$\overrightarrow a + \overrightarrow b + \overrightarrow c = 0$ , $\left| {\overrightarrow a } \right| = 1\,\,\,\left| {\overrightarrow b } \right| = 2,\,\,\,\left| {\overrightarrow c } \right| = 3,$,
then ${\overrightarrow a .\overrightarrow b + \overrightarrow b .\overrightarrow c + \overrightarrow c .\overrightarrow a }$ is equal to :
$\left( {1,b,{b^2}} \right)$ and $\left( {1,c,{c^2}} \right)\,$ are non-coplanar, then the product $abc$ equals :
vectors $7\widehat i - 4\widehat j + 7\widehat k,\widehat i - 6\widehat j + 10\widehat k, - \widehat i - 3\widehat j + 4\widehat k$ and $5\widehat i - \widehat j + 5\widehat k$ respectively. Then $ABCD$ is a :
If the distance of the point $P(43, \alpha, \beta)$, $\beta < 0$, from the line $\vec{r} = 4\hat{i} - \hat{k} + \mu (2\hat{i} + 3\hat{k}), \mu \in \mathbb{R}$ along a line with direction ratios $3, -1, 0$ is $13\sqrt{10}$, then $\alpha^2 + \beta^2$ is equal to ________
Explanation:
Point $P$ : $(43, \alpha, \beta)$ with the constraint $\beta<0$.
Target Line ( $L_1$ ) : Given by $\vec{r}=4 \hat{i}-\hat{k}+\mu(2 \hat{i}+3 \hat{k})$.
A general point $Q$ on this line is $(4+2 \mu, 0,-1+3 \mu)$.Direction of measurement : The distance is measured along a line with direction ratios $(3,-1,0)$.
Find Point $Q$ on the Target Line
Since the distance is measured along the direction $d=(3,-1,0)$, the vector $P Q$ must be parallel to $d$.
$ \begin{gathered} P Q=Q-P=(4+2 \mu-43, \quad 0-\alpha, \quad-1+3 \mu-\beta) \\ P Q=(2 \mu-39, \quad-\alpha, \quad 3 \mu-\beta-1) \\ P Q=(2 \mu-39, \quad-\alpha, \quad 3 \mu-\beta-1) \end{gathered} $
Because $P Q$ is parallel to $(3,-1,0)$, their components must be proportional :
- 1.$\frac{2 \mu-39}{3}=\frac{-\alpha}{-1} \Longrightarrow 2 \mu-39=3 \alpha \Longrightarrow 2 \mu=3 \alpha+39$
- 2.$\frac{3 \mu-\beta-1}{0}$ suggests the numerator must be zero for a finite ratio to exist: $3 \mu-\beta-1= 0 \Longrightarrow \beta=3 \mu-1$
Calculate the Distance $P Q$
The distance $P Q$ is given as $13 \sqrt{10}$. Using the distance formula on the proportional components where the ratio is $k$ :
$ \begin{gathered} P Q=k(3,-1,0)=(3 k,-k, 0) \\ |P Q|=\sqrt{(3 k)^2+(-k)^2+0^2}=\sqrt{10 k^2}=|k| \sqrt{10} \end{gathered} $
Given $|P Q|=13 \sqrt{10}$, we find $|k|=13$, so $k= \pm 13$.
Solve for $\alpha, \mu$, and $\beta$
From our proportionality equations :
$-\alpha=-k \Longrightarrow \alpha=k$
$2 \mu-39=3 k$
Case 1 : $k=13$
$\alpha=13$
$2 \mu-39=3(13)=39 \Longrightarrow 2 \mu=78 \Longrightarrow \mu=39$
$\beta=3 \mu-1=3(39)-1=116$
Constraint Check : $\beta<0$ is False. (Reject this case)
Case 2 : $k=-13$
$\alpha=-13$
$2 \mu-39=3(-13)=-39 \Longrightarrow 2 \mu=0 \Longrightarrow \mu=0$
$\beta=3(0)-1=-1$
Constraint Check : $\beta<0$ is True. (Accept this case)
$\alpha^2+\beta^2=(-13)^2+(-1)^2=169+1=170$
The value of $\alpha^2+\beta^2$ is 170 .
Let $P Q R$ be a triangle such that $\overrightarrow{P Q}=-2 \hat{i}-\hat{j}+2 \hat{k}$ and $\overrightarrow{\mathrm{PR}}=a \hat{\mathrm{i}}+b \hat{\mathrm{j}}-4 \hat{\mathrm{k}}, a, b \in \mathrm{Z}$. Let S be the point on QR , which is equidistant from the lines PQ and PR . If $|\overrightarrow{\mathrm{PR}}|=9$ and $\overrightarrow{\mathrm{PS}}=\hat{\mathrm{i}}-7 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}$, then the value of $3 a-4 b$ is $\_\_\_\_$
Explanation:
We are given that $|\overrightarrow{P R}|=9$ Using the distance formula for the vector $P R=a \hat{i}+b \hat{j}-4 \hat{k}$ :
$ \sqrt{a^2+b^2+(-4)^2}=9 $
$\Rightarrow $ $a^2+b^2+16=81$
$\Rightarrow $ $ a^2+b^2=65 $
The problem states that point $S$ lies on $Q R$ and is equidistant from lines $P Q$ and $P R$. In a triangle, the locus of points equidistant from two sides is the angle bisector. Therefore, $\overrightarrow{P S}$ is the internal angle bisector of $\angle Q P R$.
For an angle bisector $\overrightarrow{P S}$, it must be collinear with the sum of the unit vectors along the adjacent sides :
$ \overrightarrow{P S}=\lambda\left(\frac{\overrightarrow{P Q}}{|\overrightarrow{P Q}|}+\frac{\overrightarrow{P R}}{|\overrightarrow{P R}|}\right) $
where $\lambda$ is a scalar constant.
First, calculate $|\overrightarrow{P Q}|$ :
$ |\overrightarrow{P Q}|=\sqrt{(-2)^2+(-1)^2+2^2}=\sqrt{4+1+4}=3 $
Now, set up the bisector equation :
$ \overrightarrow{P S}=\lambda\left(\frac{-2 \hat{i}-\hat{j}+2 \hat{k}}{3}+\frac{a \hat{i}+b \hat{j}-4 \hat{k}}{9}\right) $
Multiply the first term by $\frac{3}{3}$ to get a common denominator:
$ \overrightarrow{P S}=\frac{\lambda}{9}((-6+a) \hat{i}+(-3+b) \hat{j}+(6-4) \hat{k}) $
$\Rightarrow $ $ \overrightarrow{P S}=\frac{\lambda}{9}((a-6) \hat{i}+(b-3) \hat{j}+2 \hat{k}) $
We are given
$ \overrightarrow{P S}=\hat{i}-7 \hat{j}+2 \hat{k} $
Comparing the $\hat{k}$-components:
$\Rightarrow $ $ \frac{\lambda}{9}(2)=2 \Rightarrow \lambda=9 $
Substitute $\lambda=9$ into other components:
For $\hat{i}$ :
$ a-6=1 \Rightarrow a=7 $
For $\hat{j}$ :
$ b-3=-7 \Rightarrow b=-4 $
Check against the constraint :
$ a^2+b^2=7^2+(-4)^2=49+16=65 $
The values are consistent. value of $3 a-4 b$ :
$ 3 a-4 b=3(7)-4(-4) $ $=21+16=\mathbf{3 7}$
Let a vector $\overrightarrow{\mathrm{a}}=\sqrt{2} \hat{i}-\hat{j}+\lambda \hat{k}, \lambda>0$, make an obtuse angle with the vector $\overrightarrow{\mathrm{b}}=-\lambda^2 \hat{i}+4 \sqrt{2} \hat{j}+4 \sqrt{2} \hat{k}$ and an angle $\theta, \frac{\pi}{6}<\theta<\frac{\pi}{2}$, with the positive $z$-axis. If the set of all possible values of $\lambda$ is $(\alpha, \beta)-\{\gamma\}$, then $\alpha+\beta+\gamma$ is equal to $\_\_\_\_$ .
Explanation:
$ \begin{aligned} & \vec{a}^{\wedge} \hat{k} \in\left(\frac{\pi}{6}, \frac{\pi}{2}\right) \\ & \frac{\lambda}{\sqrt{\left(2+1+\lambda^2\right)}} \in\left(0, \frac{\sqrt{3}}{2}\right) \Rightarrow 0<\frac{\lambda}{\sqrt{3+\lambda^2}}<\frac{\sqrt{3}}{2} \\ & \lambda>0 \& \lambda \in(-3,3) \Rightarrow \lambda \in(0,3) \ldots \ldots \ldots .(1) \end{aligned} $
$\vec{a}^{\wedge} \vec{b}$ is obtuse
$ \begin{aligned} & \vec{a} \cdot \vec{b}<0 \Rightarrow-\sqrt{2} \lambda^2-4 \sqrt{2}+4 \sqrt{2} \lambda<0 \\ & \Rightarrow \lambda^2-4 \lambda+4>0 \Rightarrow(\lambda-2)^2>0 \\ & \Rightarrow \lambda \in R-\{2\} \ldots \ldots .(2) \end{aligned} $
(1) \& (2) ⇒ $\lambda \in(0,3)-\{2\} \Rightarrow \alpha+\beta+\gamma=5$
Let the three sides of a triangle ABC be given by the vectors $2 \hat{i}-\hat{j}+\hat{k}, \hat{i}-3 \hat{j}-5 \hat{k}$ and $3 \hat{i}-4 \hat{j}-4 \hat{k}$. Let $G$ be the centroid of the triangle $A B C$. Then $6\left(|\overrightarrow{\mathrm{AG}}|^2+|\overrightarrow{\mathrm{BG}}|^2+|\overrightarrow{\mathrm{CG}}|^2\right)$ is equal to __________.
Explanation:
$\begin{aligned} &\text { Assuming Vertex } A \text { to be origin }\\ &\begin{aligned} & \vec{A}=\vec{a}_1=\overrightarrow{0} \\ & \vec{B}=\vec{a}_1+\vec{u}=\vec{u}=2 \hat{i}-\hat{j}+\hat{k} \\ & \vec{C}=\vec{a}_1+\vec{v}=\vec{v}=3 \hat{i}-4 \hat{j}-4 \hat{k} \end{aligned} \end{aligned}$
One solving
$\vec{A}=\overrightarrow{0}, \vec{B}=2 \hat{i}-\hat{j}+\hat{k}$ and $\vec{C}=3 \hat{i}-4 \hat{j}-4 \hat{k}$, are the position vector of vertices $A B$ and $C$ respectively.
$\begin{aligned} & \vec{G}=\frac{1}{3}(\vec{A}+\vec{B}+\vec{C})=\frac{1}{3}(\overrightarrow{0}+\vec{B}+\vec{C})=\frac{1}{3}(\vec{B}+\vec{C}) \\ & \Rightarrow \vec{G}=\frac{5}{3} \hat{i}-\frac{5}{3} \hat{j}-\hat{k} \\ & \overrightarrow{A G}=\vec{G}-\vec{A}=\vec{G} \\ & |\overrightarrow{A G}|^2=\left(\frac{5}{3}\right)^2+\left(\frac{5}{3}\right)^2+(1)^2=\frac{25}{9}+\frac{25}{9}+1=\frac{50}{9}+1=\frac{59}{9} \\ & \overrightarrow{B C}=\vec{G}-\vec{B} \\ & \vec{B}=2 \hat{i}-\hat{j}+\hat{k} \\ & |\overrightarrow{B G}|^2=\left(\frac{1}{3}\right)^3+\left(\frac{2}{3}\right)^2+4=\frac{1}{9}+\frac{4}{9}+4=\frac{5}{9}+4=\frac{41}{9} \\ & \overrightarrow{C G}=\vec{G}-\vec{C} \\ & \vec{C}=3 \hat{i}-4 \hat{j}-4 \hat{k} \\ & \left.\overrightarrow{C G}\right|^2=\left(\frac{4}{3}\right)^2+\left(\frac{7}{3}\right)^2+9=\frac{16}{9}+\frac{49}{9}+9=\frac{65}{9}+9=\frac{65}{9}+\frac{81}{9}=\frac{146}{9} \end{aligned}$
$6\left(|\overrightarrow{A G}|^2+|\overrightarrow{B G}|^2+|\overrightarrow{C G}|^2\right)=6 \cdot\left(\frac{59}{9}+\frac{41}{9}+\frac{146}{9}\right)=6 \cdot \frac{246}{9}=164$
Let $\vec{a}=\hat{i}+2 \hat{j}+\hat{k}, \vec{b}=3 \hat{i}-3 \hat{j}+3 \hat{k}, \vec{c}=2 \hat{i}-\hat{j}+2 \hat{k}$ and $\vec{d}$ be a vector such that $\vec{b} \times \vec{d}=\vec{c} \times \vec{d}$ and $\vec{a} \cdot \vec{d}=4$. Then $|(\vec{a} \times \vec{d})|^2$ is equal to___________.
Explanation:
Given Conditions :
The condition $\vec{b} \times \vec{d} = \vec{c} \times \vec{d}$ implies that $(\vec{b} - \vec{c}) \times \vec{d} = 0$.
This indicates that $\vec{d}$ is parallel to $\vec{b} - \vec{c}$.
Expressing $\vec{b} - \vec{c}$:
Calculate $\vec{b} - \vec{c} = (3 \hat{i} - 3 \hat{j} + 3 \hat{k}) - (2 \hat{i} - \hat{j} + 2 \hat{k}) = \hat{i} - 2 \hat{j} + \hat{k}$.
Express $\vec{d}$:
Since $\vec{d}$ is parallel to $\vec{b} - \vec{c}$, let $\vec{d} = \lambda(\hat{i} - 2 \hat{j} + \hat{k})$.
Using the condition $\vec{a} \cdot \vec{d} = 4$:
Compute $\vec{a} \cdot \vec{d} = (\hat{i} + 2 \hat{j} + \hat{k}) \cdot (\lambda(\hat{i} - 2 \hat{j} + \hat{k}))$.
This simplifies to $\lambda(1 \cdot 1 + 2 \cdot (-2) + 1 \cdot 1) = \lambda(1 - 4 + 1) = -2\lambda$.
Set $-2\lambda = 4$, which gives $\lambda = -2$.
Determine $\vec{d}$ using $\lambda$:
Therefore, $\vec{d} = -2(\hat{i} - 2 \hat{j} + \hat{k}) = -2 \hat{i} + 4 \hat{j} - 2 \hat{k}$.
Calculate $\vec{a} \times \vec{d}$:
Use the determinant form for the cross product:
$ \vec{a} \times \vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 1 \\ -2 & 4 & -2 \end{vmatrix} $
Calculate the determinant:
For $\hat{i}$: $(2 \times -2) - (1 \times 4) = -4 - 4 = -8$
For $\hat{j}$: $(-(1 \times -2) - (1 \times -2)) = 2 - 2 = 0$
For $\hat{k}$: $(1 \times 4) - (2 \times -2) = 4 + 4 = 8$
Thus, $\vec{a} \times \vec{d} = -8 \hat{i} + 0 \hat{j} + 8 \hat{k}$.
Calculate the magnitude squared:
Find $|\vec{a} \times \vec{d}|^2 = (-8)^2 + 0^2 + 8^2 = 64 + 0 + 64 = 128$.
Therefore, $|(\vec{a} \times \vec{d})|^2 = 128$.
Let $\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}=3 \hat{i}+2 \hat{j}-\hat{k}, \vec{c}=\lambda \hat{j}+\mu \hat{k}$ and $\hat{d}$ be a unit vector such that $\vec{a} \times \hat{d}=\vec{b} \times \hat{d}$ and $\vec{c} \cdot \hat{d}=1$. If $\vec{c}$ is perpendicular to $\vec{a}$, then $|3 \lambda \hat{d}+\mu \vec{c}|^2$ is equal to________
Explanation:
Given the vectors $\vec{a} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{b} = 3 \hat{i} + 2 \hat{j} - \hat{k}$, along with $\vec{c} = \lambda \hat{j} + \mu \hat{k}$, and $\hat{d}$ being a unit vector such that $\vec{a} \times \hat{d} = \vec{b} \times \hat{d}$ and $\vec{c} \cdot \hat{d} = 1$, we proceed as follows:
Determine $\hat{d}$:
Since $\vec{a} \times \hat{d} = \vec{b} \times \hat{d}$, we have:
$ (\vec{a} - \vec{b}) \times \hat{d} = 0 $
Thus, $\hat{d}$ is parallel to $\vec{a} - \vec{b}$. Calculate:
$ \vec{a} - \vec{b} = (-2 \hat{i} + \hat{j} + 2 \hat{k}) $
Therefore, we can express $\hat{d}$ as:
$ \hat{d} = \frac{-2}{3} \hat{i} - \frac{1}{3} \hat{j} + \frac{2}{3} \hat{k} $
Solve for $\lambda$ and $\mu$:
Using the condition $\vec{c} \cdot \hat{d} = 1$:
$ \frac{-\lambda}{3} + \frac{2 \mu}{3} = 1 $
Simplify to:
$ -\lambda + 2\mu = 3 \quad \text{...(i)} $
Since $\vec{c}$ is perpendicular to $\vec{a}$:
$ \vec{c} \cdot \vec{a} = 0 $
Which gives:
$ \lambda + \mu = 0 \quad \Rightarrow \lambda = -\mu $
Substitute $\lambda = -\mu$ into equation (i):
$ \mu + 2\mu = 3 \quad \Rightarrow 3\mu = 3 \quad \Rightarrow \mu = 1 $
Therefore, $\lambda = -1$.
Find $|3 \lambda \hat{d} + \mu \vec{c}|^2$:
Calculate:
$ |3 \lambda \hat{d} + \mu \vec{c}|^2 = 9 \lambda^2 |\hat{d}|^2 + \mu^2 |\vec{c}|^2 + 2 \cdot 3 \cdot \lambda \cdot \mu \cdot \vec{c} \cdot \vec{d} $
Substituting the determined values:
$ |\hat{d}|^2 = 1 $ (since $\hat{d}$ is a unit vector)
$ \vec{c} \cdot \vec{d} = 1 $
Thus:
$ = 9(-1)^2 \cdot 1 + 1^2 \cdot (1^2 + 1^2) + 2 \cdot 3 \cdot (-1) \cdot 1 \cdot 1 $
$ = 9 + 2 - 6 $
$ = 5 $
Hence, $|3 \lambda \hat{d} + \mu \vec{c}|^2 = 5$.
Let $\vec{a}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}, \overrightarrow{\mathrm{b}}=2 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}}$ and $\overrightarrow{\mathrm{d}}=\vec{a} \times \overrightarrow{\mathrm{b}}$. If $\overrightarrow{\mathrm{c}}$ is a vector such that $\vec{a} \cdot \overrightarrow{\mathrm{c}}=|\overrightarrow{\mathrm{c}}|$, $|\overrightarrow{\mathrm{c}}-2 \vec{a}|^2=8$ and the angle between $\overrightarrow{\mathrm{d}}$ and $\overrightarrow{\mathrm{c}}$ is $\frac{\pi}{4}$, then $|10-3 \overrightarrow{\mathrm{~b}} \cdot \overrightarrow{\mathrm{c}}|+|\overrightarrow{\mathrm{d}} \times \overrightarrow{\mathrm{c}}|^2$ is equal to _________.
Explanation:
$\begin{aligned} & \overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}} \\ & \overrightarrow{\mathrm{~b}}=2 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}} \\ & \overrightarrow{\mathrm{~d}}=\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}} \\ & =-\hat{\mathrm{i}}+\hat{\mathrm{j}} \\ & |\overrightarrow{\mathrm{c}}-2 \overrightarrow{\mathrm{a}}|^2=8 \\ & |\mathrm{c}|^2+4|\mathrm{a}|^2-4(\mathrm{a} \cdot \mathrm{c})=8 \\ & \mathrm{c}^2+12-4 \mathrm{c}=8 \\ & \mathrm{c}^2-4 \mathrm{c}+4=0 \\ & |\mathrm{c}|=2 \\ & \overrightarrow{\mathrm{~d}}=\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}} \\ & \overrightarrow{\mathrm{~d}} \times \overrightarrow{\mathrm{c}}=(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})^2 \times \overrightarrow{\mathrm{c}} \\ & \left(|\mathrm{~d}||\mathrm{c}| \sin \frac{\pi}{4}\right)^2=((\mathrm{a} \cdot \mathrm{c}) \cdot \mathrm{b}-(\mathrm{b} \cdot \mathrm{c}) \cdot \mathrm{a})^2 \\ & 4=4 \mathrm{~b}^2+(\mathrm{b} \cdot \mathrm{c})^2 2(\mathrm{a})^2-2(\mathrm{~b} \cdot \mathrm{c})(\mathrm{a} \cdot \mathrm{~b}) \end{aligned}$
$\begin{aligned} &\text { Let } \mathrm{b} . \mathrm{c}=\mathrm{x}\\ &\begin{aligned} & 4=36+3 x^2-20 x \\ & 3 x^2-20 x+32=0 \\ & 3 x^2-12 x-8 x+32=0 \\ & x=\frac{8}{3}, 4 \\ & \text { b.c }=\frac{8}{3}, 4 \\ & \text { b.c }=\frac{8}{3} \\ & \text { Now }|10-3 \mathrm{~b} . \mathrm{c}|+|\mathrm{d} \times \mathrm{c}|^2 \\ & |10-8|+(2)^2 \\ & \Rightarrow 6 \text { Ans. } \end{aligned} \end{aligned}$
Let $\vec{c}$ be the projection vector of $\vec{b}=\lambda \hat{i}+4 \hat{k}, \lambda>0$, on the vector $\vec{a}=\hat{i}+2 \hat{j}+2 \hat{k}$. If $|\vec{a}+\vec{c}|=7$, then the area of the parallelogram formed by the vectors $\vec{b}$ and $\vec{c}$ is _________.
Explanation:
To find the projection vector $\vec{c}$ of $\vec{b} = \lambda \hat{i} + 4 \hat{k}$ (where $\lambda > 0$) onto vector $\vec{a} = \hat{i} + 2 \hat{j} + 2 \hat{k}$, we use the formula for the projection of a vector:
$ \vec{c} = \left(\frac{\vec{b} \cdot \vec{a}}{|\vec{a}|^2}\right) \vec{a} $
Calculate the dot product $\vec{b} \cdot \vec{a}$:
$ \vec{b} \cdot \vec{a} = (\lambda \hat{i} + 4 \hat{k}) \cdot (\hat{i} + 2 \hat{j} + 2 \hat{k}) = \lambda \cdot 1 + 4 \cdot 2 = \lambda + 8 $
Calculate the magnitude of $\vec{a}$:
$ |\vec{a}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{9} = 3 $
Now substitute to find $\vec{c}$:
$ \vec{c} = \left(\frac{\lambda + 8}{9}\right)(\hat{i} + 2 \hat{j} + 2 \hat{k}) $
We know that the magnitude $|\vec{a} + \vec{c}| = 7$. Thus, substituting $\vec{c}$ in this equation, we resolve it to find $\lambda$:
$ |\vec{a} + \vec{c}| = 7 \Rightarrow (\lambda = 4) $
This indicates that $\lambda$ has a value of 4.
Next, calculating the area of the parallelogram formed by vectors $\vec{b}$ and $\vec{c}$ involves finding the cross product $|\vec{b} \times \vec{c}|$:
$ \vec{b} = 4\hat{i} + 4\hat{k}, \quad \vec{c} = \left(\frac{12}{9}\right)(\hat{i} + 2\hat{j} + 2\hat{k}) = \left(\frac{4}{3}\right)\hat{i} + \left(\frac{8}{3}\right)\hat{j} + \left(\frac{8}{3}\right)\hat{k} $
The cross product is:
$ \vec{b} \times \vec{c} = \left|\begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\ \frac{4}{3} & \frac{8}{3} & \frac{8}{3} \\ 4 & 0 & 4 \end{array}\right| $
Solving the determinant:
$ \vec{b} \times \vec{c} = \left( \left( \frac{8}{3} \times 4 - 0 \times \frac{8}{3} \right)\hat{i} - \left(\frac{4}{3} \times 4 - 4 \times \frac{8}{3} \right)\hat{j} + \left( \frac{4}{3} \times 0 - \frac{8}{3} \times 4 \right)\hat{k} \right) $
$ = \left(\frac{32}{3}\hat{i} + 0\hat{j} + (-\frac{32}{3})\hat{k}\right) $
Then, the magnitude is calculated as:
$ |\vec{b} \times \vec{c}| = \sqrt{\left(\frac{32}{3}\right)^2 + 0 + \left(-\frac{32}{3}\right)^2} = \sqrt{\left(\frac{32}{3}\right)^2 + \left(\frac{32}{3}\right)^2} = \sqrt{\frac{2048}{9}} $
$ = \frac{\sqrt{2048}}{3} = \frac{16}{3} \times 3 \approx 16 $
Therefore, the area of the parallelogram formed by $\vec{b}$ and $\vec{c}$ is 16.
Let $\vec{a}=9 \hat{i}-13 \hat{j}+25 \hat{k}, \vec{b}=3 \hat{i}+7 \hat{j}-13 \hat{k}$ and $\vec{c}=17 \hat{i}-2 \hat{j}+\hat{k}$ be three given vectors. If $\vec{r}$ is a vector such that $\vec{r} \times \vec{a}=(\vec{b}+\vec{c}) \times \vec{a}$ and $\vec{r} \cdot(\vec{b}-\vec{c})=0$, then $\frac{|593 \vec{r}+67 \vec{a}|^2}{(593)^2}$ is equal to __________.
Explanation:
$\begin{aligned} & \vec{a}=9 \hat{i}-13 \hat{j}+25 \hat{k} \\ & \vec{b}=3 \hat{i}+7 \hat{j}-13 \hat{k} \\ & \vec{c}=17 \hat{i}-2 \hat{j}+\hat{k} \\ & \vec{r} \times \vec{a}=(\vec{b}+\vec{c}) \times \vec{a} \\ & (\vec{r}-(\vec{b}+\vec{c})) \times \vec{a}=0 \\ & \Rightarrow \vec{r}=(\vec{b}+\vec{c})+\lambda \vec{a} \\ & \vec{r}=(20 \hat{i}+5 \hat{j}-12 \hat{k})+\lambda(9 \hat{i}-13 \hat{j}+25 \hat{k}) \\ & =(20+9 \lambda) \hat{i}+(5-13 \lambda) \hat{j}+(25 \lambda-12) \hat{k} \end{aligned}$
Now $\vec{r} \cdot(\vec{b}-\vec{c})=0$
$\vec{r} \cdot(-14 \hat{i}+9 \hat{j}-14 \hat{k})=0$
Now
$\begin{aligned} & -14(20+9 \lambda)+9(5-13 \lambda)-14(25 \lambda-12)=0 \\ & -593 \lambda-67=0 \\ & \lambda=-\frac{67}{593} \\ & \therefore \vec{r}=(\vec{b}+\vec{c})-\frac{67}{593} \vec{a} \\ & \frac{|593 \vec{r}+67 \vec{a}|^2}{|593|^2}=|\vec{b}+\vec{c}|^2=|20 \hat{i}+5 \hat{j}-12 \hat{k}|^2 \\ & =569 \end{aligned}$
Let $\vec{a}=2 \hat{i}-3 \hat{j}+4 \hat{k}, \vec{b}=3 \hat{i}+4 \hat{j}-5 \hat{k}$ and a vector $\vec{c}$ be such that $\vec{a} \times(\vec{b}+\vec{c})+\vec{b} \times \vec{c}=\hat{i}+8 \hat{j}+13 \hat{k}$. If $\vec{a} \cdot \vec{c}=13$, then $(24-\vec{b} \cdot \vec{c})$ is equal to _______.
Explanation:
Let $\hat{i}+8 \hat{j}+13 \hat{k}=\vec{u}$
Given $\vec{a} \times(\vec{b}+\vec{c})+\vec{b} \times \vec{c}=\vec{u}$
$\begin{gathered} \Rightarrow \quad \vec{a} \times \vec{b}+\vec{a} \times \vec{c}+\vec{b} \times \vec{c}=\vec{u} \\ (\vec{a}+\vec{b}) \times c=\vec{u}-\vec{a} \times \vec{b} \end{gathered}$
Taking cross product with $\vec{a}$ on both sides
$\vec{a} \times((\vec{a}+\vec{b}) \times \vec{c})=\vec{a} \times(\vec{u}-\vec{a} \times \vec{b})$
$\Rightarrow \quad \vec{c} \cdot\left(\vec{a}^2+\vec{a} \cdot \vec{b}\right)=13(\vec{a}+\vec{b})-\vec{a} \times \vec{u} +(\vec{a} \cdot \vec{b}) \cdot \vec{a}-\vec{a}^2 \vec{b}\quad$ $\{\because \vec{a} . \vec{c}=13\}$
Putting the values, $\vec{c}=(-1,-1,3)$
$\vec{b}.\vec{c}=-22$
$\Rightarrow 24-\vec{b} \cdot \vec{c}=46$
Let $\overrightarrow{\mathrm{a}}=\hat{i}-3 \hat{j}+7 \hat{k}, \overrightarrow{\mathrm{b}}=2 \hat{i}-\hat{j}+\hat{k}$ and $\overrightarrow{\mathrm{c}}$ be a vector such that $(\overrightarrow{\mathrm{a}}+2 \overrightarrow{\mathrm{b}}) \times \overrightarrow{\mathrm{c}}=3(\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}})$. If $\vec{a} \cdot \vec{c}=130$, then $\vec{b} \cdot \vec{c}$ is equal to __________.
Explanation:
$(\vec{a}+2 \vec{b}) \times \vec{c}=3(\vec{c} \times \vec{a})$
$\begin{aligned} \Rightarrow \quad & \vec{b} \times \vec{c}+2(\vec{a} \times \vec{c})=0 \\ & (\vec{b}+2 \vec{a}) \times \vec{c}=0 \\ & \vec{c}=\lambda(\vec{b}+2 \vec{a}) \\ & \vec{c} \cdot \vec{a}=130 \Rightarrow \lambda=1 \\ & \vec{c}=4 \hat{i}-7 \hat{j}+15 \hat{k} \\ & \vec{b} . \vec{c}=30 \end{aligned}$
Let $\mathrm{ABC}$ be a triangle of area $15 \sqrt{2}$ and the vectors $\overrightarrow{\mathrm{AB}}=\hat{i}+2 \hat{j}-7 \hat{k}, \overrightarrow{\mathrm{BC}}=\mathrm{a} \hat{i}+\mathrm{b} \hat{j}+\mathrm{c} \hat{k}$ and $\overrightarrow{\mathrm{AC}}=6 \hat{i}+\mathrm{d} \hat{j}-2 \hat{k}, \mathrm{~d}>0$. Then the square of the length of the largest side of the triangle $\mathrm{ABC}$ is _________.
Explanation:
Area of triangle $A B C=15 \sqrt{2}$
$\begin{aligned} & \Rightarrow \frac{1}{2}|\overline{A B} \times \overline{A C}|=15 \sqrt{2} \quad \text{.... (i)}\\ & \quad \overline{A B} \times \overline{A C}\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -7 \\ 6 & d & -2 \end{array}\right| \\ & =(7 d-4) \hat{i}-40 \hat{j}+(d-12) \hat{k} \quad \text{... (ii)} \end{aligned}$
From (i) and (ii) $5 d^2-8 d-4=0$
$\Rightarrow d=\frac{-}{5}$ (Rejected) or $d=2$
Also, $\overline{A B}+\overline{B C}=\overline{A C}$
$\begin{aligned} & \Rightarrow a+1=6 \Rightarrow a=5 \\ & b+2=d \Rightarrow b=0 \end{aligned}$
and $c-7=-2 \Rightarrow c=5$
$|\overline{A B}|=\sqrt{54},|\overline{A C}|=\sqrt{44},|\overline{B C}|=\sqrt{50}$
Largest side has length of $\sqrt{54}$ units
Explanation:
$\begin{aligned} & -\hat{\mathrm{i}}-8 \hat{\mathrm{j}}+2 \mathrm{k}=\left(4 \hat{\mathrm{i}}+\mathrm{c}_2 \hat{\mathrm{j}}+\mathrm{c}_3 \mathrm{k}\right)+\lambda(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\mathrm{k}) \\\\ & \lambda+4=-1 \Rightarrow \lambda=-5 \\\\ & \lambda+\mathrm{c}_2=-8 \Rightarrow \mathrm{c}_2=-3 \\\\ & \lambda+\mathrm{c}_3=2 \Rightarrow \mathrm{c}_3=7 \\\\ & \overrightarrow{\mathrm{c}}=4 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+7 \mathrm{k}\end{aligned}$
$\begin{aligned} & \cos \theta=\frac{12-12+7}{\sqrt{26} \cdot \sqrt{74}}=\frac{7}{\sqrt{26} \cdot \sqrt{74}}=\frac{7}{2 \sqrt{481}} \\\\ & \tan ^2 \theta=\frac{625 \times 3}{49} \\\\ & {\left[\tan ^2 \theta\right]=38}\end{aligned}$
Let $\vec{a}=3 \hat{i}+2 \hat{j}+\hat{k}, \vec{b}=2 \hat{i}-\hat{j}+3 \hat{k}$ and $\vec{c}$ be a vector such that $(\vec{a}+\vec{b}) \times \vec{c}=2(\vec{a} \times \vec{b})+24 \hat{j}-6 \hat{k}$ and $(\vec{a}-\vec{b}+\hat{i}) \cdot \vec{c}=-3$. Then $|\vec{c}|^2$ is equal to ________.
Explanation:
$\begin{aligned} & (\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}) \times \overrightarrow{\mathrm{c}}=2(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})+24 \hat{\mathrm{j}}-6 \hat{\mathrm{k}} \\ & (5 \hat{\mathrm{i}}+\hat{\mathrm{j}}+4 \hat{\mathrm{k}}) \times \overrightarrow{\mathrm{c}}=2(7 \hat{\mathrm{i}}-7 \hat{\mathrm{j}}-7 \hat{\mathrm{k}})+24 \hat{\mathrm{j}}-6 \hat{\mathrm{k}} \\ & \left|\begin{array}{lrr} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 5 & 1 & 4 \\ x & y & \mathrm{z} \end{array}\right|=14 \hat{\mathrm{i}}+10 \hat{\mathrm{j}}-20 \hat{\mathrm{k}} \\ & \Rightarrow \hat{\mathrm{i}}(\mathrm{z}-4 \mathrm{y})-\hat{\mathrm{j}}(5 \mathrm{z}-4 \mathrm{x})+\hat{\mathrm{k}}(5 \mathrm{y}-\mathrm{x})=14 \hat{\mathrm{i}}+10 \hat{\mathrm{j}}-20 \hat{\mathrm{k}} \\ & \mathrm{z}-4 \mathrm{y}=14,4 \mathrm{x}-5 \mathrm{z}=10,5 \mathrm{y}-\mathrm{x}=-20 \\ & (\mathrm{a}-\mathrm{b}+\mathrm{i}) \cdot \overrightarrow{\mathrm{c}}=-3 \\ & (2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-2 \hat{\mathrm{k}}) \cdot \overrightarrow{\mathrm{c}}=-3 \\ & 2 x+3 y-2 z=-3 \\ & \therefore x=5, y=-3, z=2 \\ & |\overrightarrow{\mathrm{c}}|^2=25+9+4=38 \end{aligned}$
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 $\alpha$, then $192 \sin ^2 \alpha$ is equal to ________.
Explanation:
$\begin{aligned} & \overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}=(2 \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}) \cdot \overrightarrow{\mathrm{b}}-3|\mathrm{b}|^2 \\ & |\mathrm{~b}||c| \cos \alpha=-3|\mathrm{~b}|^2 \\ & |\mathrm{c}| \cos \alpha=-12 \text {, as }|\mathrm{b}|=4 \\ & \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}=2 \\ & \cos \theta=\frac{1}{2} \Rightarrow \theta=\frac{\pi}{3} \\ & |c|^2=|(2 \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})-3 \overrightarrow{\mathrm{b}}|^2 \\ & =64 \times \frac{3}{4}+144=192 \\ & |c|^2 \cos ^2 \alpha=144 \\ & 192 \cos ^2 \alpha=144 \\ & 192 \sin ^2 \alpha=48 \end{aligned}$
Explanation:
$\begin{aligned} & \cos \theta=\frac{(\alpha \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}) \cdot(\alpha \hat{\mathrm{i}}+2 \alpha \hat{\mathrm{j}}-2 \hat{\mathrm{k}})}{\sqrt{\alpha^2+4+4} \sqrt{\alpha^2+4 \alpha^2+4}} \\ & \cos \theta=\frac{\alpha^2-4 \alpha-4}{\sqrt{\alpha^2+8} \sqrt{5 \alpha^2+4}} \\ & \Rightarrow \alpha^2-4 \alpha-4>0 \quad \Rightarrow(\alpha-2)^2>8 \\ & \Rightarrow \alpha^2-4 \alpha+4>8 \quad \alpha-2<-2 \sqrt{2} \\ & \Rightarrow \alpha-2>2 \sqrt{2} \text { or } \alpha-2<2 \sqrt{2} \\ & \alpha>2+2 \sqrt{2} \text { or } \alpha<2-2 \sqrt{2} \\ & \alpha \in(-\infty,-0.82) \cup(4.82, \infty) \end{aligned}$
Least positive integral value of $\alpha \Rightarrow 5$
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 $|72-| \vec{b}+\left.\vec{c}\right|^{2} \mid$ is equal to __________.
Explanation:
Given that $\vec{a} = \vec{b} \times \vec{c}$, we can find the magnitudes of $\vec{a}$ and $\vec{c}$:
$|\vec{a}| = \sqrt{3^2 + 1^2 + (-1)^2} = \sqrt{11}$
$|\vec{c}| = \sqrt{2^2 + (-3)^2 + 3^2} = \sqrt{22}$
We know that the magnitude of the cross product of two vectors is equal to the product of the magnitudes of the vectors and the sine of the angle between them :
$|\vec{a}| = |\vec{b} \times \vec{c}| = |\vec{b}||\vec{c}|\sin\theta$
Plugging in the known values :
$\sqrt{11} = \sqrt{50}\sqrt{22}\sin\theta$
Solving for the sine of the angle between the vectors :
$\sin\theta = \frac{1}{10}$
Now we can find $|\vec{b} + \vec{c}|^2$ using the formula :
$|\vec{b} + \vec{c}|^2 = |\vec{b}|^2 + |\vec{c}|^2 + 2\vec{b} \cdot \vec{c}$
We have the dot product $\vec{b} \cdot \vec{c} = |\vec{b}||\vec{c}|\cos\theta$, and we can use the relationship between sine and cosine: $\cos\theta = \sqrt{1 - \sin^2\theta} = \frac{\sqrt{99}}{10}$.
Substitute the values into the formula :
$|\vec{b} + \vec{c}|^2 = 50 + 22 + 2\sqrt{50}\sqrt{22}\frac{\sqrt{99}}{10} = 72 + 66$
Finally, we need to find the absolute value of the difference :
$|72 - |\vec{b} + \vec{c}|^2| = |72 - (72 + 66)| = 66$
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 _________.
Explanation:
$ \begin{aligned} & \vec{a}=\hat{i}+2 \hat{j}+3 \hat{k} \\\\ & \vec{b}=\hat{i}+\hat{j}-\hat{k} \\\\ & \vec{a} \cdot \vec{c}=11 \\\\ & \vec{b} \cdot(\vec{a} \times \vec{c})=27 \\\\ & \vec{b} \cdot \vec{c}=-\sqrt{3}|\vec{b}| \\\\ & (\vec{b} \times \vec{a}) \cdot \vec{c}=27 \end{aligned} $
$ \text { Let } \vec{c}=c_1 \hat{i}+c_2 \hat{j}+c_3 \hat{k} $
As $\vec{a} \cdot \vec{c}=11$
$ \therefore $ $ c_1+2 c_2+3 c_3=11 $ ......(i)
Also, $\vec{b} \cdot \vec{c}=-\sqrt{3}|\vec{b}|$
$ \begin{aligned} & \therefore c_1+c_2-c_3=-\sqrt{3} \sqrt{3} \\\\ & \Rightarrow c_1+c_2-c_3=-3 ......(ii) \end{aligned} $
Also, $\vec{b} \cdot(\vec{a} \times \vec{c})=27$
$ \therefore $ $ 5 c_1-4 c_2+c_3=27 $ ...........(iii)
From (i), (ii) & (iii)
$ \vec{c}=3 \hat{i}-2 \hat{j}+4 \hat{k} $
$ \begin{aligned} & |\vec{a} \times \vec{c}|^2=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 3 & -2 & +4 \end{array}\right|^2 \\\\ & =|14 \hat{i}+5 \hat{j}-8 \hat{k}|^2 \\\\ & =14^2+5^2+8^2=285 \end{aligned} $
Let $\vec{a}=6 \hat{i}+9 \hat{j}+12 \hat{k}, \vec{b}=\alpha \hat{i}+11 \hat{j}-2 \hat{k}$ and $\vec{c}$ be vectors such that $\vec{a} \times \vec{c}=\vec{a} \times \vec{b}$. If
$\vec{a} \cdot \vec{c}=-12, \vec{c} \cdot(\hat{i}-2 \hat{j}+\hat{k})=5$, then $\vec{c} \cdot(\hat{i}+\hat{j}+\hat{k})$ is equal to _______________.
Explanation:
Now, $\vec{a} \cdot \vec{c}=-12$
$ \Rightarrow 6 c_1+9 c_2+12 c_3=-12 $ ..............(i)
Also, $\vec{c} \cdot(\hat{i}-2 \hat{j}+\hat{k})=5$
$ \Rightarrow c_1-2 c_2+c_3=5 $ ................(ii)
$ \begin{aligned} & \text { Now, } \vec{a} \times \vec{c}=\vec{a} \times \vec{b} \\\\ & \Rightarrow \vec{a} \times(\vec{c}-\vec{b})=0 \\\\ & \Rightarrow \vec{a} \text { is parallel to }(\vec{c}-\vec{b}) \\\\ & \Rightarrow \vec{a}=\lambda(\vec{c}-\vec{b}) \\\\ & \Rightarrow 6 \hat{i}+9 \hat{j}+12 \hat{k}=\lambda\left(c_1-\alpha\right) \hat{i}+\lambda\left(c_2-11\right) \hat{j}+\lambda\left(c_3+2\right) \hat{k} \end{aligned} $
On comparing, we get
$ c_1=\frac{6}{\lambda}+\alpha, c_2=\frac{9}{\lambda}+11, c_3=\frac{12}{\lambda}-2 $
Put there values in (ii), we get
$ \begin{aligned} & \frac{6}{\lambda}+\alpha-\frac{18}{\lambda}-22+\frac{12}{\lambda}-2=5 \\\\ & \Rightarrow \alpha=29 \end{aligned} $
From (i) and values of $\mathrm{c}_1, \mathrm{c}_2, \mathrm{c}_3$, and $\alpha$ we have
$ \begin{aligned} & 6\left(\frac{6}{\lambda}+29\right)+9\left(\frac{9}{\lambda}+11\right)+12\left(\frac{12}{\lambda}-2\right)=-12 \\\\ & \Rightarrow \frac{261}{\lambda}=-261 \Rightarrow \lambda=-1 \end{aligned} $
$ \begin{aligned} & \text { So, } c_1=23, c_2=2, c_3=-14 \\\\ & \therefore \vec{c} \cdot(\hat{i}+\hat{j}+\hat{k})=(23 \hat{i}+2 \hat{j}+-14 \hat{k}) \cdot(\hat{i}+\hat{j}+\hat{k}) \\\\ & =23+2-14=11 \end{aligned} $
Let $\vec{v}=\alpha \hat{i}+2 \hat{j}-3 \hat{k}, \vec{w}=2 \alpha \hat{i}+\hat{j}-\hat{k}$ and $\vec{u}$ be a vector such that $|\vec{u}|=\alpha>0$. If the minimum value of the scalar triple product $\left[ {\matrix{ {\overrightarrow u } & {\overrightarrow v } & {\overrightarrow w } \cr } } \right]$ is $-\alpha \sqrt{3401}$, and $|\vec{u} \cdot \hat{i}|^{2}=\frac{m}{n}$ where $m$ and $n$ are coprime natural numbers, then $m+n$ is equal to ____________.
Explanation:
$\left[ {\matrix{ {\overrightarrow u } & {\overrightarrow v } & {\overrightarrow w } \cr } } \right] = \overrightarrow u .\left( {\overrightarrow v \times \overrightarrow w } \right)$
$=|\vec{u}||\vec{v} \times \vec{w}| \times \cos \theta$
$=\alpha \sqrt{34 \alpha^{2}+1} \cos \theta$
$[\vec{u} \vec{v} \vec{w}]_{\min }=-\alpha \sqrt{3401}$
$\alpha \sqrt{34 \alpha^{2}+1} \times(-1)=-\alpha \sqrt{3401}$
(taking $\cos \theta=1$ )
$\Rightarrow \alpha=10$
$\vec{v} \times \vec{w}=\hat{i}-50 \hat{j}-30 \hat{k}$
$\cos \theta=-1 \Rightarrow \vec{u}$ is antiparallel to $\vec{v} \times \vec{w}$
$\vec{u}=-|\vec{u}| \cdot \frac{\vec{v} \times \vec{w}}{|\vec{v} \times \vec{w}|}=\frac{-10(\hat{i}-50 \hat{j}-30 \hat{k})}{\sqrt{3401}}$
$|\vec{u} \cdot \hat{i}|^{2}=\left|\frac{-10}{\sqrt{3401}}\right|^{2}=\frac{100}{3401}=\frac{m}{n}$
$m+n=3501$
$A(2,6,2), B(-4,0, \lambda), C(2,3,-1)$ and $D(4,5,0),|\lambda| \leq 5$ are the vertices of a quadrilateral $A B C D$. If its area is 18 square units, then $5-6 \lambda$ is equal to __________.
Explanation:
$ \begin{aligned} & =(3 \lambda+15) \hat{i}-\hat{j}(-24)+\hat{k}(-24) \\\\ & \overrightarrow{A C} \times \overrightarrow{B D}=(3 \lambda+15) \hat{i}+24 \hat{j}-24 \hat{k} \\\\ & =\sqrt{(3 \lambda+15)^2+(24)^2+(24)^2}=36 \\\\ & =\lambda^2+10 \lambda+9=0 \\\\ & =\lambda=-1,-9 \\\\ & |\lambda| \leq 5 \Rightarrow \lambda=-1 \\\\ & 5-6 \lambda=5-6(-1)=11 \end{aligned} $
$|\vec{a}|=\sqrt{31}, 4|\vec{b}|=|\vec{c}|=2$ and $2(\vec{a} \times \vec{b})=3(\vec{c} \times \vec{a})$.
If the angle between $\vec{b}$ and $\vec{c}$ is $\frac{2 \pi}{3}$, then $\left(\frac{\vec{a} \times \vec{c}}{\vec{a} \cdot \vec{b}}\right)^{2}$ is equal to __________.
Explanation:
$\vec{a} \times(2 \vec{b}+3 \vec{c})=0$
$ \begin{aligned} & \vec{a}=\lambda(2 \vec{b}+3 \vec{c}) \\\\ & |\vec{a}|^{2}=\lambda^{2}\left(4|b|^{2}+9|c|^{2}+12 \vec{b} \cdot \vec{c}\right) \\\\ & 31=31 \lambda^{2} \\\\ & \lambda=\pm 1 \\\\ & \vec{a}=\pm(2 \vec{b}+3 \vec{c}) \\\\ & \frac{|\vec{a} \times \vec{c}|}{|\vec{a} \cdot \vec{b}|}=\frac{2|\vec{b} \times \vec{c}|}{2 \vec{b} \cdot \vec{b}+3 \vec{c} \cdot \vec{b}} \\\\ & |\vec{b} \times \vec{c}|^{2}=\frac{1}{4} \cdot 4-\left(1-\frac{1}{2}\right)^{2} \\\\ & =\frac{3}{4} \\\\ & \therefore \frac{|\vec{a} \times \vec{c}|}{|\vec{a} \cdot \vec{b}|}=\frac{\sqrt{3}}{2 \cdot \frac{1}{4}-\frac{3}{2}}=\frac{\sqrt{3}}{-1} \\\\ & \left(\frac{|\vec{a} \times \vec{c}|}{|\vec{a} \cdot \vec{b}|}\right)^{2}=3 \end{aligned} $
Let $\vec{a}$ and $\vec{b}$ be two vectors 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 ___________.
Explanation:
$ \begin{aligned} & \Rightarrow |\vec{a} \times \vec{b}|^{2}+(\vec{a} \cdot \vec{b})^{2}=|\vec{a}|^{2}|\vec{b}|^{2} \\\\ & \Rightarrow 48+(\vec{a} \cdot \vec{b})^{2}=6 \times 14 \\\\ & \Rightarrow (\vec{a} \cdot \vec{b})^{2}=84-48 \\\\ &=36 \end{aligned} $
Let $\overrightarrow a $, $\overrightarrow b $ and $\overrightarrow c $ be three non-zero non-coplanar vectors. Let the position vectors of four points $A,B,C$ and $D$ be $\overrightarrow a - \overrightarrow b + \overrightarrow c ,\lambda \overrightarrow a - 3\overrightarrow b + 4\overrightarrow c , - \overrightarrow a + 2\overrightarrow b - 3\overrightarrow c $ and $2\overrightarrow a - 4\overrightarrow b + 6\overrightarrow c $ respectively. If $\overrightarrow {AB} ,\overrightarrow {AC} $ and $\overrightarrow {AD} $ are coplanar, then $\lambda$ is equal to __________.
Explanation:
$ \overline{A C}=2 \bar{a}+3 \bar{b}-4 \bar{c} $
$ \overline{A D}=\bar{a}-3 \bar{b}+5 \bar{c} $
$ \left|\begin{array}{ccc} \lambda-1 & -2 & 3 \\ -2 & 3 & -4 \\ 1 & -3 & 5 \end{array}\right|=0 $
$ \Rightarrow(\lambda-1)(15-12)+2(-10+4)+3(6-3)=0 $
$ \Rightarrow(\lambda-1)=1 \Rightarrow \lambda=2 $
Let $\overrightarrow a = \widehat i + 2\widehat j + \lambda \widehat k,\overrightarrow b = 3\widehat i - 5\widehat j - \lambda \widehat k,\overrightarrow a \,.\,\overrightarrow c = 7,2\overrightarrow b \,.\,\overrightarrow c + 43 = 0,\overrightarrow a \times \overrightarrow c = \overrightarrow b \times \overrightarrow c $. Then $\left| {\overrightarrow a \,.\,\overrightarrow b } \right|$ is equal to :
Explanation:
Now $\vec{a} \cdot \overrightarrow{\mathbf{c}}=7$ gives $2 \lambda^2+12=7 \mu$
And $\vec{b} \cdot \vec{c}=-\frac{43}{2}$ gives $4 \lambda^2+82=43 \mu$
$\mu=2$ and $\lambda^2=1$
$ |\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}|=8 $
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 __________.
Explanation:
or $|\vec{a}|^{2}+|\vec{b}|^{2}+2 \vec{a} \cdot \vec{b}=|\vec{a}|^{2}+2|\vec{b}|^{2}$
$\therefore|\vec{b}|^{2}=6$
Now $|\vec{a} \times \vec{b}|^{2}=|\vec{a}|^{2}|\vec{b}|^{2}-(\vec{a} \cdot \vec{b})^{2}$
$ 75=|\vec{a}|^{2} \cdot 6-9 $
$\therefore|\vec{a}|^{2}=14$
Let $\overrightarrow a $, $\overrightarrow b $, $\overrightarrow c $ be three non-coplanar vectors such that $\overrightarrow a $ $\times$ $\overrightarrow b $ = 4$\overrightarrow c $, $\overrightarrow b $ $\times$ $\overrightarrow c $ = 9$\overrightarrow a $ and $\overrightarrow c $ $\times$ $\overrightarrow a $ = $\alpha$$\overrightarrow b $, $\alpha$ > 0. If $\left| {\overrightarrow a } \right| + \left| {\overrightarrow b } \right| + \left| {\overrightarrow c } \right| = {1 \over {36}}$, then $\alpha$ is equal to __________.
Explanation:
Given,
$\overrightarrow a \times \overrightarrow b = 4\,.\,\overrightarrow c $ ..... (i)
$\overrightarrow b \times \overrightarrow c = 9\,.\,\overrightarrow a $ ..... (ii)
$\overrightarrow c \times \overrightarrow a = \alpha \,.\,\overrightarrow b $ .... (iii)
Taking dot products with $\overrightarrow c ,\overrightarrow a ,\overrightarrow b $ we get
$\overrightarrow a \,.\,\overrightarrow b = \overrightarrow b \,.\,\overrightarrow c = \overrightarrow c \,.\,\overrightarrow a = 0$
Hence,
(i) $ \Rightarrow |\overrightarrow a |\,.\,|\overrightarrow b | = 4\,.\,|\overrightarrow c |$ ..... (iv)
(ii) $ \Rightarrow |\overrightarrow b |\,.\,|\overrightarrow c | = 9\,.\,|\overrightarrow a |$ ..... (v)
(iii) $ \Rightarrow |\overrightarrow c |\,.\,|\overrightarrow a | = \alpha \,.\,|\overrightarrow b |$ .... (vi)
Multiplying (iv), (v) and (vi)
$ \Rightarrow |\overrightarrow a |\,.\,|\overrightarrow b |\,.\,|\overrightarrow c | = 36\alpha $ ..... (vii)
Dividing (vii) by (iv) $ \Rightarrow |\overrightarrow c {|^2} = 9\alpha \Rightarrow |\overrightarrow c | = 3\sqrt \alpha $ ..... (viii)
Dividing (vii) by (v) $ \Rightarrow |\overrightarrow a {|^2} = 4\alpha \Rightarrow |\overrightarrow a | = 2\sqrt \alpha $
Dividing (viii) by (vi) $ \Rightarrow |\overrightarrow b {|^2} = 36 \Rightarrow |\overrightarrow b | = 6$
Now, as given, $3\sqrt \alpha + 2\sqrt \alpha + 6 = {1 \over {36}} \Rightarrow \sqrt \alpha = {{ - 43} \over {36}}$
Let $\overrightarrow a = \widehat i - 2\widehat j + 3\widehat k$, $\overrightarrow b = \widehat i + \widehat j + \widehat k$ and $\overrightarrow c $ be a vector such that $\overrightarrow a + \left( {\overrightarrow b \times \overrightarrow c } \right) = \overrightarrow 0 $ and $\overrightarrow b \,.\,\overrightarrow c = 5$. Then the value of $3\left( {\overrightarrow c \,.\,\overrightarrow a } \right)$ is equal to _________.
Explanation:
Given: $\vec{a}+(\vec{b} \times \vec{c})=0$
$ \Rightarrow \vec{a} \cdot \vec{b}=0 $ ........(ii)
Equation (i) and equation (ii) are contradicting.
If $\overrightarrow a = 2\widehat i + \widehat j + 3\widehat k$, $\overrightarrow b = 3\widehat i + 3\widehat j + \widehat k$ and $\overrightarrow c = {c_1}\widehat i + {c_2}\widehat j + {c_3}\widehat k$ are coplanar vectors and $\overrightarrow a \,.\,\overrightarrow c = 5$, $\overrightarrow b \bot \overrightarrow c $, then $122({c_1} + {c_2} + {c_3})$ is equal to ___________.
Explanation:
$2{C_1} + {C_2} + 3{C_3} = 5$ ...... (i)
$3{C_1} + 3{C_2} + {C_3} = 0$ ...... (ii)
$\left[ {\overrightarrow a \overrightarrow b \overrightarrow c } \right] = \left| {\matrix{ 2 & 1 & 3 \cr 3 & 3 & 1 \cr {{C_1}} & {{C_2}} & {{C_3}} \cr } } \right|$
$ = 2(3{C_3} - {C_2}) - 1(3{C_3} - {C_1}) + 3(3{C_2} - 3{C_1})$
$ = 3{C_3} + 7{C_2} - 8{C_1}$
$ \Rightarrow 8{C_1} - 7{C_2} - 3{C_3} = 0$ ...... (iii)
${C_1} = {{10} \over {122}},{C_2} = {{ - 85} \over {122}},{C_3} = {{225} \over {122}}$
So $122({C_1} + {C_2} + {C_3}) = 150$
Let $\overrightarrow b = \widehat i + \widehat j + \lambda \widehat k$, $\lambda$ $\in$ R. If $\overrightarrow a $ is a vector such that $\overrightarrow a \times \overrightarrow b = 13\widehat i - \widehat j - 4\widehat k$ and $\overrightarrow a \,.\,\overrightarrow b + 21 = 0$, then $\left( {\overrightarrow b - \overrightarrow a } \right).\,\left( {\widehat k - \widehat j} \right) + \left( {\overrightarrow b + \overrightarrow a } \right).\,\left( {\widehat i - \widehat k} \right)$ is equal to _____________.
Explanation:
Let $\overrightarrow a = x\widehat i = y\widehat j + z\widehat k$
So, $\left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr x & y & z \cr 1 & 1 & \lambda \cr } } \right| = \widehat i(\lambda y - z) + \widehat j(z - \lambda x) + \widehat k(x - y)$
$ \Rightarrow \lambda y - z = 13,\,z - \lambda x = - 1,\,x - y = - 4$
and $x + y + \lambda z = - 21$
$\Rightarrow$ Clearly, $\lambda = 3$, $x = - 2$, $y = 2$ and $z = - 7$
So, $\overrightarrow b - \overrightarrow a = 3\widehat i - \widehat j + 10\widehat k$
and $\overrightarrow b + \overrightarrow a = - \widehat i + 3\widehat j - 4\widehat k$
$ \Rightarrow \left( {\overrightarrow b - \overrightarrow a } \right)\,.\,\left( {\widehat k - \widehat j} \right) + \left( {\overrightarrow b + \overrightarrow a } \right)\,.\,\left( {\widehat i - \widehat k} \right) = 11 + 3 = 14$
Let $\theta$ be the angle between the vectors $\overrightarrow a $ and $\overrightarrow b $, where $|\overrightarrow a | = 4,$ $|\overrightarrow b | = 3$ and $\theta \in \left( {{\pi \over 4},{\pi \over 3}} \right)$. Then ${\left| {\left( {\overrightarrow a - \overrightarrow b } \right) \times \left( {\overrightarrow a + \overrightarrow b } \right)} \right|^2} + 4{\left( {\overrightarrow a \,.\,\overrightarrow b } \right)^2}$ is equal to __________.
Explanation:
${\left| {\left( {\overrightarrow a - \overrightarrow b } \right) \times \left( {\overrightarrow a + \overrightarrow b } \right)} \right|^2} + 4{\left( {\overrightarrow a \,.\,\overrightarrow b } \right)^2}$
$ \Rightarrow {\left| {\overrightarrow a \times \overrightarrow a + \overrightarrow a \times \overrightarrow b - \overrightarrow b \times \overrightarrow a - \overrightarrow b \times \overrightarrow b } \right|^2} + 4{\left( {\overrightarrow a \,.\,\overrightarrow b } \right)^2}$
$ \Rightarrow {\left| {2\left( {\overrightarrow a \times \overrightarrow b } \right)} \right|^2} + 4{\left( {\overrightarrow a \,.\,\overrightarrow b } \right)^2}$
$ \Rightarrow 4{\left( {\overrightarrow a \times \overrightarrow b } \right)^2} + {\left( {\overrightarrow a \,.\,\overrightarrow b } \right)^2}$
$ \Rightarrow 4{\left| {\overrightarrow a } \right|^2}{\left| {\overrightarrow b } \right|^2} = 4\,.\,16\,.\,9 = 576$
Explanation:
$\overrightarrow b = \widehat i + 2\widehat j - \widehat k$
$\overrightarrow c = 3\widehat i + 2\widehat j - \widehat k$
$\overrightarrow v = x\overrightarrow a + y\overrightarrow b $
$\overrightarrow v \left( {3\widehat i + 2\widehat j - \widehat k} \right) = 0$
$\overrightarrow v .\widehat a = 19$
$\overrightarrow v = \lambda \overrightarrow c \times \left( {\overrightarrow a \times \overrightarrow b } \right)$
$\overrightarrow v = \lambda \left[ {\left( {\overrightarrow c .\overrightarrow b } \right)\overrightarrow a - \left( {\overrightarrow c .\overrightarrow a } \right)\overrightarrow b } \right]$
$ = \lambda [(3 + 4 + 1)\left( {2\widehat i - \widehat j + 2\widehat k} \right) - \left( {{{6 - 2 - 2} \over 2}} \right)\left( {\widehat i + 2\widehat j + \widehat k} \right)$
$ = \lambda [16\widehat i - 8\widehat j + 16\widehat k - 2\widehat i - 4\widehat j + 2\widehat k]$
$\overrightarrow v = \lambda \left[ {14\widehat i - 12\widehat j + 18\widehat k} \right]$
$\lambda [14\widehat i - 12\widehat j + 18\widehat k].{{\left( {2\widehat i - \widehat j + 2\widehat k} \right)} \over {\sqrt {4 + 1 + 4} }} = 19$
$\lambda {{[28 + 12 + 36]} \over 3} = 19$
$\lambda \left( {{{76} \over 3}} \right) = 19$
$4\lambda = 3 \Rightarrow \lambda = {3 \over 4}$
$|2{v^2}| = {\left| {2 \times {3 \over 4}(14\widehat i - 12\widehat j + 18\widehat k)} \right|^2}$
${9 \over 4} \times 4{\left( {7\widehat i - 6\widehat j + 9\widehat k} \right)^2}$
$ = 9(49 + 36 + 81)$
$ = 9(166)$
$ = 1494$