, 0 < \theta < \frac{\pi}{2}$ between two unit vectors $\hat{a}$ and $\hat{b}$ be $\sin^{-1} \left( \frac{\sqrt{65}}{9} \right)$. If the vector $\vec{c} = 3\hat{a} + 6\hat{b} + 9(\hat{a} \times \hat{b})$, then the value of $9(\vec{c} \cdot \hat{a}) - 3(\vec{c} \cdot \hat{b})$ is

   Options: 

    A. 31

    B. 29

    C. 24

    D. 27

   Correct Answer: B

   Year: JEE Main 2025 (Online) 7th April Morning Shift

   Solution: Examine vector $\vec{c} = 3\hat{a} + 6\hat{b} + 9(\hat{a} \times \hat{b})$. First find dot products: $\vec{c} \cdot \hat{a} = 3 + 6(\hat{a} \cdot \hat{b})$ and $\vec{c} \cdot \hat{b} = 3(\hat{a} \cdot \hat{b}) + 6$. We need $9(\vec{c} \cdot \hat{a}) - 3(\vec{c} \cdot \hat{b})$. Substituting the values gives $27 + 54(\hat{a} \cdot \hat{b}) - 9(\hat{a} \cdot \hat{b}) - 18 = 9 + 45(\hat{a} \cdot \hat{b})$. Since $\sin \theta = \frac{\sqrt{65}}{9}$, $\cos^2 \theta = 1 - \frac{65}{81} = \frac{16}{81}$, so $\hat{a} \cdot \hat{b} = \frac{4}{9}$. Final value: $9 + 45 \times \frac{4}{9} = 9 + 20 = 29$.

   Step Solution:

    1.  Determine $\hat{a} \cdot \hat{b}$ from $\sin \theta$: $\cos \theta = \sqrt{1 - \left(\frac{\sqrt{65}}{9}\right)^2} = \frac{4}{9}$.

    2.  Compute the scalar projection of $\vec{c}$ on $\hat{a}$: $\vec{c} \cdot \hat{a} = 3|\hat{a}|^2 + 6(\hat{a} \cdot \hat{b}) + 9(\hat{a} \times \hat{b}) \cdot \hat{a} = 3 + 6(\frac{4}{9}) = \frac{17}{3}$.

    3.  Compute the scalar projection of $\vec{c}$ on $\hat{b}$: $\vec{c} \cdot \hat{b} = 3(\hat{a} \cdot \hat{b}) + 6|\hat{b}|^2 + 9(\hat{a} \times \hat{b}) \cdot \hat{b} = 3(\frac{4}{9}) + 6 = \frac{22}{3}$.

    4.  Substitute into the target expression: $9(\frac{17}{3}) - 3(\frac{22}{3}) = 51 - 22$.

    5.  Result: $29$.

   Difficulty level: Medium

   Concept Name: Scalar Product (Dot Product)

   Short cut solution: Expand the full expression $9(3 + 6\hat{a} \cdot \hat{b}) - 3(3\hat{a} \cdot \hat{b} + 6)$ to get $9 + 45(\hat{a} \cdot \hat{b})$ and immediately plug in $\hat{a} \cdot \hat{b} = \cos \theta = \frac{4}{9}$.

 Question 22

   Question: Let $\vec{a}$ and $\vec{b}$ be the vectors of the same magnitude such that $\frac{|\vec{a} + \vec{b}| + |\vec{a} - \vec{b}|}{|\vec{a} + \vec{b}| - |\vec{a} - \vec{b}|} = \sqrt{2} + 1$. Then $\frac{|\vec{a} + \vec{b}|^2}{|\vec{a}|^2}$ is :

   Options: 

    A. $2 + \sqrt{2}$

    B. $2 + 4\sqrt{2}$

    C. $4 + 2\sqrt{2}$

    D. $1 + \sqrt{2}$

   Correct Answer: A

   Year: JEE Main 2025 (Online) 7th April Evening Shift

   Solution: Apply componendo and dividendo to the given ratio to find $\frac{|\vec{a} + \vec{b}|}{|\vec{a} - \vec{b}|} = \frac{\sqrt{2} + 2}{\sqrt{2}} = 1 + \sqrt{2}$. Squaring both sides: $|\vec{a} + \vec{b}|^2 = (3 + 2\sqrt{2}) |\vec{a} - \vec{b}|^2$. Expanding magnitudes (with $|\vec{a}| = |\vec{b}|$): $2|\vec{a}|^2 + 2\vec{a} \cdot \vec{b} = (3 + 2\sqrt{2})(2|\vec{a}|^2 - 2\vec{a} \cdot \vec{b})$. Simplifying yields $\frac{\vec{a} \cdot \vec{b}}{|\vec{a}|^2} = \frac{1}{\sqrt{2}}$. The required expression is $1 + \frac{|\vec{b}|^2}{|\vec{a}|^2} + \frac{2\vec{a} \cdot \vec{b}}{|\vec{a}|^2} = 2 + 2(\frac{1}{\sqrt{2}}) = 2 + \sqrt{2}$.

   Step Solution:

    1.  Apply Componendo and Dividendo: $\frac{|\vec{a} + \vec{b}|}{|\vec{a} - \vec{b}|} = \frac{(\sqrt{2} + 1) + 1}{(\sqrt{2} + 1) - 1} = 1 + \sqrt{2}$.

    2.  Square the result: $|\vec{a} + \vec{b}|^2 = (3 + 2\sqrt{2}) |\vec{a} - \vec{b}|^2$.

    3.  Express magnitudes in terms of dot products (let $|\vec{a}|=|\vec{b}|=k$): $2k^2 + 2\vec{a} \cdot \vec{b} = (3 + 2\sqrt{2})(2k^2 - 2\vec{a} \cdot \vec{b})$.

    4.  Solve for the ratio: $2\vec{a} \cdot \vec{b} (4 + 2\sqrt{2}) = 2k^2 (2 + 2\sqrt{2}) \Rightarrow \frac{2\vec{a} \cdot \vec{b}}{|\vec{a}|^2} = \sqrt{2}$.

    5.  Final Calculation: $\frac{|\vec{a} + \vec{b}|^2}{|\vec{a}|^2} = \frac{2|\vec{a}|^2 + 2\vec{a} \cdot \vec{b}}{|\vec{a}|^2} = 2 + \sqrt{2}$.

   Difficulty level: Hard

   Concept Name: Componendo and Dividendo / Magnitude of Vector Sum

   Short cut solution: Let $x = |\vec{a} + \vec{b}|^2$ and $y = |\vec{a} - \vec{b}|^2$. Since $|\vec{a}| = |\vec{b}|$, $|\vec{a}|^2 = \frac{x+y}{4}$. The ratio $\frac{x}{y} = (1+\sqrt{2})^2 = 3+2\sqrt{2}$. Then $\frac{x}{(x+y)/4} = \frac{4(3+2\sqrt{2})}{4+2\sqrt{2}} = 2+\sqrt{2}$.

Question 25

   Question: The least positive integral value of $\alpha$, for which the angle between the vectors $\alpha\hat{i} - 2\hat{j} + 2\hat{k}$ and $\alpha\hat{i} + 2\alpha\hat{j} - 2\hat{k}$ is acute, is

   Options: (None explicitly listed in source text, however standard JEE options for such integer types usually include 3, 4, 5, or 6).

   Correct Answer: 5

   Year: JEE Main 2024 (Online) 27th January Shift 1

   Solution: For the angle to be acute, the dot product must be greater than zero. $(\alpha\hat{i} - 2\hat{j} + 2\hat{k}) \cdot (\alpha\hat{i} + 2\alpha\hat{j} - 2\hat{k}) > 0 \Rightarrow \alpha^2 - 4\alpha - 4 > 0$. The roots of the quadratic $\alpha^2 - 4\alpha - 4 = 0$ are $2 \pm \sqrt{8}$. Thus, $\alpha > 2 + 2.82 = 4.82$ or $\alpha < 2 - 2.82 = -0.82$. The least positive integer is 5.

   Step Solution:

    1.  Set up condition for acute angle: $\text{Vector}_1 \cdot \text{Vector}_2 > 0$.

    2.  Calculate dot product: $\alpha(\alpha) + (-2)(2\alpha) + (2)(-2) = \alpha^2 - 4\alpha - 4$.

    3.  Apply inequality: $\alpha^2 - 4\alpha - 4 > 0 \Rightarrow (\alpha - 2)^2 > 8$.

    4.  Identify range: $\alpha - 2 > \sqrt{8} \approx 2.82 \Rightarrow \alpha > 4.82$.

    5.  Select smallest positive integer: $\alpha = 5$.

   Difficulty level: Easy

   Concept Name: Scalar Product (Dot Product)

   Short cut solution: Quickly evaluate the quadratic $\alpha^2 - 4\alpha - 4 > 0$ for small integers: for $\alpha=4, 16-16-4 < 0$; for $\alpha=5, 25-20-4 = 1 > 0$. The first positive integer to satisfy it is 5.

Question 29

   Question: Let $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$ and $\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}$ be two vectors such that $|\vec{a}| = 1, \vec{a} \cdot \vec{b} = 2$ and $|\vec{b}| = 4$. If $\vec{c} = 2(\vec{a} \times \vec{b}) - 3\vec{b}$, then the angle between $\vec{b}$ and $\vec{c}$ is equal to :

   Options: 

    A. $\cos^{-1}\left(\frac{2}{\sqrt{3}}\right)$

    B. $\cos^{-1}\left(-\frac{1}{\sqrt{3}}\right)$

    C. $\cos^{-1}\left(-\frac{\sqrt{3}}{2}\right)$

    D. $\cos^{-1}\left(\frac{2}{3}\right)$

   Correct Answer: C

   Year: 30-Jan-2024 Shift 1

   Solution: Given $|\vec{a}|=1, |\vec{b}|=4, \vec{a}\cdot\vec{b}=2$. $\vec{c} = 2(\vec{a} \times \vec{b}) - 3\vec{b}$. Dot product with $\vec{b}$ gives $\vec{b}\cdot\vec{c} = -3|\vec{b}|^2 = -48$. Magnitude $|\vec{c}|^2 = 4|\vec{a}\times\vec{b}|^2 + 9|\vec{b}|^2$. Since $|\vec{a}\times\vec{b}|^2 = |\vec{a}|^2|\vec{b}|^2 - (\vec{a}\cdot\vec{b})^2 = 16-4=12$, we get $|\vec{c}|^2 = 192$. Then $\cos \theta = \frac{\vec{b}\cdot\vec{c}}{|\vec{b}||\vec{c}|} = \frac{-48}{4\sqrt{192}} = -\frac{\sqrt{3}}{2}$.

   Step Solution:

    1.  List knowns: $|\vec{a}|=1, |\vec{b}|=4, \vec{a}\cdot\vec{b}=2$. Expression: $\vec{c} = 2(\vec{a} \times \vec{b}) - 3\vec{b}$.

    2.  Calculate $\vec{b} \cdot \vec{c}$: $\vec{b} \cdot (2(\vec{a} \times \vec{b}) - 3\vec{b}) = 0 - 3(4)^2 = -48$.

    3.  Find $|\vec{c}|^2$: $4|\vec{a} \times \vec{b}|^2 + 9|\vec{b}|^2 = 4(|\vec{a}|^2|\vec{b}|^2 - (\vec{a}\cdot\vec{b})^2) + 9(16) = 4(16-4) + 144 = 48 + 144 = 192$.

    4.  Compute $\cos \theta$: $\frac{\vec{b} \cdot \vec{c}}{|\vec{b}||\vec{c}|} = \frac{-48}{4 \cdot \sqrt{192}} = \frac{-12}{\sqrt{64 \cdot 3}} = \frac{-12}{8\sqrt{3}} = -\frac{\sqrt{3}}{2}$.

    5.  Determine angle: $\theta = \cos^{-1}(-\sqrt{3}/2) = 150^\circ$.

   Difficulty level: Medium

   Concept Name: Scalar Product (Dot Product) and Angle Between Vectors

   Short cut solution: Recognize that $\vec{c}$ is a linear combination of a vector perpendicular to $\vec{b}$ and $\vec{b}$ itself. The dot product $\vec{b}\cdot\vec{c}$ only depends on the $-3\vec{b}$ term.

 Question 35

   Question: Let $\vec{a}$ and $\vec{b}$ be two vectors such that $|\vec{a}| = 1, |\vec{b}| = 4$ and $\vec{a} \cdot \vec{b} = 2$. If $\vec{c} = (2\vec{a} \times \vec{b}) - 3\vec{b}$ and the angle between $\vec{b}$ and $\vec{c}$ is $\delta q$, then $192\sin^2 \delta q$ is equal to:

   Options: (Numerical Answer Type)

   Correct Answer: 48

   Year: 31-Jan-2024 Shift 1

   Solution: $\vec{b} \cdot \vec{c} = -3|\vec{b}|^2$, so $|\vec{b}||\vec{c}|\cos \alpha = -3|\vec{b}|^2$, which means $|\vec{c}|\cos \alpha = -12$. Squaring gives $|\vec{c}|^2\cos^2 \alpha = 144$. We also found $|\vec{c}|^2 = 192$. Thus $192\cos^2 \alpha = 144$. Therefore $192\sin^2 \alpha = 192 - 144 = 48$.

   Step Solution:

    1.  Compute dot product: $\vec{b} \cdot \vec{c} = \vec{b} \cdot (2\vec{a} \times \vec{b} - 3\vec{b}) = -3|\vec{b}|^2 = -48$.

    2.  Compute $|\vec{c}|^2$: $|2\vec{a}\times\vec{b}|^2 + |-3\vec{b}|^2 = 4(12) + 9(16) = 192$.

    3.  Set up the identity: $|\vec{b} \times \vec{c}|^2 = |\vec{b}|^2|\vec{c}|^2 - (\vec{b}\cdot\vec{c})^2$.

    4.  Substitute values: $|\vec{b} \times \vec{c}|^2 = (16)(192) - (-48)^2 = 3072 - 2304 = 768$.

    5.  Solve for target: $192\sin^2 \alpha = 192 \cdot \frac{|\vec{b} \times \vec{c}|^2}{|\vec{b}|^2|\vec{c}|^2} = 192 \cdot \frac{768}{16 \cdot 192} = \frac{768}{16} = 48$.

   Difficulty level: Medium

   Concept Name: Vector Algebra Identities

   Short cut solution: Since $|\vec{c}|^2 = 192$, the expression $192\sin^2 \delta q$ is exactly $|\vec{c}|^2(1 - \cos^2 \delta q)$. Using $\cos^2 \delta q = 3/4$ from previous problem logic, $192(1/4) = 48$.

 Question 39

   Question: Let $\vec{a} = \hat{i} + \hat{j} + \hat{k}$, $\vec{b} = -\hat{i} - 8\hat{j} + 2\hat{k}$ and $\vec{c} = 4\hat{i} + c_2\hat{j} + c_3\hat{k}$ be three vectors such that $\vec{b} \times \vec{a} = \vec{c} \times \vec{a}$. If the angle between the vector $\vec{c}$ and the vector $3\hat{i} + 4\hat{j} + \hat{k}$ is $\theta$, then the greatest integer less than or equal to $\tan^2 \theta$ is :

   Options: (Numerical Answer Type)

   Correct Answer: 38

   Year: 1-Feb-2024 Shift 2

   Solution: From $\vec{b} \times \vec{a} = \vec{c} \times \vec{a}$, we have $(\vec{c} - \vec{b}) \parallel \vec{a}$, so $\vec{c} = \vec{b} + \lambda\vec{a}$. Comparing components with $\vec{c} = 4\hat{i} + c_2\hat{j} + c_3\hat{k}$, we find $\lambda = 5$ and $\vec{c} = 4\hat{i} - 3\hat{j} + 7\hat{k}$. The angle $\theta$ with $\vec{v} = 3\hat{i} + 4\hat{j} + \hat{k}$ gives $\cos \theta = \frac{7}{\sqrt{74}\sqrt{26}}$. Then $\tan^2 \theta = \frac{1875}{49} \approx 38.26$.

   Step Solution:

    1.  Relate vectors: $\vec{b} \times \vec{a} = \vec{c} \times \vec{a} \Rightarrow (\vec{c}-\vec{b}) \times \vec{a} = 0 \Rightarrow \vec{c} = \vec{b} + \lambda\vec{a}$.

    2.  Find $\lambda$: $4\hat{i} + c_2\hat{j} + c_3\hat{k} = (-\hat{i}-8\hat{j}+2\hat{k}) + \lambda(\hat{i}+\hat{j}+\hat{k})$. $4 = -1 + \lambda \Rightarrow \lambda = 5$.

    3.  Determine $\vec{c}$ and magnitudes: $\vec{c} = 4\hat{i} - 3\hat{j} + 7\hat{k}$. $|\vec{c}|^2 = 74$. Vector $\vec{v} = 3\hat{i} + 4\hat{j} + \hat{k}$ has $|\vec{v}|^2 = 26$.

    4.  Find $\cos^2 \theta$: $\cos \theta = \frac{\vec{c} \cdot \vec{v}}{|\vec{c}||\vec{v}|} = \frac{12-12+7}{\sqrt{74 \cdot 26}} = \frac{7}{\sqrt{1924}}$. So $\cos^2 \theta = \frac{49}{1924}$.

    5.  Calculate $\tan^2 \theta$: $\frac{1}{\cos^2 \theta} - 1 = \frac{1924}{49} - 1 = \frac{1875}{49} \approx 38.26$. Greatest integer is 38.

   Difficulty level: Hard

   Concept Name: Vector Cross Product Equation and Directional Angles

   Short cut solution: Use the component comparison to quickly find $\lambda=5$, then immediately calculate the dot product and magnitudes to find $\cos^2 \theta$. Final calculation $1875/49$ leads directly to the integer part 38.

Question 53

   Question: Let $\lambda \in \mathbb{R}, \vec{a} = \lambda \hat{i} + 2 \hat{j} - 3 \hat{k}, \vec{b} = \hat{i} - \lambda \hat{j} + 2 \hat{k}.$ If $((\vec{a} + \vec{b}) \times (\vec{a} \times \vec{b})) \times (\vec{a} - \vec{b}) = 8 \hat{i} - 40 \hat{j} - 24 \hat{k}$ , then $|\lambda((\vec{a} + \vec{b}) \times (\vec{a} - \vec{b}))|^2$ is equal to.

   Options: 

    A. 140

    B. 132

    C. 144

    D. 136

   Correct Answer: A

   Year: 2023 (Online) 30th January Shift 2

   Solution: Given $\vec{a} = \lambda \hat{i} + 2 \hat{j} - 3 \hat{k}$ and $\vec{b} = \hat{i} - \lambda \hat{j} + 2 \hat{k}$. The equation $((\vec{a} + \vec{b}) \times (\vec{a} \times \vec{b})) \times (\vec{a} - \vec{b}) = 8 \hat{i} - 40 \hat{j} - 24 \hat{k}$ simplifies to $8(\vec{a} \times \vec{b}) = 8\hat{i} - 40\hat{j} - 24\hat{k}$. Thus $\vec{a} \times \vec{b} = \hat{i} - 5\hat{j} - 3\hat{k}$. By calculating the cross product of $\vec{a}$ and $\vec{b}$ and equating components, we find $\lambda = 1$. Then $(\vec{a} + \vec{b}) \times (\vec{a} - \vec{b}) = -2(\vec{a} \times \vec{b}) = -2(\hat{i} - 5\hat{j} - 3\hat{k})$. The magnitude squared of $\lambda$ times this vector is $4(1^2 + 5^2 + 3^2) = 4(35) = 140$.

   Step Solution:

    1.  Use Vector Triple Product to simplify the LHS: $((\vec{a}+\vec{b}) \cdot \vec{b} \vec{a} - (\vec{a}+\vec{b}) \cdot \vec{a} \vec{b}) \times (\vec{a}-\vec{b})$.

    2.  Identify resulting vector: From source simplification, the equation becomes $8(\vec{a} \times \vec{b}) = 8\hat{i} - 40\hat{j} - 24\hat{k}$, so $\vec{a} \times \vec{b} = \hat{i} - 5\hat{j} - 3\hat{k}$.

    3.  Find $\lambda$: Compute $\vec{a} \times \vec{b} = (4-3\lambda)\hat{i} - (2\lambda+3)\hat{j} + (-\lambda^2-2)\hat{k}$. Equating $\hat{j}$ components: $-(2\lambda+3) = -5 \Rightarrow 2\lambda = 2 \Rightarrow \lambda = 1$.

    4.  Simplify required expression: $(\vec{a} + \vec{b}) \times (\vec{a} - \vec{b}) = \vec{a}\times\vec{a} - \vec{a}\times\vec{b} + \vec{b}\times\vec{a} - \vec{b}\times\vec{b} = -2(\vec{a} \times \vec{b})$.

    5.  Final calculation: $|\lambda(-2)(\vec{a} \times \vec{b})|^2 = |1 \cdot (-2)(\hat{i}-5\hat{j}-3\hat{k})|^2 = 4(1+25+9) = 140$.

   Difficulty level: Hard

   Concept Name: Vector Triple Product Expansion (Lagrange's Formula)

   Short cut solution: Recognize that $(\vec{a}+\vec{b}) \times (\vec{a}-\vec{b}) = -2(\vec{a} \times \vec{b})$ immediately to avoid expanding the sum vectors.

Question 55

   Question: Let $\vec{a} = 2 \hat{i} + \hat{j} + \hat{k}$ , and $\vec{b}$ and $\vec{c}$ be two nonzero vectors such that $|\vec{a} + \vec{b} + \vec{c}| = |\vec{a} + \vec{b} - \vec{c}|$ and $\vec{b} \cdot \vec{c} = 0$ . Consider the following two statements: (A) $|\vec{a} + \lambda \vec{c}| \geq |\vec{a}|$ for all $\lambda \in \mathbb{R}$ . (B) $\vec{a}$ and $\vec{c}$ are always parallel.

   Options: 

    A. only (B) is correct 

    B. neither (A) nor (B) is correct

    C. only (A) is correct 

    D. both (A) and (B) are correct

   Correct Answer: C

   Year: 2023 (Online) 31st January Shift 1

   Solution: Squaring $|\vec{a} + \vec{b} + \vec{c}|^2 = |\vec{a} + \vec{b} - \vec{c}|^2$ and simplifying gives $4(\vec{a} + \vec{b}) \cdot \vec{c} = 0$. Since $\vec{b} \cdot \vec{c} = 0$, this implies $\vec{a} \cdot \vec{c} = 0$. Statement (B) is incorrect because $\vec{a} \cdot \vec{c} = 0$ means they are perpendicular, not parallel. For statement (A), $|\vec{a} + \lambda \vec{c}|^2 = |\vec{a}|^2 + \lambda^2 |\vec{c}|^2 + 2\lambda(\vec{a} \cdot \vec{c}) = |\vec{a}|^2 + \lambda^2 |\vec{c}|^2$. Since $\lambda^2 |\vec{c}|^2 \geq 0$, statement (A) is true for all $\lambda \in \mathbb{R}$.

   Step Solution:

    1.  Apply the property $|X+Y|^2 = |X-Y|^2 \Rightarrow X \cdot Y = 0$. Here $X = \vec{a}+\vec{b}$ and $Y = \vec{c}$.

    2.  Establish orthogonality: $(\vec{a} + \vec{b}) \cdot \vec{c} = 0 \Rightarrow \vec{a} \cdot \vec{c} + \vec{b} \cdot \vec{c} = 0$.

    3.  Substitute given $\vec{b} \cdot \vec{c} = 0$: We get $\vec{a} \cdot \vec{c} = 0$ (implying Statement B is false).

    4.  Expand the expression in Statement A: $|\vec{a} + \lambda \vec{c}|^2 = |\vec{a}|^2 + \lambda^2 |\vec{c}|^2 + 2\lambda(\vec{a} \cdot \vec{c})$.

    5.  Simplify and conclude: Since $\vec{a} \cdot \vec{c} = 0$, the expression is $|\vec{a}|^2 + \lambda^2 |\vec{c}|^2$, which is $\geq |\vec{a}|^2$ as $|\vec{c}| \neq 0$ and $\lambda^2 \geq 0$. Statement A is true.

   Difficulty level: Medium

   Concept Name: Dot Product and Vector Magnitudes

   Short cut solution: The magnitude condition $|(\vec{a}+\vec{b}) + \vec{c}| = |(\vec{a}+\vec{b}) - \vec{c}|$ immediately implies $(\vec{a}+\vec{b}) \perp \vec{c}$. With $\vec{b} \perp \vec{c}$ already given, $\vec{a}$ must also be perpendicular to $\vec{c}$.

Question 56

   Question: Let $\vec{a}$ and $\vec{b}$ be two vector such that $|\vec{a}| = \sqrt{14}$ , $|\vec{b}| = \sqrt{6}$ and $|\vec{a} \times \vec{b}| = \sqrt{48}$ . Then $(\vec{a} \cdot \vec{b})^2$ is equal to.

   Options: (Numerical Answer Type)

   Correct Answer: 36

   Year: 2023 (Online) 31st January Shift 1

   Solution: Use the identity relating dot and cross products: $|\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2 |\vec{b}|^2$. Substituting the values: $(\sqrt{48})^2 + (\vec{a} \cdot \vec{b})^2 = (\sqrt{14})^2 (\sqrt{6})^2$. This results in $48 + (\vec{a} \cdot \vec{b})^2 = 14 \times 6 = 84$. Therefore, $(\vec{a} \cdot \vec{b})^2 = 84 - 48 = 36$.

   Step Solution:

    1.  State Lagrange's Identity: $|\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2 |\vec{b}|^2$.

    2.  Substitute magnitudes: $(\sqrt{48})^2 + (\vec{a} \cdot \vec{b})^2 = (\sqrt{14})^2 (\sqrt{6})^2$.

    3.  Calculate squares of roots: $48 + (\vec{a} \cdot \vec{b})^2 = 14 \cdot 6$.

    4.  Multiply constants: $14 \cdot 6 = 84$.

    5.  Subtract to find target: $(\vec{a} \cdot \vec{b})^2 = 84 - 48 = 36$.

   Difficulty level: Easy

   Concept Name: Lagrange's Identity

   Short cut solution: Directly substitute into $(\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2 |\vec{b}|^2 - |\vec{a} \times \vec{b}|^2$ to get $14 \times 6 - 48 = 36$.

Question 72

   Question: For any vector $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$, with $|a_i| < 1, i = 1, 2, 3$ consider the following statements : 

    (A) : $\max\{|a_1|, |a_2|, |a_3|\} \le |\vec{a}|$

    (B) : $|\vec{a}| \le 3 \max\{|a_1|, |a_2|, |a_3|\}$

   Options: 

    A. Only (B) is true 

    B. Both (A) and (B) are true 

    C. Neither (A) nor (B) is true 

    D. Only (A) is true

   Correct Answer: B

   Year: 11-Apr-2023 shift 1

   Solution: Without loss of generality, let $|a_1| \le |a_2| \le |a_3|$. Then $|\vec{a}|^2 = |a_1|^2 + |a_2|^2 + |a_3|^2 \ge (a_3)^2$, which implies $|\vec{a}| \ge |a_3| = \max\{|a_1|, |a_2|, |a_3|\}$ (Statement A is true). Also, $|\vec{a}|^2 = |a_1|^2 + |a_2|^2 + |a_3|^2 \le 3|a_3|^2 \Rightarrow |\vec{a}| \le \sqrt{3}|a_3| = \sqrt{3}\max\{|a_1|, |a_2|, |a_3|\} \le 3\max\{|a_1|, |a_2|, |a_3|\}$ (Statement B is true).

   Step Solution:

    1. Define the magnitude squared: $|\vec{a}|^2 = a_1^2 + a_2^2 + a_3^2$.

    2. Let $M = \max\{|a_1|, |a_2|, |a_3|\}$. Therefore, $M^2 \ge a_1^2$, $M^2 \ge a_2^2$, and $M^2 \ge a_3^2$.

    3. To verify (A): Since $a_1^2 + a_2^2 + a_3^2 \ge M^2$ (as the sum of non-negative squares is greater than any single term), taking square roots gives $|\vec{a}| \ge M$.

    4. To verify (B): Summing the inequalities from Step 2, we get $a_1^2 + a_2^2 + a_3^2 \le 3M^2$, so $|\vec{a}|^2 \le 3M^2$.

    5. Final conclusion: $|\vec{a}| \le \sqrt{3}M \approx 1.732M$. Since $1.732M < 3M$, the condition $|\vec{a}| \le 3M$ is satisfied.

   Difficulty level: Easy

   Concept Name: Vector Magnitude and Inequality Relations

   Short cut solution: Use the property that the magnitude of a vector is always at least as large as its largest component, and at most $\sqrt{3}$ times the largest component. Both given statements fall within these bounds.

Question 75

   Question: Let $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$ and $\vec{b} = \hat{i} + \hat{j} - \hat{k}$. If $\vec{c}$ is a vector such that $\vec{a} \cdot \vec{c} = 11, \vec{b} \cdot (\vec{a} \times \vec{c}) = 27$ and $\vec{b} \cdot \vec{c} = -\sqrt{3}|\vec{b}|$, then $|\vec{a} \times \vec{c}|^2$ is equal to

   Options: (Numerical Answer Type)

   Correct Answer: 285

   Year: 11-Apr-2023 shift 2

   Solution: Given $\vec{b} \cdot (\vec{a} \times \vec{c}) = 27$. Also $\vec{a} \cdot \vec{b} = 1+2-3=0$, so $\vec{a} \perp \vec{b}$. Let $\theta$ be the angle between $\vec{b}$ and $(\vec{a} \times \vec{c})$. We find $\sin \theta = \sqrt{14}/\sqrt{95}$ and $|\vec{b}| \times |\vec{a} \times \vec{c}| = 3\sqrt{95}$. Solving leads to $|\vec{a} \times \vec{c}| = \sqrt{3} \times \sqrt{95}$, making the square 285.

   Step Solution:

    1. Calculate known magnitudes and products: $|\vec{b}| = \sqrt{3}$, $|\vec{a}| = \sqrt{14}$, and $\vec{a} \cdot \vec{b} = 0$.

    2. Process given condition: $\vec{b} \cdot \vec{c} = -\sqrt{3}(\sqrt{3}) = -3$.

    3. Use the scalar triple product: $\vec{b} \cdot (\vec{a} \times \vec{c}) = 27$. Let $\vec{v} = \vec{a} \times \vec{c}$. Then $|\vec{b}||\vec{v}| \cos \phi = 27$ where $\phi$ is the angle between $\vec{b}$ and $\vec{v}$.

    4. Apply vector triple product identity: $|\vec{b} \times \vec{v}|^2 = |\vec{b} \times (\vec{a} \times \vec{c})|^2 = |(\vec{b} \cdot \vec{c})\vec{a} - (\vec{b} \cdot \vec{a})\vec{c}|^2$. With $\vec{b} \cdot \vec{a} = 0$, this is $|(-3)\vec{a}|^2 = 9(14) = 126$.

    5. Use the identity $|\vec{b}|^2|\vec{v}|^2 = (\vec{b} \cdot \vec{v})^2 + |\vec{b} \times \vec{v}|^2$: $3|\vec{v}|^2 = (27)^2 + 126 = 729 + 126 = 855$. Thus $|\vec{v}|^2 = 285$.

   Difficulty level: Hard

   Concept Name: Vector Triple Product Expansion

   Short cut solution: Recognize that $|\vec{b} \times (\vec{a} \times \vec{c})|^2 + (\vec{b} \cdot (\vec{a} \times \vec{c}))^2 = |\vec{b}|^2 |\vec{a} \times \vec{c}|^2$. Substitute the triple product expansion for the first term to solve for $|\vec{a} \times \vec{c}|^2$ quickly.

 Question 78

   Question: Let $\vec{a} = 3\hat{i} + \hat{j} - \hat{k}$ and $\vec{c} = 2\hat{i} - 3\hat{j} + 3\hat{k}$. If $\vec{b}$ is a vector such that $\vec{a} = \vec{b} \times \vec{c}$ and $|\vec{b}|^2 = 50$, then $|72 - |\vec{b} + \vec{c}|^2|$ is equal to

   Options: (Numerical Answer Type)

   Correct Answer: 66

   Year: 13-Apr-2023 shift 1

   Solution: Calculate $|\vec{a}|^2 = 11$ and $|\vec{c}|^2 = 22$. Given $\vec{a} = \vec{b} \times \vec{c}$, so $\vec{a} \perp \vec{b}$ and $\vec{a} \perp \vec{c}$. Using Lagrange's identity: $|\vec{b} \times \vec{c}|^2 = |\vec{b}|^2|\vec{c}|^2 - (\vec{b} \cdot \vec{c})^2$. Substituting values: $11 = 50(22) - (\vec{b} \cdot \vec{c})^2$, so $(\vec{b} \cdot \vec{c})^2 = 1089$. Thus $2\vec{b} \cdot \vec{c} = \pm 66$. The expression $|72 - (|\vec{b}|^2 + |\vec{c}|^2 + 2\vec{b} \cdot \vec{c})|$ simplifies to $|72 - (50 + 22 \pm 66)| = 66$.

   Step Solution:

    1. Find magnitudes: $|\vec{a}|^2 = 11$ and $|\vec{c}|^2 = 22$.

    2. Use the cross product magnitude formula: $|\vec{a}|^2 = |\vec{b} \times \vec{c}|^2 = |\vec{b}|^2|\vec{c}|^2 - (\vec{b} \cdot \vec{c})^2$.

    3. Calculate the dot product: $11 = 50(22) - (\vec{b} \cdot \vec{c})^2 \Rightarrow (\vec{b} \cdot \vec{c})^2 = 1100 - 11 = 1089$, so $\vec{b} \cdot \vec{c} = \pm 33$.

    4. Expand the sum magnitude: $|\vec{b} + \vec{c}|^2 = |\vec{b}|^2 + |\vec{c}|^2 + 2\vec{b} \cdot \vec{c} = 50 + 22 \pm 66 = 72 \pm 66$.

    5. Final target: $|72 - (72 \pm 66)| = | \mp 66 | = \mathbf{66}$.

   Difficulty level: Hard

   Concept Name: Lagrange's Identity

   Short cut solution: Recognize that $|72 - |\vec{b} + \vec{c}|^2| = |72 - (|\vec{b}|^2 + |\vec{c}|^2 + 2\vec{b} \cdot \vec{c})|$. Since $|\vec{b}|^2 + |\vec{c}|^2 = 50 + 22 = 72$, the expression reduces exactly to $|2\vec{b} \cdot \vec{c}|$. Use Lagrange's Identity to find $\vec{b} \cdot \vec{c}$ and multiply by 2.

Question 80

   Question: Let $|\vec{a}| = 2, |\vec{b}| = 3$ and the angle between the vectors $\vec{a}$ and $\vec{b}$ be $\pi/4$. Then $|(\vec{a} + 2\vec{b}) \times (2\vec{a} - 3\vec{b})|^2$ is equal to:

   Options: 

    A. 482

    B. 841

    C. 882

    D. 441

   Correct Answer: C

   Year: 13-Apr-2023 shift 2

   Solution: The source provides the final answer as 882.

   Step Solution:

    1.  Expand the cross product: $(\vec{a} + 2\vec{b}) \times (2\vec{a} - 3\vec{b}) = 2(\vec{a} \times \vec{a}) - 3(\vec{a} \times \vec{b}) + 4(\vec{b} \times \vec{a}) - 6(\vec{b} \times \vec{b})$.

    2.  Simplify using identities: Since $\vec{v} \times \vec{v} = 0$ and $\vec{b} \times \vec{a} = -(\vec{a} \times \vec{b})$, the expression becomes $0 - 3(\vec{a} \times \vec{b}) - 4(\vec{a} \times \vec{b}) - 0 = -7(\vec{a} \times \vec{b})$.

    3.  Square the magnitude: $|-7(\vec{a} \times \vec{b})|^2 = 49 |\vec{a} \times \vec{b}|^2$.

    4.  Calculate the cross product magnitude: $|\vec{a} \times \vec{b}|^2 = |\vec{a}|^2|\vec{b}|^2 \sin^2(\pi/4) = (2)^2(3)^2(1/\sqrt{2})^2 = 4 \cdot 9 \cdot 1/2 = 18$.

    5.  Final Calculation: $49 \times 18 = 882$.

   Difficulty level: Medium

   Concept Name: Vector Cross Product Properties

   Short cut solution: Recognize that $(\vec{a} + m\vec{b}) \times (n\vec{a} + p\vec{b}) = (p - mn)(\vec{a} \times \vec{b})$. Here, $(1 \cdot -3 - 2 \cdot 2) = -7$. Thus, find $49|\vec{a} \times \vec{b}|^2$ directly.

 Question 85

   Question: Let $\theta$ be the angle between the vectors $\vec{a}$ and $\vec{b}$, where $|\vec{a}| = 4, |\vec{b}| = 3$ and $\theta \in (\pi/4, \pi/3)$. Then $|(\vec{a} - \vec{b}) \times (\vec{a} + \vec{b})|^2 + 4(\vec{a} \cdot \vec{b})^2$ is equal to:

   Options: (Numerical Answer Type)

   Correct Answer: 576

   Year: 25-Jun-2022-Shift-1

   Solution: $((\vec{a} - \vec{b}) \times (\vec{a} + \vec{b})) = 2(\vec{a} \times \vec{b})$. The expression becomes $|2(\vec{a} \times \vec{b})|^2 + 4(\vec{a} \cdot \vec{b})^2 = 4|\vec{a} \times \vec{b}|^2 + 4(\vec{a} \cdot \vec{b})^2 = 4(|\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2) = 4|\vec{a}|^2|\vec{b}|^2$. Substituting values: $4(16)(9) = 576$.

   Step Solution:

    1.  Expand the cross product: $(\vec{a} - \vec{b}) \times (\vec{a} + \vec{b}) = \vec{a} \times \vec{a} + \vec{a} \times \vec{b} - \vec{b} \times \vec{a} - \vec{b} \times \vec{b} = 2(\vec{a} \times \vec{b})$.

    2.  Square the magnitude: $|2(\vec{a} \times \vec{b})|^2 = 4|\vec{a} \times \vec{b}|^2$.

    3.  Apply Lagrange's Identity: $|\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2|\vec{b}|^2$.

    4.  Factor the expression: $4|\vec{a} \times \vec{b}|^2 + 4(\vec{a} \cdot \vec{b})^2 = 4(|\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2) = 4(|\vec{a}|^2|\vec{b}|^2)$.

    5.  Final Calculation: $4 \cdot (4)^2 \cdot (3)^2 = 4 \cdot 16 \cdot 9 = 576$.

   Difficulty level: Easy

   Concept Name: Lagrange's Identity

   Short cut solution: Use the property that $(\vec{a}-\vec{b}) \times (\vec{a}+\vec{b}) = 2(\vec{a} \times \vec{b})$. The sum $4|\vec{a} \times \vec{b}|^2 + 4(\vec{a} \cdot \vec{b})^2$ is simply $4$ times the square of the product of the magnitudes, independent of $\theta$.

Question 90

   Question: Let $\vec{a}$ and $\vec{b}$ be the vectors along the diagonals of a parallelogram having area $2\sqrt{2}$. Let the angle between $\vec{a}$ and $\vec{b}$ be acute, $|\vec{a}| = 1$, and $|\vec{a} \cdot \vec{b}| = |\vec{a} \times \vec{b}|$. If $\vec{c} = 2\sqrt{2}(\vec{a} \times \vec{b}) - 2\vec{b}$, then an angle between $\vec{b}$ and $\vec{c}$ is:

   Options: 

    A. $\pi/4$

    B. $-\pi/4$

    C. $5\pi/6$

    D. $3\pi/4$

   Correct Answer: D

   Year: 27-Jun-2022-Shift-2

   Solution: Area of parallelogram is $\frac{1}{2}|\vec{a} \times \vec{b}| = 2\sqrt{2} \Rightarrow |\vec{a} \times \vec{b}| = 4\sqrt{2}$. Given $|\vec{a} \cdot \vec{b}| = |\vec{a} \times \vec{b}|$, then $\cos \theta = \sin \theta \Rightarrow \theta = \pi/4$. Magnitude $|\vec{b}| = 8$. Dot product $\vec{b} \cdot \vec{c} = -2|\vec{b}|^2 = -128$. This leads to $\cos \alpha = -1/\sqrt{2}$.

   Step Solution:

    1.  Find cross product magnitude: Area $= \frac{1}{2}|\vec{a} \times \vec{b}| = 2\sqrt{2} \Rightarrow |\vec{a} \times \vec{b}| = 4\sqrt{2}$.

    2.  Determine angle $\theta$ and $|\vec{b}|$: $|\vec{a} \cdot \vec{b}| = |\vec{a} \times \vec{b}| \Rightarrow \cos\theta = \sin\theta \Rightarrow \theta = \pi/4$. Thus $1 \cdot |\vec{b}| \cdot \frac{1}{\sqrt{2}} = 4\sqrt{2} \Rightarrow |\vec{b}| = 8$.

    3.  Compute $\vec{b} \cdot \vec{c}$: $\vec{b} \cdot (2\sqrt{2}(\vec{a} \times \vec{b}) - 2\vec{b}) = 0 - 2|\vec{b}|^2 = -2(64) = -128$.

    4.  Find magnitude of $\vec{c}$: Since $\vec{a} \times \vec{b}$ is perpendicular to $\vec{b}$, $|\vec{c}|^2 = (2\sqrt{2} \cdot 4\sqrt{2})^2 + (2 \cdot 8)^2 = 16^2 + 16^2 = 512$. Thus $|\vec{c}| = 16\sqrt{2}$.

    5.  Calculate angle $\alpha$: $\cos \alpha = \frac{\vec{b} \cdot \vec{c}}{|\vec{b}||\vec{c}|} = \frac{-128}{8 \cdot 16\sqrt{2}} = \frac{-128}{128\sqrt{2}} = -\frac{1}{\sqrt{2}} \Rightarrow \alpha = 3\pi/4$.

   Difficulty level: Hard

   Concept Name: Parallelogram Area (Diagonals) and Dot Product

   Short cut solution: Since $\vec{c}$ is composed of a vector perpendicular to $\vec{b}$ (the cross product part) and $\vec{b}$ itself, the angle $\alpha$ satisfies $\tan \alpha = \frac{|\text{perpendicular component}|}{|\text{parallel component}|}$. Here $\tan \alpha = \frac{2\sqrt{2}(4\sqrt{2})}{-2(8)} = \frac{16}{-16} = -1 \Rightarrow \alpha = 135^\circ$ or $3\pi/4$.

Question 106

   Question: Let a vector $\vec{a}$ has magnitude 9. Let a vector $\vec{b}$ be such that for every $(x, y) \in \mathbb{R} \times \mathbb{R} - \{(0, 0)\}$, the vector $(x\vec{a} + y\vec{b})$ is perpendicular to the vector $(6y\vec{a} - 18x\vec{b})$. Then the value of $|\vec{a} \times \vec{b}|$ is equal to:

   Options: 

    A. $9\sqrt{3}$

    B. $27\sqrt{3}$

    C. 9

    D. 81

   Correct Answer: B

   Year: 28-Jul-2022-Shift-1

   Solution: $(x\vec{a} + y\vec{b}) \cdot (6y\vec{a} - 18x\vec{b}) = 0$. $\Rightarrow (6xy|\vec{a}|^2 - 18xy|\vec{b}|^2) + (6y^2 - 18x^2)\vec{a} \cdot \vec{b} = 0$. As given equation is identity, coefficient of $xy = 0$, so $6|\vec{a}|^2 - 18|\vec{b}|^2 = 0 \Rightarrow |\vec{a}|^2 = 3|\vec{b}|^2 \Rightarrow |\vec{b}| = 3\sqrt{3}$. Also coefficient of $y^2$ and $x^2$ must involve $\vec{a} \cdot \vec{b} = 0$. Then $|\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}|\sin 90^\circ = 9 \cdot 3\sqrt{3} \cdot 1 = 27\sqrt{3}$.

   Step Solution:

    1.  Apply the perpendicularity condition: $(x\vec{a} + y\vec{b}) \cdot (6y\vec{a} - 18x\vec{b}) = 0$.

    2.  Expand the dot product: $6xy(\vec{a} \cdot \vec{a}) - 18x^2(\vec{a} \cdot \vec{b}) + 6y^2(\vec{b} \cdot \vec{a}) - 18xy(\vec{b} \cdot \vec{b}) = 0$.

    3.  Group terms as an identity in $x$ and $y$: $xy(6|\vec{a}|^2 - 18|\vec{b}|^2) + \vec{a} \cdot \vec{b}(6y^2 - 18x^2) = 0$.

    4.  Solve for magnitudes: Since it holds for all $(x, y)$, coefficients must be zero. $6(9)^2 = 18|\vec{b}|^2 \Rightarrow 486 = 18|\vec{b}|^2 \Rightarrow |\vec{b}|^2 = 27$, so $|\vec{b}| = 3\sqrt{3}$ and $\vec{a} \cdot \vec{b} = 0$.

    5.  Calculate the cross product magnitude: $|\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}|\sin(90^\circ) = 9 \times 3\sqrt{3} = \mathbf{27\sqrt{3}}$.

   Difficulty level: Hard

   Concept Name: Scalar Product (Dot Product) Identity

   Short cut solution: Treat the dot product as a polynomial in $x$ and $y$. For it to be zero for all values, the coefficient of $xy$ ($6a^2 - 18b^2$) must be zero. Directly find $b = a/\sqrt{3}$ and multiply by $a$.

 Question 107

   Question: Let S be the set of all $a \in \mathbb{R}$ for which the angle between the vectors $\vec{u} = a(\log_e b)\hat{i} - 6\hat{j} + 3\hat{k}$ and $\vec{v} = (\log_e b)\hat{i} + 2\hat{j} + 2a(\log_e b)\hat{k}$, $(b > 1)$ is acute. Then S is equal to:

   Options: 

    A. $(-\infty, -4/3)$ 

    B. $\Phi$ 

    C. $(-4/3, 0)$ 

    D. $(12/7, \infty)$

   Correct Answer: B

   Year: 28-Jul-2022-Shift-2

   Solution: For acute angle $\vec{u} \cdot \vec{v} > 0$. $\Rightarrow a(\log_e b)^2 - 12 + 6a(\log_e b) > 0$. Since $b > 1$, let $\log_e b = t \Rightarrow t > 0$. The inequality $at^2 + 6at - 12 > 0$ must hold for all $t > 0$. This leads to $a \in \phi$.

   Step Solution:

    1.  Establish the dot product condition for an acute angle: $\vec{u} \cdot \vec{v} > 0$.

    2.  Calculate the product: $(a \log_e b)(\log_e b) + (-6)(2) + (3)(2a \log_e b) = a(\log_e b)^2 + 6a\log_e b - 12$.

    3.  Define $t = \log_e b$. Since $b > 1$, then $t > 0$.

    4.  Analyze the quadratic inequality $at^2 + 6at - 12 > 0$.

    5.  Conclusion: For this to hold $\forall t > 0$, testing values shows no such real '$a$' exists that satisfies the condition for the entire range of $t$, thus $S = \Phi$.

   Difficulty level: Medium

   Concept Name: Scalar Product (Acute Angle Condition)

   Short cut solution: Observe that for $t \to 0^+$, the expression $at^2 + 6at - 12$ approaches $-12$. Since $-12$ is not $> 0$, there is no value of $a$ that can make the expression positive for all $t > 0$.

 Question 112

   Question: Let $\vec{a}$ and $\vec{b}$ be two vectors such that $|\vec{a} + \vec{b}|^2 = |\vec{a}|^2 + 2|\vec{b}|^2, \vec{a} \cdot \vec{b} = 3$ and $|\vec{a} \times \vec{b}|^2 = 75$. Then $|\vec{a}|^2$ is equal to

   Options: (Numerical Answer Type)

   Correct Answer: 14

   Year: 29-Jul-2022-Shift-2

   Solution: $|\vec{a} + \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + 2\vec{a} \cdot \vec{b}$. Given $|\vec{a} + \vec{b}|^2 = |\vec{a}|^2 + 2|\vec{b}|^2$, so $|\vec{b}|^2 = 2\vec{a} \cdot \vec{b} = 6$. Using Lagrange's identity $|\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2 |\vec{b}|^2$. $75 + 9 = |\vec{a}|^2 \cdot 6 \Rightarrow |\vec{a}|^2 = 14$.

   Step Solution:

    1.  Expand the sum magnitude: $|\vec{a}|^2 + |\vec{b}|^2 + 2\vec{a} \cdot \vec{b} = |\vec{a}|^2 + 2|\vec{b}|^2$.

    2.  Simplify to find $|\vec{b}|^2$: $|\vec{b}|^2 = 2\vec{a} \cdot \vec{b}$. Given $\vec{a} \cdot \vec{b} = 3$, so $|\vec{b}|^2 = 6$.

    3.  Apply Lagrange's Identity: $|\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2 |\vec{b}|^2$.

    4.  Substitute the known values: $75 + (3)^2 = |\vec{a}|^2 \cdot (6)$.

    5.  Final calculation: $84 = 6 |\vec{a}|^2 \Rightarrow \mathbf{|\vec{a}|^2 = 14}$.

   Difficulty level: Easy

   Concept Name: Lagrange's Identity

   Short cut solution: From the first equation, notice $|\vec{b}|^2 = 2(\vec{a} \cdot \vec{b}) = 6$. Immediately use $|\vec{a}|^2 = (|\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2) / |\vec{b}|^2 = (75 + 9) / 6 = 14$.

Question 122

   Question: Let a and b be two non-zero vectors perpendicular to each other and $| \mathbf{a} | = | \mathbf{b} |$ . If $| \mathbf{a} \times \mathbf{b} | = | \mathbf{a} | ,$ then the angle between the vectors $[\mathbf{a} + \mathbf{b} + (\mathbf{a} \times \mathbf{b}) ]$ and a is equal to

   Options: 

    A. $\sin^{-1} \left( \frac{1}{\sqrt{3}} \right)$ 

    B. $\cos^{-1} \left( \frac{1}{\sqrt{3}} \right)$ 

    C. $\cos^{-1} \left( \frac{1}{\sqrt{2}} \right)$ 

    D. $\sin^{-1} \left( \frac{1}{\sqrt{6}} \right)$

   Correct Answer: B

   Year: JEE Main 18 Mar 2021 Shift 2

   Solution: Given, $a \perp b$ and $|a| = |b|$. Also $|a \times b| = |a| \Rightarrow |a||b|\sin 90^\circ = |a|$. This implies $|b| = 1$, and since $|a|=|b|$, then $|a|=1$. Thus $a, b, a \times b$ are mutually perpendicular unit vectors. Let $a = i$ and $b = j$, then $a \times b = k$. The vector is $(i + j + k)$. The angle $\theta$ with vector $a$ is $\cos \theta = \frac{(i+j+k) \cdot i}{\sqrt{3} \sqrt{1}} = \frac{1}{\sqrt{3}}$.

   Step Solution:

    1.  Use $|a \times b| = |a|$ with $a \perp b$: $|a||b|\sin 90^\circ = |a| \Rightarrow |b|=1$.

    2.  Use $|a| = |b|$ to find $|a| = 1$. Vectors $a, b, (a \times b)$ are now known to be mutually perpendicular unit vectors.

    3.  Let $v = a + b + (a \times b)$. Calculate the dot product $v \cdot a = (a \cdot a) + (b \cdot a) + ((a \times b) \cdot a) = 1 + 0 + 0 = 1$.

    4.  Calculate the magnitude $|v| = \sqrt{|a|^2 + |b|^2 + |a \times b|^2} = \sqrt{1 + 1 + 1} = \sqrt{3}$.

    5.  Find the angle: $\cos \theta = \frac{v \cdot a}{|v||a|} = \frac{1}{\sqrt{3} \cdot 1} \Rightarrow \theta = \cos^{-1} \left( \frac{1}{\sqrt{3}} \right)$.

   Difficulty level: Medium

   Concept Name: Scalar Product and Angle between Vectors

   Short cut solution: Since $a, b, c$ (where $c=a \times b$) are orthonormal, the angle any vector $v = x a + y b + z c$ makes with $a$ is simply $\cos \theta = \frac{x}{\sqrt{x^2+y^2+z^2}}$. Here $x=y=z=1$, so $\cos \theta = 1/\sqrt{3}$.

Question 128

   Question: If $(\vec{a} + 3\vec{b})$ is perpendicular to $(7\vec{a} - 5\vec{b})$ and $(\vec{a} - 4\vec{b})$ is perpendicular to $(7\vec{a} - 2\vec{b})$ , then the angle between a and b (in degrees) is

   Options: (Numerical Answer Type)

   Correct Answer: 60

   Year: JEE Main 25 Jul 2021 Shift 2

   Solution: $(a+3b) \cdot (7a-5b) = 0 \Rightarrow 7|a|^2 - 15|b|^2 + 16(a \cdot b) = 0$. Also $(a-4b) \cdot (7a-2b) = 0 \Rightarrow 7|a|^2 + 8|b|^2 - 30(a \cdot b) = 0$. Solving these equations gives $|a| = |b|$ and $\cos \theta = \frac{|b|}{2|a|} = \frac{1}{2}$, so $\theta = 60^\circ$.

   Step Solution:

    1.  Expand the first perpendicularity condition: $(a + 3b) \cdot (7a - 5b) = 7|a|^2 - 5a \cdot b + 21b \cdot a - 15|b|^2 = 7|a|^2 - 15|b|^2 + 16a \cdot b = 0$.

    2.  Expand the second condition: $(a - 4b) \cdot (7a - 2b) = 7|a|^2 - 2a \cdot b - 28b \cdot a + 8|b|^2 = 7|a|^2 + 8|b|^2 - 30a \cdot b = 0$.

    3.  Subtract the equations to eliminate $|a|^2$: $(7|a|^2 + 8|b|^2 - 30a \cdot b) - (7|a|^2 - 15|b|^2 + 16a \cdot b) = 23|b|^2 - 46a \cdot b = 0$.

    4.  Solve for the dot product relation: $46a \cdot b = 23|b|^2 \Rightarrow a \cdot b = \frac{1}{2}|b|^2$.

    5.  Find the angle: Substitute $a \cdot b$ into the first equation to find $|a|=|b|$, then $\cos \theta = \frac{a \cdot b}{|a||b|} = \frac{1/2 |b|^2}{|b|^2} = \frac{1}{2} \Rightarrow \theta = 60^\circ$.

   Difficulty level: Medium

   Concept Name: Dot Product and Orthogonality

   Short cut solution: Treat $|a|^2$, $|b|^2$, and $a \cdot b$ as variables $X, Y, Z$. Solving the system $7X - 15Y + 16Z = 0$ and $7X + 8Y - 30Z = 0$ quickly yields $X=Y$ and $Z=X/2$, leading directly to $\cos \theta = 1/2$.

 Question 129

   Question: Let $\vec{a}, \vec{b}, \vec{c}$ be three mutually perpendicular vectors of the same magnitude and equally inclined at an angle $\theta$, with the vector $\vec{a} + \vec{b} + \vec{c}$ . Then $36 \cos^2 2\theta$ is equal to

   Options: (Numerical Answer Type)

   Correct Answer: 4

   Year: JEE Main 20 Jul 2021 Shift 1

   Solution: $|a+b+c|^2 = |a|^2 + |b|^2 + |c|^2 + 2(0) = 3|a|^2$ (since they are mutually perpendicular and have the same magnitude). So $|a+b+c| = \sqrt{3}|a|$. The dot product $a \cdot (a+b+c) = |a|^2 + 0 + 0 = |a|^2$. Also $a \cdot (a+b+c) = |a||a+b+c|\cos \theta$. This gives $1 = \sqrt{3}\cos \theta \Rightarrow \cos \theta = 1/\sqrt{3}$. Then $\cos 2\theta = 2\cos^2 \theta - 1 = 2/3 - 1 = -1/3$. Finally, $36 \cos^2 2\theta = 36(1/9) = 4$.

   Step Solution:

    1.  Determine the magnitude of the sum vector: $|a+b+c| = \sqrt{|a|^2 + |b|^2 + |c|^2} = \sqrt{3}|a|$ (let $|a|=k$, so $|a+b+c| = \sqrt{3}k$).

    2.  Set up the dot product for inclination: $a \cdot (a+b+c) = |a||a+b+c|\cos \theta$.

    3.  Solve for $\cos \theta$: $k^2 = k(\sqrt{3}k)\cos \theta \Rightarrow \cos \theta = \frac{1}{\sqrt{3}}$.

    4.  Apply the double angle formula: $\cos 2\theta = 2\cos^2 \theta - 1 = 2(1/3) - 1 = -1/3$.

    5.  Calculate the final expression: $36 \cos^2 2\theta = 36(-1/3)^2 = 36(1/9) = 4$.

   Difficulty level: Medium

   Concept Name: Dot Product and Directional Inclination

   Short cut solution: For $n$ mutually perpendicular vectors of equal magnitude, the angle $\theta$ they make with their sum satisfies $\cos^2 \theta = 1/n$. Here $n=3$, so $\cos^2 \theta = 1/3$. Immediate calculation of $\cos 2\theta = -1/3$ follows.

Question 133

   Question: If $|\vec{a}| = 2, |\vec{b}| = 5$ and $|\vec{a} \times \vec{b}| = 8$, then $|\vec{a} \cdot \vec{b}|$ is equal to :

   Options: 

    A. 6

    B. 4

    C. 3

    D. 5

   Correct Answer: A

   Year: 25 Jul 2021 Shift 2

   Solution: Given $|\vec{a}| = 2, |\vec{b}| = 5$. We have $|\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}|\sin \theta = 10\sin \theta = \pm 8$. Thus $\sin \theta = \pm \frac{4}{5}$. Then $\cos \theta = \pm \sqrt{1 - \sin^2 \theta} = \pm \frac{3}{5}$. Therefore, $|\vec{a} \cdot \vec{b}| = |\vec{a}||\vec{b}||\cos \theta| = 10 \cdot \frac{3}{5} = 6$.

   Step Solution:

    1.  Use the magnitude of the cross product: $|\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}| \sin \theta = (2)(5) \sin \theta = 10 \sin \theta = 8$.

    2.  Find $\sin \theta$: $\sin \theta = 8/10 = 4/5$.

    3.  Calculate $\cos \theta$ using the identity $\cos^2 \theta = 1 - \sin^2 \theta$: $\cos^2 \theta = 1 - (4/5)^2 = 9/25 \Rightarrow |\cos \theta| = 3/5$.

    4.  Apply the dot product formula: $|\vec{a} \cdot \vec{b}| = |\vec{a}||\vec{b}| |\cos \theta|$.

    5.  Substitute and calculate: $2 \cdot 5 \cdot (3/5) = \mathbf{6}$.

   Difficulty level: Easy

   Concept Name: Scalar and Vector Product Magnitudes (Lagrange's Identity)

   Short cut solution: Use Lagrange’s Identity: $|\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2 |\vec{b}|^2$. Directly solve: $8^2 + (\vec{a} \cdot \vec{b})^2 = 2^2 \cdot 5^2 \Rightarrow 64 + X^2 = 100 \Rightarrow X^2 = 36 \Rightarrow X = 6$.

 Question 141

   Question: Let a and b be two vectors such that $|2\mathbf{a}+3\mathbf{b}| = |3\mathbf{a}+\mathbf{b}|$ and the angle between a and b is $60^\circ$. If $\frac{1}{8}\mathbf{a}$ is a unit vector, then $|\mathbf{b}|$ is equal to

   Options: 

    A. 4

    B. 6

    C. 5

    D. 8

   Correct Answer: C

   Year: 31 Aug 2021 Shift 1

   Solution: $|2a + 3b|^2 = |3a + b|^2 \Rightarrow 4|a|^2 + 9|b|^2 + 12a \cdot b = 9|a|^2 + |b|^2 + 6a \cdot b$. This simplifies to $8|b|^2 + 6a \cdot b - 5|a|^2 = 0$. Since $\frac{a}{8}$ is a unit vector, $|a| = 8$. Substituting $|a|=8$ and $a \cdot b = |a||b|\cos 60^\circ$ into the equation: $8|b|^2 + 6(8)|b|(1/2) - 5(64) = 0 \Rightarrow 8|b|^2 + 24|b| - 320 = 0$. Solving the quadratic gives $|b| = 5$.

   Step Solution:

    1.  Determine magnitude of $\mathbf{a}$: Since $\frac{1}{8}\mathbf{a}$ is a unit vector, $|\mathbf{a}|/8 = 1 \Rightarrow |\mathbf{a}| = 8$.

    2.  Square the given equality: $|2\mathbf{a}+3\mathbf{b}|^2 = |3\mathbf{a}+\mathbf{b}|^2 \Rightarrow 4|\mathbf{a}|^2 + 9|\mathbf{b}|^2 + 12\mathbf{a} \cdot \mathbf{b} = 9|\mathbf{a}|^2 + |\mathbf{b}|^2 + 6\mathbf{a} \cdot \mathbf{b}$.

    3.  Simplify the equation: $8|\mathbf{b}|^2 + 6\mathbf{a} \cdot \mathbf{b} - 5|\mathbf{a}|^2 = 0$.

    4.  Substitute known values ($|\mathbf{a}|=8, \mathbf{a}\cdot\mathbf{b}=8|\mathbf{b}| \cos 60^\circ$): $8|\mathbf{b}|^2 + 6(8)|\mathbf{b}|(1/2) - 5(64) = 0 \Rightarrow 8|\mathbf{b}|^2 + 24|\mathbf{b}| - 320 = 0$.

    5.  Solve the quadratic $|\mathbf{b}|^2 + 3|\mathbf{b}| - 40 = 0$: $(|\mathbf{b}| + 8)(|\mathbf{b}| - 5) = 0$. Since magnitude is positive, $|\mathbf{b}| = 5$.

   Difficulty level: Medium

   Concept Name: Magnitude of Vector Sum / Scalar Product

   Short cut solution: Reduce the simplified quadratic $8|b|^2 + 24|b| - 320 = 0$ immediately to $|b|^2 + 3|b| - 40 = 0$ and mentally check factors of 40 that differ by 3 (8 and 5).

 Question 145

   Question: Let $\mathbf{a} = \hat{i} + 5\hat{j} + \alpha\hat{k}, \mathbf{b} = \hat{i} + 3\hat{j} + \beta\hat{k}$ and $\mathbf{c} = -\hat{i} + 2\hat{j} - 3\hat{k}$ be three vectors such that, $|\mathbf{b} \times \mathbf{c}| = 5\sqrt{3}$ and a is perpendicular to b. Then, the greatest amongst the values of $|\mathbf{a}|^2$ is

   Options: (Numerical Answer Type, Answer provided as 90)

   Correct Answer: 90

   Year: 27 Aug 2021 Shift 1

   Solution: $\mathbf{a} \perp \mathbf{b} \Rightarrow 1 + 15 + \alpha\beta = 0 \Rightarrow \alpha\beta = -16$. Next, calculate $\mathbf{b} \times \mathbf{c} = \hat{i}(-9-2\beta) - \hat{j}(-3+\beta) + 5\hat{k}$. Given $|\mathbf{b} \times \mathbf{c}|^2 = 75$, so $(-9-2\beta)^2 + (3-\beta)^2 + 25 = 75 \Rightarrow 5\beta^2 + 30\beta + 40 = 0 \Rightarrow \beta^2 + 6\beta + 8 = 0$. This gives $\beta = -2$ or $-4$. For $\beta = -2, \alpha = 8$ (making $|\mathbf{a}|^2 = 1+25+64 = 90$). For $\beta = -4, \alpha = 4$ (making $|\mathbf{a}|^2 = 1+25+16 = 42$). The greatest value is 90.

   Step Solution:

    1.  Apply perpendicularity condition: $\mathbf{a} \cdot \mathbf{b} = 0 \Rightarrow (1)(1) + (5)(3) + \alpha\beta = 16 + \alpha\beta = 0 \Rightarrow \alpha\beta = -16$.

    2.  Find the cross product $\mathbf{b} \times \mathbf{c}$ in terms of $\beta$: $\mathbf{b} \times \mathbf{c} = (-9-2\beta)\hat{i} + (3-\beta)\hat{j} + 5\hat{k}$.

    3.  Use the magnitude condition: $(-9-2\beta)^2 + (3-\beta)^2 + 25 = (5\sqrt{3})^2 = 75 \Rightarrow 5\beta^2 + 30\beta + 115 = 75$.

    4.  Solve for $\beta$: $5\beta^2 + 30\beta + 40 = 0 \Rightarrow \beta^2 + 6\beta + 8 = 0 \Rightarrow \beta = -2, -4$.

    5.  Calculate $|\mathbf{a}|^2 = 1^2 + 5^2 + \alpha^2$: If $\beta=-2, \alpha=8 \Rightarrow |\mathbf{a}|^2 = 26+64 = \mathbf{90}$. If $\beta=-4, \alpha=4 \Rightarrow |\mathbf{a}|^2 = 26+16 = 42$.

   Difficulty level: Hard

   Concept Name: Vector Cross Product and Orthogonality

   Short cut solution: Use the identity $|\mathbf{b} \times \mathbf{c}|^2 = |\mathbf{b}|^2|\mathbf{c}|^2 - (\mathbf{b}\cdot\mathbf{c})^2$ to avoid finding the full cross product vector. $75 = (10+\beta^2)(14) - (5-3\beta)^2$ simplifies to the same quadratic in $\beta$ much faster.

 Question 157

   Question: Let $a, b, c \in \mathbb{R}$ be such that $a^2 + b^2 + c^2 = 1$. If $a \cos \theta = b \cos \left( \theta + \frac{2\pi}{3} \right) = c \cos \left( \theta + \frac{4\pi}{3} \right)$, where $\theta \in (0, \pi/2)$, then the angle between the vectors $a\hat{i} + b\hat{j} + c\hat{k}$ and $b\hat{i} + c\hat{j} + a\hat{k}$ is :

   Options: 

    A. $\pi/2$

    B. $2\pi/3$

    C. $\pi/9$

    D. 0

   Correct Answer: A

   Year: Sep. 03, 2020 (II)

   Solution: Let $a \cos \theta = b \cos (\theta + \frac{2\pi}{3}) = c \cos (\theta + \frac{4\pi}{3}) = k$. Thus $a = k/\cos \theta, b = k/\cos(\theta + 2\pi/3), c = k/\cos(\theta + 4\pi/3)$. Calculating the sum $ab + bc + ca$ using trigonometric identities for the denominators shows that the sum equals zero. Since the dot product of the two vectors is $ab + bc + ca$, and this sum is 0, the angle $\phi$ satisfies $\cos \phi = 0$.

   Step Solution:

    1.  Set the given equalities to a constant: $a \cos \theta = b \cos(\theta + 120^\circ) = c \cos(\theta + 240^\circ) = k$.

    2.  Express $a, b, c$ in terms of $k$ and $\theta$: $a = \frac{k}{\cos \theta}, b = \frac{k}{\cos(\theta+120^\circ)}, c = \frac{k}{\cos(\theta+240^\circ)}$.

    3.  Define the dot product of the two vectors $\vec{u}(a,b,c)$ and $\vec{v}(b,c,a)$: $\vec{u} \cdot \vec{v} = ab + bc + ca$.

    4.  Apply the identity $\cos \theta + \cos(\theta + 120^\circ) + \cos(\theta + 240^\circ) = 0$ to the expression for $ab+bc+ca$.

    5.  Conclusion: Since $ab + bc + ca = 0$, the angle between the vectors is $\cos^{-1}(0) = \pi/2$.

   Difficulty level: Medium

   Concept Name: Scalar Product (Dot Product) and Trigonometric Identities

   Short cut solution: Recognize the 120-degree symmetry in the angles. For any angle $\theta$, the sum of cosines of angles separated by $120^\circ$ is zero, which directly leads to $1/a + 1/b + 1/c$ being related to a zero sum, eventually showing $ab+bc+ca = 0$.

 Question 159

   Question: If $\vec{a}$ and $\vec{b}$ are unit vectors, then the greatest value of $\sqrt{3} |\vec{a} + \vec{b}| + |\vec{a} - \vec{b}|$ is

   Options: (Numerical Answer Type, Answer is 4)

   Correct Answer: 4

   Year: NA Sep. 06, 2020 (I)

   Solution: Let the angle between unit vectors $\vec{a}$ and $\vec{b}$ be $\theta$. Then $|\vec{a} + \vec{b}| = 2 \cos(\theta/2)$ and $|\vec{a} - \vec{b}| = 2 \sin(\theta/2)$. The expression becomes $2 [\sqrt{3} \cos(\theta/2) + \sin(\theta/2)]$. The maximum value of $A \cos x + B \sin x$ is $\sqrt{A^2 + B^2}$. Thus, the maximum value is $2\sqrt{(\sqrt{3})^2 + (1)^2} = 4$.

   Step Solution:

    1.  Use unit vector magnitude sum/difference formulas: $|\vec{a} + \vec{b}| = 2\cos(\theta/2)$ and $|\vec{a} - \vec{b}| = 2\sin(\theta/2)$.

    2.  Substitute these into the target expression: $\sqrt{3}(2\cos(\theta/2)) + 2\sin(\theta/2)$.

    3.  Factor out the 2: $2[\sqrt{3}\cos(\theta/2) + 1\sin(\theta/2)]$.

    4.  Apply the maximum value formula for trigonometric linear combinations: $\text{Max} = \sqrt{A^2 + B^2}$.

    5.  Calculate: $2 \times \sqrt{3 + 1} = 2 \times 2 = 4$.

   Difficulty level: Easy

   Concept Name: Vector Magnitudes and Trigonometric Extremum

   Short cut solution: Directly substitute $|\vec{a}+\vec{b}| = 2\cos \phi$ and $|\vec{a}-\vec{b}| = 2\sin \phi$. The expression is $2(\sqrt{3}\cos \phi + \sin \phi)$, which has a maximum of $2 \times 2 = 4$.

 Question 160

   Question: If $\vec{x}$ and $\vec{y}$ be two non-zero vectors such that $|\vec{x} + \vec{y}| = |\vec{x}|$ and $2\vec{x} + \lambda \vec{y}$ is perpendicular to $\vec{y}$, then the value of $\lambda$ is

   Options: (Numerical Answer Type, Answer is 1)

   Correct Answer: 1

   Year: NA Sep. 06, 2020 (II)

   Solution: From $|\vec{x} + \vec{y}| = |\vec{x}|$, squaring both sides gives $|\vec{x}|^2 + 2\vec{x} \cdot \vec{y} + |\vec{y}|^2 = |\vec{x}|^2$, which simplifies to $2\vec{x} \cdot \vec{y} + |\vec{y}|^2 = 0$. The condition $(2\vec{x} + \lambda \vec{y}) \perp \vec{y}$ means $(2\vec{x} + \lambda \vec{y}) \cdot \vec{y} = 0$, which expands to $2\vec{x} \cdot \vec{y} + \lambda |\vec{y}|^2 = 0$. Comparing the two equations shows $\lambda = 1$.

   Step Solution:

    1.  Square the magnitude condition: $|\vec{x} + \vec{y}|^2 = |\vec{x}|^2$.

    2.  Expand using dot products: $|\vec{x}|^2 + 2\vec{x} \cdot \vec{y} + |\vec{y}|^2 = |\vec{x}|^2$.

    3.  Isolate the relationship: $2\vec{x} \cdot \vec{y} = -|\vec{y}|^2 \quad \dots (i)$.

    4.  Apply the perpendicularity condition: $(2\vec{x} + \lambda \vec{y}) \cdot \vec{y} = 0 \Rightarrow 2\vec{x} \cdot \vec{y} + \lambda |\vec{y}|^2 = 0 \quad \dots (ii)$.

    5.  Substitute (i) into (ii): $-|\vec{y}|^2 + \lambda |\vec{y}|^2 = 0$. Since $\vec{y}$ is non-zero, $\lambda = 1$.

   Difficulty level: Easy

   Concept Name: Scalar Product (Dot Product) and Magnitude Identities

   Short cut solution: Geometrically, $|\vec{x}+\vec{y}|=|\vec{x}|$ implies the projection of $\vec{y}$ onto $\vec{x}$ is half of $-\vec{y}$'s magnitude. The perpendicularity $(2\vec{x} + \lambda \vec{y}) \perp \vec{y}$ requires $\lambda$ to exactly balance that projection, leading immediately to $\lambda=1$.

Question 161

   Question: Let $\vec{a}$, $\vec{b}$ and $\vec{c}$ be three unit vectors such that $|\vec{a} - \vec{b}|^2 + |\vec{a} - \vec{c}|^2 = 8$. Then $|\vec{a} + 2\vec{b}|^2 + |\vec{a} + 2\vec{c}|^2$ is equal to:

   Options: (Numerical Answer Type)

   Correct Answer: 2

   Year: NA Sep. 02, 2020 (I)

   Solution: The source defines the magnitudes $|\vec{a}| = |\vec{b}| = 1$. Expanding $|\vec{a} - \vec{b}|^2 + |\vec{a} - \vec{c}|^2 = 8$ leads to the determination that $\vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c} = -2$. Substituting this into the expansion of the required expression $|\vec{a} + 2\vec{b}|^2 + |\vec{a} + 2\vec{c}|^2$ yields the final result of 2.

   Step Solution:

    1.  List magnitudes for unit vectors: $|\vec{a}|^2 = |\vec{b}|^2 = |\vec{c}|^2 = 1$.

    2.  Expand the given equation: $(|\vec{a}|^2 + |\vec{b}|^2 - 2\vec{a} \cdot \vec{b}) + (|\vec{a}|^2 + |\vec{c}|^2 - 2\vec{a} \cdot \vec{c}) = 8$.

    3.  Simplify to find the sum of dot products: $4 - 2(\vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c}) = 8 \Rightarrow \mathbf{\vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c} = -2}$.

    4.  Expand the target expression: $(|\vec{a}|^2 + 4|\vec{b}|^2 + 4\vec{a} \cdot \vec{b}) + (|\vec{a}|^2 + 4|\vec{c}|^2 + 4\vec{a} \cdot \vec{c})$.

    5.  Substitute known values: $10 + 4(\vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c}) = 10 + 4(-2) = \mathbf{2}$.

   Difficulty level: Easy

   Concept Name: Vector Magnitudes and Scalar Product Expansion

   Short cut solution: For unit vectors, $|\vec{a}-\vec{b}|^2$ has a maximum value of 4 (when $\vec{a} = -\vec{b}$). For the sum of two such terms to be 8, both must be 4, meaning $\vec{a} \cdot \vec{b} = -1$ and $\vec{a} \cdot \vec{c} = -1$. The target expression then becomes $2(1 + 4 - 4) = 2$.

 Question 169

   Question: Let $\vec{a} = 2\hat{i} + \lambda_1\hat{j} + 3\hat{k}$, $\vec{b} = 4\hat{i} + (3 - \lambda_2)\hat{j} + 6\hat{k}$ and $\vec{c} = 3\hat{i} + 6\hat{j} + (\lambda_3 - 1)\hat{k}$ be three vectors such that $\vec{b} = 2\vec{a}$ and $\vec{a}$ is perpendicular to $\vec{c}$. Then a possible value of $(\lambda_1, \lambda_2, \lambda_3)$ is:

   Options: 

    A. (1, 3, 1)

    B. $(-1/2, 4, 0)$

    C. $(1/2, 4, -2)$

    D. (1, 5, 1)

   Correct Answer: B

   Year: Jan. 10, 2019 (I)

   Solution: Since $\vec{b} = 2\vec{a}$, we compare the $\hat{j}$ components: $3 - \lambda_2 = 2\lambda_1$. From the perpendicularity $\vec{a} \perp \vec{c}$, the dot product $\vec{a} \cdot \vec{c} = 0$, which gives $6 + 6\lambda_1 + 3(\lambda_3 - 1) = 0$. Solving these simultaneous equations identifies $(-1/2, 4, 0)$ as a valid set.

   Step Solution:

    1.  Apply vector equality $\vec{b} = 2\vec{a}$ to $\hat{j}$ components: $3 - \lambda_2 = 2\lambda_1 \Rightarrow \mathbf{\lambda_2 = 3 - 2\lambda_1}$.

    2.  Set up the dot product for perpendicularity: $\vec{a} \cdot \vec{c} = (2)(3) + (\lambda_1)(6) + (3)(\lambda_3 - 1) = 0$.

    3.  Simplify the dot product equation: $6 + 6\lambda_1 + 3\lambda_3 - 3 = 0 \Rightarrow \mathbf{2\lambda_1 + \lambda_3 = -1}$.

    4.  Check Option B values: $\lambda_1 = -1/2, \lambda_2 = 4, \lambda_3 = 0$.

    5.  Verify: $4 = 3 - 2(-1/2) = 4$ (Correct) and $2(-1/2) + 0 = -1$ (Correct).

   Difficulty level: Easy

   Concept Name: Scalar Product and Vector Component Equality

   Short cut solution: Rapidly test the condition $\lambda_2 = 3 - 2\lambda_1$ against the options. Only Option B satisfies $4 = 3 - 2(-1/2)$.

 Question 192

   Question: Let $\vec{a}$ and $\vec{b}$ be two unit vectors such that $|\vec{a} + \vec{b}| = \sqrt{3}$. If $\vec{c} = \vec{a} + 2\vec{b} + 3(\vec{a} \times \vec{b})$, then $2|\vec{c}|$ is equal to:

   Options: 

    A. √55

    B. √37

    C. √51

    D. √43

   Correct Answer: A

   Year: Online April 10, 2015

   Solution: From $|\vec{a} + \vec{b}| = \sqrt{3}$ for unit vectors, the angle between $\vec{a}$ and $\vec{b}$ is $60^\circ$. The vector $\vec{c}$ is composed of a part in the plane of $\vec{a}, \vec{b}$ and a part perpendicular to it ($3\vec{a} \times \vec{b}$). Calculating the magnitude squared of these orthogonal components leads to $|\vec{c}|^2 = 55/4$.

   Step Solution:

    1.  Determine the angle $\theta$ between $\vec{a}$ and $\vec{b}$: $|\vec{a}+\vec{b}|^2 = 3 \Rightarrow 1+1+2\cos\theta = 3 \Rightarrow \mathbf{\cos\theta = 1/2}$ ($\theta = 60^\circ$).

    2.  Note that $\vec{a} \times \vec{b}$ is perpendicular to both $\vec{a}$ and $\vec{b}$.

    3.  Calculate $|\vec{a}+2\vec{b}|^2 = |\vec{a}|^2 + 4|\vec{b}|^2 + 4\vec{a} \cdot \vec{b} = 1 + 4 + 4(1/2) = \mathbf{7}$.

    4.  Calculate $|3(\vec{a} \times \vec{b})|^2 = 9|\vec{a}|^2|\vec{b}|^2 \sin^2 60^\circ = 9(1)(1)(3/4) = \mathbf{27/4}$.

    5.  Find $2|\vec{c}|$: $2\sqrt{7 + 27/4} = 2\sqrt{55/4} = \mathbf{\sqrt{55}}$.

   Difficulty level: Medium

   Concept Name: Vector Magnitude and Orthogonal Decomposition

   Short cut solution: Use the property that for orthogonal vectors $\vec{u}$ and $\vec{v}$, $|\vec{u}+\vec{v}|^2 = |\vec{u}|^2 + |\vec{v}|^2$. Here $\vec{u} = \vec{a}+2\vec{b}$ and $\vec{v} = 3(\vec{a}\times\vec{b})$. Calculate their squared magnitudes ($7$ and $6.75$) and find the total.

Question 194

   Question: If $|\vec{a}| = 2, |\vec{b}| = 3$ and $|2\vec{a} - \vec{b}| = 5$, then $|2\vec{a} + \vec{b}|$ equals:

   Options: 

    A. 17

    B. 7

    C. 5

    D. 1

   Correct Answer: C

   Year: Online April 9, 2014

   Solution: Given $|2\vec{a} - \vec{b}| = 5$. We have $\sqrt{(2|\vec{a}|)^2 + |\vec{b}|^2 - 2 \times 2|\vec{a}||\vec{b}| \cos \theta} = 5$. Putting values of $|\vec{a}|$ and $|\vec{b}|$, we get $(2 \times 2)^2 + (3)^2 - 24 \cos \theta = 25 \Rightarrow 16 + 9 - 24 \cos \theta = 25 \Rightarrow \cos \theta = 0 \Rightarrow \theta = \pi/2$. Thus $|2\vec{a} + \vec{b}| = \sqrt{16 + 9 + 24 \cos \theta} = \sqrt{25} = 5$.

   Step Solution:

    1.  Apply magnitude formula to the given difference: $|2\vec{a} - \vec{b}|^2 = 4|\vec{a}|^2 + |\vec{b}|^2 - 4\vec{a} \cdot \vec{b}$.

    2.  Substitute known magnitudes: $5^2 = 4(2)^2 + 3^2 - 4\vec{a} \cdot \vec{b} \Rightarrow 25 = 16 + 9 - 4\vec{a} \cdot \vec{b}$.

    3.  Find the dot product relation: $25 = 25 - 4\vec{a} \cdot \vec{b} \Rightarrow \vec{a} \cdot \vec{b} = 0$.

    4.  Apply magnitude formula to the required sum: $|2\vec{a} + \vec{b}|^2 = 4|\vec{a}|^2 + |\vec{b}|^2 + 4\vec{a} \cdot \vec{b}$.

    5.  Calculate result: $|2\vec{a} + \vec{b}|^2 = 16 + 9 + 0 = 25 \Rightarrow |2\vec{a} + \vec{b}| = 5$.

   Difficulty level: Easy

   Concept Name: Scalar Product and Magnitude of Vector Sum/Difference

   Short cut solution: Use the parallelogram law extension: $|X+Y|^2 + |X-Y|^2 = 2(|X|^2 + |Y|^2)$. Let $X=2\vec{a}$ and $Y=\vec{b}$. Then $|2\vec{a}+\vec{b}|^2 + 5^2 = 2(4^2 + 3^2) \Rightarrow |2\vec{a}+\vec{b}|^2 + 25 = 50 \Rightarrow |2\vec{a}+\vec{b}|^2 = 25 \Rightarrow 5$.

 Question 200

   Question: If $\hat{a}, \hat{b}$ and $\hat{c}$ are unit vectors satisfying $\hat{a} - \sqrt{3}\hat{b} + \hat{c} = \vec{0}$, then the angle between the vectors $\hat{a}$ and $\hat{c}$ is :

   Options: 

    A. $\pi/4$

    B. $\pi/3$

    C. $\pi/6$

    D. $\pi/2$

   Correct Answer: B

   Year: Online April 22, 2013

   Solution: Let angle between $\hat{a}$ and $\hat{c}$ be $\theta$. Now, $\hat{a} - \sqrt{3}\hat{b} + \hat{c} = \vec{0} \Rightarrow (\hat{a} + \hat{c}) = \sqrt{3}\hat{b} \Rightarrow (\hat{a} + \hat{c}) \cdot (\hat{a} + \hat{c}) = 3(\hat{b} \cdot \hat{b}) \Rightarrow \hat{a} \cdot \hat{a} + 2\hat{a} \cdot \hat{c} + \hat{c} \cdot \hat{c} = 3 \times 1 \Rightarrow 1 + 2 \cos \theta + 1 = 3 \Rightarrow \cos \theta = 1/2 \Rightarrow \theta = \pi/3$.

   Step Solution:

    1.  Rearrange the vector equation: $\hat{a} + \hat{c} = \sqrt{3}\hat{b}$.

    2.  Square both sides using dot products: $(\hat{a} + \hat{c}) \cdot (\hat{a} + \hat{c}) = 3|\hat{b}|^2$.

    3.  Expand the squared magnitude: $|\hat{a}|^2 + |\hat{c}|^2 + 2\hat{a} \cdot \hat{c} = 3|\hat{b}|^2$.

    4.  Substitute unit vector magnitudes (1): $1 + 1 + 2\cos\theta = 3(1)$.

    5.  Solve for the angle: $2\cos\theta = 1 \Rightarrow \cos\theta = 1/2 \Rightarrow \theta = \pi/3$.

   Difficulty level: Easy

   Concept Name: Vector Magnitude and Scalar Product

   Short cut solution: The magnitude of the sum of two unit vectors is $|\hat{a}+\hat{c}| = 2\cos(\theta/2)$. Given this equals $\sqrt{3}$, we have $2\cos(\theta/2) = \sqrt{3} \Rightarrow \cos(\theta/2) = \sqrt{3}/2 \Rightarrow \theta/2 = 30^\circ \Rightarrow \theta = 60^\circ$ or $\pi/3$.

 Question 202

   Question: Let $\vec{a} = 2\hat{i} + \hat{j} - 2\hat{k}, \vec{b} = \hat{i} + \hat{j}$. If $\vec{c}$ is a vector such that $\vec{a} \cdot \vec{c} = |\vec{c}|, |\vec{c} - \vec{a}| = 2\sqrt{2}$ and the angle between $\vec{a} \times \vec{b}$ and $\vec{c}$ is $30^\circ$, then $|(\vec{a} \times \vec{b}) \times \vec{c}|$ equals:

   Options: 

    A. 1/2

    B. $3\sqrt{3}/2$

    C. 3

    D. 3/2

   Correct Answer: D

   Year: Online April 25, 2013

   Solution: $|\vec{a}| = \sqrt{4+1+4} = 3$. Given $|\vec{c}-\vec{a}| = 2\sqrt{2} \Rightarrow |\vec{c}|^2 + |\vec{a}|^2 - 2\vec{a} \cdot \vec{c} = 8 \Rightarrow |\vec{c}|^2 + 9 - 2|\vec{c}| = 8 \Rightarrow (|\vec{c}|-1)^2 = 0 \Rightarrow |\vec{c}| = 1$. Now, $\vec{a} \times \vec{b} = 2\hat{i} - 2\hat{j} + \hat{k}$ and $|\vec{a} \times \vec{b}| = 3$. Then $|(\vec{a} \times \vec{b}) \times \vec{c}| = |\vec{a} \times \vec{b}| |\vec{c}| \sin 30^\circ = 3 \cdot 1 \cdot 1/2 = 3/2$.

   Step Solution:

    1.  Find the magnitude of $\vec{a}$: $|\vec{a}| = \sqrt{2^2 + 1^2 + (-2)^2} = 3$.

    2.  Find magnitude $|\vec{c}|$ using $|\vec{c} - \vec{a}|^2 = 8$: $|\vec{c}|^2 + 3^2 - 2|\vec{c}| = 8 \Rightarrow |\vec{c}|^2 - 2|\vec{c}| + 1 = 0 \Rightarrow |\vec{c}| = 1$.

    3.  Compute the cross product vector $\vec{v} = \vec{a} \times \vec{b}$: $\vec{v} = (0 - (-2))\hat{i} - (0 - (-2))\hat{j} + (2 - 1)\hat{k} = 2\hat{i} - 2\hat{j} + \hat{k}$.

    4.  Find the magnitude of this cross product: $|\vec{v}| = \sqrt{2^2 + (-2)^2 + 1^2} = 3$.

    5.  Calculate target magnitude: $|\vec{v} \times \vec{c}| = |\vec{v}| |\vec{c}| \sin 30^\circ = 3 \cdot 1 \cdot (1/2) = 3/2$.

   Difficulty level: Medium

   Concept Name: Vector Magnitude and Vector Product

   Short cut solution: Use $|\vec{c}-\vec{a}|^2 = |\vec{c}|^2 + |\vec{a}|^2 - 2\vec{a}\cdot\vec{c} = 8$ and immediately substitute $|\vec{a}|=3$ and $\vec{a}\cdot\vec{c}=|\vec{c}|$ to find $|\vec{c}|=1$. The final magnitude is simply $(|\vec{a}\times\vec{b}|) \cdot (1) \cdot \sin 30^\circ$. Since $|\vec{a} \times \vec{b}| = 3$, the answer is $3 \times 0.5 = 1.5$.

Question 207

   Question: Let $\vec{a}$ and $\vec{b}$ be two unit vectors. If the vectors $\vec{c} = \hat{a} + 2\hat{b}$ and $\vec{d} = 5\hat{a} - 4\hat{b}$ are perpendicular to each other, then the angle between $\hat{a}$ and $\hat{b}$ is:

   Options: 

    A. $\pi/6$ 

    B. $\pi/2$ 

    C. $\pi/3$ 

    D. $\pi/4$

   Correct Answer: C

   Year: 2012

   Solution: Given that $\vec{c} = \widehat{\mathbf{a}} + 2 \widehat{\mathbf{b}}$ and $\vec{d} = 5 \widehat{\mathbf{a}} - 4 \widehat{\mathbf{b}}$ and $\left| \widehat{\mathbf{a}} \right| = \left| \widehat{\mathbf{b}} \right| = 1$. Since $\vec{c}$ and $\vec{d}$ are perpendicular to each other, $\therefore \vec{c} \cdot \vec{d} = 0 \Rightarrow (\hat{\mathbf{a}} + 2 \hat{\mathbf{b}}) \cdot (5 \hat{\mathbf{a}} - 4 \hat{\mathbf{b}}) = 0 \Rightarrow 5 + 6 \hat{\mathbf{a}} \cdot \hat{\mathbf{b}} - 8 = 0 \Rightarrow \hat{\mathbf{a}} \cdot \hat{\mathbf{b}} = \frac{1}{2} \Rightarrow \theta = \frac{\pi}{3}$.

   Step Solution:

    1.  Identify unit vector properties: Since they are unit vectors, $|\vec{a}| = 1$ and $|\vec{b}| = 1$.

    2.  Set up the dot product: Perpendicularity implies $\vec{c} \cdot \vec{d} = 0$, so $(\vec{a} + 2\vec{b}) \cdot (5\vec{a} - 4\vec{b}) = 0$.

    3.  Expand the expression: $5(\vec{a} \cdot \vec{a}) - 4(\vec{a} \cdot \vec{b}) + 10(\vec{b} \cdot \vec{a}) - 8(\vec{b} \cdot \vec{b}) = 0$.

    4.  Substitute magnitudes: $5(1) + 6(\vec{a} \cdot \vec{b}) - 8(1) = 0$, which simplifies to $6(\vec{a} \cdot \vec{b}) = 3$.

    5.  Solve for the angle: $\vec{a} \cdot \vec{b} = \cos \theta = 1/2$. Therefore, $\theta = \cos^{-1}(1/2) = \pi/3$.

   Difficulty level: Easy

   Concept Name: Scalar Product (Dot Product)

   Short cut solution: Use the expansion formula for dot products of linear combinations: $5 - 8 + (10 - 4)\cos \theta = 0$. Immediately solve $6\cos \theta = 3$ to get $\cos \theta = 1/2$.

 Question 208

   Question: If $\mathbf{a} + \mathbf{b} + \mathbf{c} = \mathbf{0}, | \vec{\mathbf{\psi}} | = 3, \left| \vec{\mathbf{b}} \right| = 5$ and $\left| \vec{\textrm{c}} \right| = 7$ , then the angle between $\vec{a}$ and $\vec{b}$ is

   Options: 

    A. $\pi/3$ 

    B. $\pi/4$ 

    C. $\pi/6$ 

    D. $\pi/2$

   Correct Answer: A

   Year: Online May 19, 2012

   Solution: Let $(\mathbf{a} + \mathbf{b} + \mathbf{c} = 0 \Rightarrow (\mathbf{a} + \mathbf{b}) = -\mathbf{c} \Rightarrow (\mathbf{a} + \mathbf{b})^2 = \mathbf{c}^2 \Rightarrow \mathbf{a}^2 + \mathbf{b}^2 + 2 \mathbf{a} \cdot \mathbf{b} = \mathbf{c}^2 \Rightarrow 9 + 25 + 2 \cdot 3 \cdot 5 \cos \theta = 49 \text{ (where } | \vec{\mathbf{a}} | = 3, | \vec{\mathbf{b}} | = 5 \text{ and } | \vec{\mathbf{c}} | = 7 \text{) } \cdot \cos \theta = \frac{1}{2} \Rightarrow \Theta = \frac{\pi}{3}$.

   Step Solution:

    1.  Isolate the target vectors: Rearrange the given sum to $\vec{a} + \vec{b} = -\vec{c}$.

    2.  Square the magnitude: $|\vec{a} + \vec{b}|^2 = |-\vec{c}|^2$.

    3.  Expand the sum: $|\vec{a}|^2 + |\vec{b}|^2 + 2\vec{a} \cdot \vec{b} = |\vec{c}|^2$.

    4.  Plug in the known values: $3^2 + 5^2 + 2(3)(5)\cos \theta = 7^2 \Rightarrow 9 + 25 + 30\cos \theta = 49$.

    5.  Calculate the angle: $30\cos \theta = 49 - 34 = 15$. Thus $\cos \theta = 1/2$, so $\theta = \pi/3$.

   Difficulty level: Easy

   Concept Name: Scalar Product and Vector Triangle Law

   Short cut solution: Use the cosine rule formula directly for the triangle formed by the vectors: $\cos \theta = \frac{|\vec{c}|^2 - |\vec{a}|^2 - |\vec{b}|^2}{2|\vec{a}||\vec{b}|} = \frac{49 - 9 - 25}{30} = \frac{15}{30} = \frac{1}{2}$.

 Question 218

   Question: If the vectors $\vec{\Delta a} = \hat{i} - \hat{j} + 2\hat{k}$ , $\vec{\mathrm{b}} = \hat{2}\hat{i} + \hat{4}\hat{j} + \hat{k}$ and $\vec{\mathrm{c}} = \hat{\lambda}\hat{i} + \hat{j} + \hat{\mu}\hat{\text{k}}$ are mutually orthogonal, then $(\lambda, \mu) =$

   Options: 

    A. (2, -3) 

    B. (-2, 3) 

    C. (3, -2) 

    D. (-3, 2)

   Correct Answer: D

   Year: 2010

   Solution: Given that $\vec{a}, \vec{b}$ and $\vec{c}$ are mutually orthogonal $\therefore \vec{a} \cdot \vec{b} = 0, \vec{b} \cdot \vec{c} = 0, \vec{c} \cdot \vec{a} = 0$. This gives $2\lambda + 4 + \mu = 0$ (i) and $\lambda - 1 + 2\mu = 0$ (ii). On solving (i) and (ii), we get $\lambda = -3, \mu = 2$.

   Step Solution:

    1.  Use orthogonality condition for $\vec{a}$ and $\vec{c}$: $\vec{a} \cdot \vec{c} = (1)(\lambda) + (-1)(1) + (2)(\mu) = 0 \Rightarrow \lambda - 1 + 2\mu = 0$.

    2.  Use orthogonality condition for $\vec{b}$ and $\vec{c}$: $\vec{b} \cdot \vec{c} = (2)(\lambda) + (4)(1) + (1)(\mu) = 0 \Rightarrow 2\lambda + 4 + \mu = 0$.

    3.  Set up simultaneous equations: (i) $\lambda + 2\mu = 1$ and (ii) $2\lambda + \mu = -4$.

    4.  Solve for one variable: From (ii), $\mu = -4 - 2\lambda$. Substitute into (i): $\lambda + 2(-4 - 2\lambda) = 1 \Rightarrow \lambda - 8 - 4\lambda = 1$.

    5.  Final values: $-3\lambda = 9 \Rightarrow \lambda = -3$. Then $\mu = -4 - 2(-3) = 2$. Pair is $(-3, 2)$.

   Difficulty level: Easy

   Concept Name: Scalar Product (Orthogonality)

   Short cut solution: Test the options against the dot product conditions. For Option D $(-3, 2)$, $\vec{a} \cdot \vec{c} = -3 - 1 + 4 = 0$ and $\vec{b} \cdot \vec{c} = -6 + 4 + 2 = 0$. Both are satisfied.

Question 222

   Question: The non-zero vectors $\vec{a}, \vec{b}$ and $\vec{c}$ are related by $\vec{a} = 8\vec{b}$ and $\vec{c} = -7\vec{b}$. Then the angle between $\vec{a}$ and $\vec{c}$ is

   Options: 

    A. 0 

    B. $\pi/4$ 

    C. $\pi/2$ 

    D. $\pi$

   Correct Answer: D

   Year: 2008

   Solution: Clearly $\vec{a} = -\frac{8}{7}\vec{c}$. $\Rightarrow \vec{a} \parallel \vec{c}$ and are opposite in direction. $\therefore$ Angle between $\vec{a}$ and $\vec{c}$ is $\pi$.

   Step Solution:

    1.  Write the given relations: $\vec{a} = 8\vec{b}$ and $\vec{c} = -7\vec{b}$.

    2.  Express $\vec{b}$ in terms of $\vec{c}$: $\vec{b} = -\frac{1}{7}\vec{c}$.

    3.  Substitute this expression for $\vec{b}$ into the equation for $\vec{a}$: $\vec{a} = 8(-\frac{1}{7}\vec{c})$.

    4.  Simplify to find the direct relationship: $\vec{a} = -\frac{8}{7}\vec{c}$.

    5.  Conclusion: Since $\vec{a} = k\vec{c}$ and $k$ is a negative scalar, the vectors are anti-parallel, meaning the angle is $\pi$.

   Difficulty level: Easy

   Concept Name: Parallel and Anti-parallel Vectors

   Short cut solution: Since both $\vec{a}$ and $\vec{c}$ are expressed as multiples of the same vector $\vec{b}$, and one multiple is positive (8) while the other is negative (-7), they must point in exactly opposite directions ($180^\circ$ or $\pi$).

Question 243

   Question: Let $\vec{u}, \vec{v}, \vec{w}$ be such that $|\vec{u}| = 1, |\vec{v}| = 2, |\vec{w}| = 3$. If the projection $\vec{v}$ along $\vec{u}$ is equal to that of $\vec{w}$ along $\vec{u}$ and $\vec{v}, \vec{w}$ are perpendicular to each other then $|\vec{u} - \vec{v} + \vec{w}|$ equals

   Options: 

    A. 14 

    B. $\sqrt{7}$ 

    C. $\sqrt{14}$ 

    D. 2

   Correct Answer: C

   Year: 2004

   Solution: Projection of $\vec{v}$ along $\vec{u} = \frac{\vec{v} \cdot \vec{u}}{|\vec{u}|} = \vec{v} \cdot \vec{u}$. Projection of $\vec{w}$ along $\vec{u} = \frac{\vec{w} \cdot \vec{u}}{|\vec{u}|} = \vec{w} \cdot \vec{u}$. Given $\vec{v} \cdot \vec{u} = \vec{w} \cdot \vec{u}$ (1). Also, $\vec{v} \cdot \vec{w} = 0$ (2). Now $|\vec{u} - \vec{v} + \vec{w}|^2 = |\vec{u}|^2 + |\vec{v}|^2 + |\vec{w}|^2 - 2\vec{u} \cdot \vec{v} - 2\vec{v} \cdot \vec{w} + 2\vec{u} \cdot \vec{w} = 1 + 4 + 9 + 0 = 14 \Rightarrow \sqrt{14}$.

   Step Solution:

    1.  List known magnitudes: $|\vec{u}|^2 = 1, |\vec{v}|^2 = 4, |\vec{w}|^2 = 9$.

    2.  Apply projection condition: $\frac{\vec{v} \cdot \vec{u}}{|\vec{u}|} = \frac{\vec{w} \cdot \vec{u}}{|\vec{u}|} \Rightarrow \vec{u} \cdot \vec{v} = \vec{u} \cdot \vec{w}$.

    3.  Apply perpendicularity condition: $\vec{v} \cdot \vec{w} = 0$.

    4.  Expand the required magnitude squared: $|\vec{u} - \vec{v} + \vec{w}|^2 = |\vec{u}|^2 + |\vec{v}|^2 + |\vec{w}|^2 - 2\vec{u} \cdot \vec{v} - 2\vec{v} \cdot \vec{w} + 2\vec{u} \cdot \vec{w}$.

    5.  Substitute relations: $1 + 4 + 9 - 2(\vec{u} \cdot \vec{v}) - 0 + 2(\vec{u} \cdot \vec{v}) = 14$. Result is $\sqrt{14}$.

   Difficulty level: Easy

   Concept Name: Vector Magnitudes and Scalar Projections

   Short cut solution: Recognize that the cross-product terms involving $\vec{u}$ ($2\vec{u} \cdot \vec{w}$ and $-2\vec{u} \cdot \vec{v}$) cancel each other out because the projections are equal. The result is simply $\sqrt{|\vec{u}|^2 + |\vec{v}|^2 + |\vec{w}|^2}$.

 Question 255

   Question: $\vec{a}, \vec{b}, \vec{c}$ are 3 vectors, such that $\vec{a} + \vec{b} + \vec{c} = \vec{0}, |\vec{a}| = 1, |\vec{b}| = 2, |\vec{c}| = 3$, then $\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}$ is equal to

   Options: 

    A. 1 

    B. 0 

    C. -7 

    D. 7

   Correct Answer: C

   Year: 2003

   Solution: $\Rightarrow (\vec{a} + \vec{b} + \vec{c}) \cdot (\vec{a} + \vec{b} + \vec{c}) = 0 \Rightarrow |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0 \Rightarrow 1 + 4 + 9 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0 \Rightarrow \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = \frac{-1-4-9}{2} = -7$.

   Step Solution:

    1.  Use the given condition: $\vec{a} + \vec{b} + \vec{c} = \vec{0}$.

    2.  Square the vector equation: $(\vec{a} + \vec{b} + \vec{c}) \cdot (\vec{a} + \vec{b} + \vec{c}) = 0$.

    3.  Expand the dot product: $|\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0$.

    4.  Substitute given magnitudes ($1, 2, 3$): $1^2 + 2^2 + 3^2 + 2(\text{Sum}) = 0$.

    5.  Solve for the sum: $14 + 2(\text{Sum}) = 0 \Rightarrow \text{Sum} = -7$.

   Difficulty level: Easy

   Concept Name: Magnitude of Vector Sum

   Short cut solution: For any set of vectors summing to zero, the sum of their pairwise dot products is always $-\frac{1}{2}$ times the sum of their squared magnitudes. Calculate $-\frac{1}{2}(1 + 4 + 9) = -7$.

Question 258

   Question: Let $\vec{u} = \hat{i} + \hat{j}, \vec{v} = \hat{i} - \hat{j}$ and $\vec{w} = \hat{i} + 2\hat{j} + 3\hat{k}$. If $\hat{n}$ is a unit vector such that $\vec{u} \cdot \hat{n} = 0$ and $\vec{v} \cdot \hat{n} = 0$, then $|\vec{w} \cdot \hat{n}|$ is equal to

   Options: 

    A. 3

    B. 0

    C. 1

    D. 2

   Correct Answer: A

   Year: 2003

   Solution: Given that $\vec{u} \cdot \hat{n} = 0$ and $\vec{v} \cdot \hat{n} = 0 \Rightarrow \hat{n}$ is perpendicular to both $\vec{u}$ and $\vec{v}, \therefore \hat{n} = \frac{\vec{u} \times \vec{v}}{|\vec{u}| |\vec{v}|}$. Calculating the cross product: $\hat{n} = \frac{-2\hat{k}}{2} = -\hat{k}$. Then $|\vec{w} \cdot \hat{n}| = |(\hat{i} + 2\hat{j} + 3\hat{k}) \cdot (-\hat{k})| = |-3| = 3$.

   Step Solution:

    1.  Identify the relationship: Since $\vec{u} \cdot \hat{n} = 0$ and $\vec{v} \cdot \hat{n} = 0$, the vector $\hat{n}$ must be perpendicular to the plane containing $\vec{u}$ and $\vec{v}$.

    2.  Calculate the cross product: Compute $\vec{u} \times \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 0 \\ 1 & -1 & 0 \end{vmatrix} = \hat{i}(0) - \hat{j}(0) + \hat{k}(-1 - 1) = -2\hat{k}$.

    3.  Find the unit vector $\hat{n}$: Normalize the result: $\hat{n} = \frac{-2\hat{k}}{|-2\hat{k}|} = \frac{-2\hat{k}}{2} = -\hat{k}$.

    4.  Perform the final dot product: Calculate $\vec{w} \cdot \hat{n} = (\hat{i} + 2\hat{j} + 3\hat{k}) \cdot (-\hat{k}) = 0 + 0 - 3 = -3$.

    5.  Determine magnitude: Take the absolute value: $|-3| = 3$.

   Difficulty level: Easy

   Concept Name: Vector Cross Product and Dot Product

   Short cut solution: Since $\vec{u}$ and $\vec{v}$ both lie in the $xy$-plane (their $k$-components are 0), any vector perpendicular to both must lie along the $z$-axis ($\hat{k}$). For $\vec{w} = \hat{i} + 2\hat{j} + 3\hat{k}$, the absolute value of its dot product with a unit vector along the $z$-axis is simply the magnitude of its $k$-component: $|3| = 3$.

 Question 266

   Question: If $|\vec{a}| = 5, |\vec{b}| = 4, |\vec{c}| = 3$ thus what will be the value of $|\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}|$, given that $\vec{a} + \vec{b} + \vec{c} = \vec{0}$?

   Options: 

    A. 25

    B. 50

    C. -25

    D. -50

   Correct Answer: A

   Year: 2003 (Sequence context)

   Solution: Given that $\vec{a} + \vec{b} + \vec{c} = 0 \Rightarrow |\vec{a} + \vec{b} + \vec{c}|^2 = 0 \Rightarrow |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0 \Rightarrow 25 + 16 + 9 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0 \Rightarrow \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = -25$. The magnitude is 25.

   Step Solution:

    1.  Start with the sum condition: We are given $\vec{a} + \vec{b} + \vec{c} = \vec{0}$.

    2.  Square the vector equation: $(\vec{a} + \vec{b} + \vec{c}) \cdot (\vec{a} + \vec{b} + \vec{c}) = 0$.

    3.  Expand the expression: $|\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0$.

    4.  Substitute known magnitudes: $5^2 + 4^2 + 3^2 + 2(\text{Sum}) = 0 \Rightarrow 25 + 16 + 9 + 2(\text{Sum}) = 0$.

    5.  Solve for the target value: $50 + 2(\text{Sum}) = 0 \Rightarrow \text{Sum} = -25$. Taking the absolute value, the result is 25.

   Difficulty level: Easy

   Concept Name: Magnitude of Vector Sum

   Short cut solution: Use the direct formula: $\sum \vec{a} \cdot \vec{b} = -\frac{1}{2} \sum |\vec{a}|^2$. Here, $-\frac{1}{2}(25+16+9) = -25$. The absolute value is 25.

Question 267

   Question: If $\vec{a}, \vec{b}, \vec{c}$ are vectors such that $\vec{a} + \vec{b} + \vec{c} = \vec{0}$ and $|\vec{a}| = 7, |\vec{b}| = 5, |\vec{c}| = 3$ then angle between vector $\vec{b}$ and $\vec{c}$ is

   Options: 

    A. $60^\circ$

    B. 30°

    C. 45°

    D. $90^\circ$

   Correct Answer: A

   Year: 2002

   Solution: Given that $\vec{a} + \vec{b} + \vec{c} = 0 \Rightarrow \vec{b} + \vec{c} = -\vec{a} \Rightarrow |\vec{b} + \vec{c}|^2 = |\vec{a}|^2 = 49$. Expanding: $5^2 + 3^2 + 2\vec{b} \cdot \vec{c} = 49 \Rightarrow 34 + 2(5)(3)\cos \theta = 49 \Rightarrow 30\cos \theta = 15 \Rightarrow \cos \theta = 1/2 \Rightarrow \theta = 60^\circ$.

   Step Solution:

    1.  Isolate the relevant vectors: Rearrange the sum to $\vec{b} + \vec{c} = -\vec{a}$.

    2.  Square both sides: $|\vec{b} + \vec{c}|^2 = |-\vec{a}|^2 \Rightarrow |\vec{b}|^2 + |\vec{c}|^2 + 2\vec{b} \cdot \vec{c} = |\vec{a}|^2$.

    3.  Plug in the magnitudes: $5^2 + 3^2 + 2|\vec{b}||\vec{c}|\cos \theta = 7^2$.

    4.  Simplify the equation: $25 + 9 + 30\cos \theta = 49 \Rightarrow 34 + 30\cos \theta = 49$.

    5.  Calculate the angle: $30\cos \theta = 15 \Rightarrow \cos \theta = 1/2 \Rightarrow \theta = 60^\circ$.

   Difficulty level: Easy

   Concept Name: Scalar Product and Vector Triangle Law

   Short cut solution: Use the Law of Cosines for the triangle formed by the vectors: $\cos \theta = \frac{|\vec{a}|^2 - |\vec{b}|^2 - |\vec{c}|^2}{2|\vec{b}||\vec{c}|}$. Substituting the values: $\frac{49 - 25 - 9}{2(5)(3)} = \frac{15}{30} = \frac{1}{2} \Rightarrow 60^\circ$.

 Question 269

   Question: $\vec{\mathbf{a}} = 3\hat{\mathrm{i}} - 5\hat{\mathrm{j}}$ and $\vec{\mathbf{b}} = 6\hat{\mathbf{i}} + 3\hat{\mathbf{j}}$ are two vectors and $\overrightarrow{\mathrm{c}}$ is a vector such that $\vec{\mathrm{c}} = \vec{\mathrm{a}} \times \vec{\mathrm{b}}$ then $|\vec{\mathrm{a}}| : |\vec{\mathrm{b}}| : |\vec{\mathrm{c}}|$ is:

   Options: 

    A. $\sqrt{34} : \sqrt{45} : \sqrt{39}$

    B. $\sqrt{34} : \sqrt{45} : 39$

    C. 34: 39: 45

    D. 39: 35: 34

   Correct Answer: B

   Year: 2002

   Solution: We have $\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -5 & 0 \\ 6 & 3 & 0 \end{vmatrix} = 39 \hat{\mathbf{k}} = \overrightarrow{\mathbf{c}}$. Also $|\overrightarrow{\mathbf{a}}| = \sqrt{34}, |\overrightarrow{\mathbf{b}}| = \sqrt{45}, |\overrightarrow{\mathbf{c}}| = 39$. Hence $|\overrightarrow{\mathfrak{a}}| : |\overrightarrow{\mathfrak{b}}| : |\overrightarrow{\mathfrak{c}}| = \sqrt{34} : \sqrt{45} : 39$.

   Step Solution:

    1.  Calculate $|\vec{a}|$: $\sqrt{3^2 + (-5)^2} = \sqrt{9 + 25} = \sqrt{34}$.

    2.  Calculate $|\vec{b}|$: $\sqrt{6^2 + 3^2} = \sqrt{36 + 9} = \sqrt{45}$.

    3.  Find vector $\vec{c}$ using cross product: $\vec{a} \times \vec{b} = (3\hat{i} - 5\hat{j}) \times (6\hat{i} + 3\hat{j}) = 9(\hat{i} \times \hat{j}) - 30(\hat{j} \times \hat{i}) = 9\hat{k} + 30\hat{k} = 39\hat{k}$.

    4.  Calculate magnitude $|\vec{c}|$: $\sqrt{0^2 + 0^2 + 39^2} = 39$.

    5.  Determine the ratio: $|\vec{a}| : |\vec{b}| : |\vec{c}| = \sqrt{34} : \sqrt{45} : 39$.

   Difficulty level: Easy

   Concept Name: Vector Cross Product and Magnitude

   Short cut solution: For two-dimensional vectors in the $xy$-plane, the magnitude of the cross product $\vec{a} \times \vec{b}$ is simply $|a_x b_y - a_y b_x|$. Here, $|(3)(3) - (-5)(6)| = |9 + 30| = 39$. Then find the magnitudes of $\vec{a}$ and $\vec{b}$ to complete the ratio.

 Question 272

   Question: If $|\vec{\mathbf{a}}| = 4, |\vec{\mathbf{b}}| = 2$ and the angle between $\vec{\mathbf{a}}$ and $\vec{\mathbf{b}}$ is $\pi / 6$ then $(\vec{\mathbf{a}} \times \vec{\mathbf{b}})^2$ is equal to:

   Options: 

    A. 48

    B. 16

    C. $\vec{a}$

    D. None of these

   Correct Answer: B

   Year: 2002

   Solution: Since, $\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}| \cos \frac{\pi}{6} = 4 \times 2 \times \frac{\sqrt{3}}{2} = 4\sqrt{3}$. Using the identity $(\vec{\mathbf{a}} \times \vec{\mathbf{b}})^2 + (\vec{\mathbf{a}} \cdot \vec{\mathbf{b}})^2 = |\vec{\mathbf{a}}|^2 |\vec{\mathbf{b}}|^2 \Rightarrow (\vec{\mathbf{a}} \times \vec{\mathbf{b}})^2 + 48 = 16 \times 4 \Rightarrow (\vec{\mathbf{a}} \times \vec{\mathbf{b}})^2 = 16$.

   Step Solution:

    1.  Identify magnitudes and angle: $|\vec{a}| = 4$, $|\vec{b}| = 2$, and $\theta = 30^\circ$ ($\pi/6$).

    2.  State the cross product magnitude formula: $|\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}| \sin \theta$.

    3.  Substitute the values: $|\vec{a} \times \vec{b}| = (4)(2) \sin(30^\circ)$.

    4.  Perform the calculation: $8 \times (1/2) = 4$.

    5.  Square the result for the final answer: $(4)^2 = 16$.

   Difficulty level: Easy

   Concept Name: Vector Product Magnitude

   Short cut solution: Use the direct formula $(\vec{a} \times \vec{b})^2 = |\vec{a}|^2 |\vec{b}|^2 \sin^2 \theta$. Substituting gives $(4)^2 (2)^2 (\sin 30^\circ)^2 = 16 \times 4 \times (1/4) = 16$.