Vector Algebra

Given data

Each vector has magnitude $ a = 3\sqrt{1.5} = 3\sqrt{\frac{3}{2}} $.

Angle between any pair of vectors $ = \frac{\pi}{3} $.

Let the three vectors be $ \vec{A}, \vec{B}, \vec{C} $.

We want the magnitude of $ \vec{R} = \vec{A} + \vec{B} + \vec{C} $.

Step 1: Compute $ R^2 = |\vec{A} + \vec{B} + \vec{C}|^2 $

$ R^2 = A^2 + B^2 + C^2 + 2(\vec{A}\cdot\vec{B} + \vec{B}\cdot\vec{C} + \vec{C}\cdot\vec{A}) $

Since all magnitudes are equal:

$ A^2 = B^2 = C^2 = a^2, $

and

$ \vec{A}\cdot\vec{B} = a^2 \cos\left(\frac{\pi}{3}\right) = a^2 \left(\frac{1}{2}\right). $

Step 2: Substitute

$ R^2 = 3a^2 + 2 \times 3 \times a^2 \left(\frac{1}{2}\right) = 3a^2 + 3a^2 = 6a^2. $

Therefore,

$ R = a\sqrt{6}. $

Step 3: Substitute value of $ a $

$ a = 3\sqrt{1.5} = 3\sqrt{\frac{3}{2}}. $

So,

$ R = 3\sqrt{\frac{3}{2}} \times \sqrt{6} = 3 \sqrt{ \frac{3 \times 6}{2} } = 3 \sqrt{9} = 3 \times 3 = 9. $

✅ Final Result:

$ \boxed{R = 9 \text{ units}} $

Correct Option: B) 9 units

To find a vector that is perpendicular to $(4 \hat{\mathbf{i}}-3 \hat{\mathbf{j}})$, remember that two vectors are perpendicular if their dot product is zero.

Let's check with $7 \hat{\mathbf{k}}$. Here, $\hat{\mathbf{k}}$ is perpendicular to both $\hat{\mathbf{i}}$ and $\hat{\mathbf{j}}$.

Compute the dot product: $(4 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}) \cdot 7 \hat{\mathbf{k}} = 4 \times 0 - 3 \times 0 + 0 \times 7 = 0$.

So, $7 \hat{\mathbf{k}}$ is perpendicular to $(4 \hat{\mathbf{i}}-3 \hat{\mathbf{j}})$.

$ \begin{aligned} &\text { We know that, }\\ &\begin{aligned} & \cos ^2 \alpha+\cos ^2 \beta+\cos ^2 \gamma=1 \\ & \Rightarrow 1-\sin ^2 \alpha+1-\sin ^2 \beta+\cos ^2 \gamma=1 \\ & \Rightarrow 1+\cos ^2 \gamma-\left(\sin ^2 \alpha+\sin ^2 \beta\right)=0 \\ & \Rightarrow 1+\cos ^2 \gamma=\sin ^2 \alpha+\sin ^2 \beta \\ & \Rightarrow \sin ^2 \alpha+\sin ^2 \beta=1+\cos ^2 \gamma \end{aligned} \end{aligned} $

$P=|P|=25$ units

$ P_y=7 $

Since, $\sqrt{P_x^2+P_y^2}=P$

$ \begin{array}{cc} \Rightarrow & P_x^2+P_y^2=P^2 \\ \Rightarrow & P_x^2+7^2=25^2 \\ \Rightarrow & P_x=\sqrt{25^2-7^2} \\ & =\sqrt{625-49} \\ & =\sqrt{576}=24 \text { units } \end{array} $

We know that,

$ \begin{aligned} R_{\max } & =A+B \text { and } R_{\min }=A-B \\ \text { Since, } R_{\max } & =n R_{\min } \\ \Rightarrow \quad A+B & =n(A-B) \\ \Rightarrow \quad \frac{A+B}{A-B} & =\frac{n}{1} \end{aligned} $

Applying componendo and dividendo rule,

$ \begin{aligned} & \frac{(A+B)+(A-B)}{(A+B)-(A-B)}=\frac{n+1}{n-1} \\ \Rightarrow \quad & \frac{2 A}{2 B}=\frac{n+1}{n-1} \Rightarrow \frac{A}{B}=\frac{n+1}{n-1} \end{aligned} $

The task is to determine the angle the resultant vector of two given vectors makes with the X-axis.

Given vectors are:

$ \begin{align*} \mathbf{A} &= 2 \hat{\mathbf{i}} + 3 \hat{\mathbf{j}} + 4 \hat{\mathbf{k}} \\ \mathbf{B} &= 2 \hat{\mathbf{i}} - 7 \hat{\mathbf{j}} - 4 \hat{\mathbf{k}} \end{align*} $

First, calculate the resultant vector $\mathbf{R}$:

$ \begin{align*} \mathbf{R} &= \mathbf{A} + \mathbf{B} \\ &= (2 \hat{\mathbf{i}} + 3 \hat{\mathbf{j}} + 4 \hat{\mathbf{k}}) + (2 \hat{\mathbf{i}} - 7 \hat{\mathbf{j}} - 4 \hat{\mathbf{k}}) \\ &= (2+2) \hat{\mathbf{i}} + (3-7) \hat{\mathbf{j}} + (4-4) \hat{\mathbf{k}} \\ &= 4 \hat{\mathbf{i}} - 4 \hat{\mathbf{j}} + 0 \hat{\mathbf{k}} \\ &= 4 \hat{\mathbf{i}} - 4 \hat{\mathbf{j}} \end{align*} $

To find the angle $\theta$ between the resultant vector $\mathbf{R}$ and the X-axis, we use the dot product:

$ \cos(\theta) = \frac{\mathbf{R} \cdot \hat{\mathbf{i}}}{|\mathbf{R}| \cdot |\hat{\mathbf{i}}|} $

Calculate the dot product $\mathbf{R} \cdot \hat{\mathbf{i}}$:

$ \mathbf{R} \cdot \hat{\mathbf{i}} = (4 \hat{\mathbf{i}} - 4 \hat{\mathbf{j}}) \cdot \hat{\mathbf{i}} = 4 $

Calculate the magnitude of $\mathbf{R}$:

$ |\mathbf{R}| = \sqrt{(4)^2 + (-4)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2} $

Now, substitute these values into the cosine equation:

$ \begin{align*} \cos \theta &= \frac{4}{4 \sqrt{2}} = \frac{1}{\sqrt{2}} \\ \theta &= \cos^{-1}\left(\frac{1}{\sqrt{2}}\right) \\ \theta &= 45^{\circ} \end{align*} $

Therefore, the resultant vector makes an angle of $45^{\circ}$ with the X-axis.

Given that $ |\mathbf{P}+\mathbf{Q}| = |\mathbf{P}| = |\mathbf{Q}| $, let's analyze the problem step-by-step:

Begin with the equation:

$ |\mathbf{P}+\mathbf{Q}|^2 = |\mathbf{P}|^2 $

Expanding $(\mathbf{P}+\mathbf{Q}) \cdot (\mathbf{P}+\mathbf{Q})$:

$ (\mathbf{P}+\mathbf{Q}) \cdot (\mathbf{P}+\mathbf{Q}) = \mathbf{P} \cdot \mathbf{P} + 2 \mathbf{P} \cdot \mathbf{Q} + \mathbf{Q} \cdot \mathbf{Q} = P^2 + 2 \mathbf{P} \cdot \mathbf{Q} + Q^2 $

Since $ |\mathbf{P}| = |\mathbf{Q}| $, let them both be $ P $ (i.e., $ Q = P $). Thus, the equation becomes:

$ P^2 + 2 \mathbf{P} \cdot \mathbf{Q} + P^2 = P^2 $

Simplify and rearrange:

$ 2 \mathbf{P} \cdot \mathbf{Q} + P^2 = 0 $

Substitute $\mathbf{P} \cdot \mathbf{Q} = PQ \cos \theta$:

$ 2PQ \cos \theta + P^2 = 0 $

Solve for $\cos \theta$:

$ 2P^2 \cos \theta = -P^2 $

$ \cos \theta = -\frac{1}{2} $

Recognize that $\cos \theta = -\frac{1}{2}$ corresponds to:

$ \theta = 120^\circ $

Thus, the angle between $\mathbf{P}$ and $\mathbf{Q}$ is $120^\circ$.

Given, $\mathbf{P}=3 \hat{i}+4 \hat{j}$

Let $Q=\hat{i}+2 \hat{j}$

The component of $\mathbf{P}$ along the $\mathbf{Q}$ is given as $|\mathbf{P}| \cos \theta$.

Hence, we know that,

$\begin{aligned} \text { Hence, } \mathbf{P} \cdot \mathbf{Q} & =|\mathbf{P}||\mathbf{Q}| \cos \theta \\ \Rightarrow|P| \cos \theta & =\frac{\mathbf{P} \cdot \mathbf{Q}}{|\mathbf{Q}|} \\ & =\frac{(3 \hat{i}+4 \hat{j}) \cdot(\hat{i}+2 \hat{j})}{\sqrt{1^2+2^2}}=\frac{3+8}{\sqrt{5}}=\frac{11}{\sqrt{5}} \end{aligned}$

The two vectors $\mathbf{A}$ and $\mathbf{B}$ are mutually perpendicular to each other

$\begin{array}{ll} & \mathbf{A} \cdot \mathbf{B}=0 \\ \Rightarrow & A B \cos \theta=0 \\ \Rightarrow & \cos \theta=0 \Rightarrow \theta=90^{\circ} \\ |\mathbf{A}-\mathbf{B}| & \\ & =\sqrt{A^2+B^2-2 A B \cos 90^{\circ}} \\ & =\sqrt{A^2+B^2} \end{array}$

Unit vector along $\mathbf{A}-\mathbf{B}$

$=\frac{\mathbf{A}-\mathbf{B}}{|\mathbf{A}-\mathbf{B}|}=\frac{\mathbf{A}-\mathbf{B}}{\sqrt{A^2+B^2}}$

component of $(\mathbf{A}+\mathbf{B})$ in direction of $(\mathbf{A}-\mathbf{B})$ will be given as,

$\begin{aligned} & =\frac{(\mathbf{A}+\mathbf{B}) \cdot(\mathbf{A}-\mathbf{B})}{|\mathbf{A}-\mathbf{B}|} \\ & =\frac{\mathbf{A} \cdot \mathbf{A}-\mathbf{A} \cdot \mathbf{B}+\mathbf{B} \cdot \mathbf{A}-\mathbf{B} \cdot \mathbf{B}}{\sqrt{A^2+B^2}} \\ & =\frac{A^2-B^2}{\sqrt{A^2+B^2}}=\frac{|\mathbf{A}|^2-|\mathbf{B}|^2}{\sqrt{|\mathbf{A}|^2+|\mathbf{B}|^2}} \end{aligned}$

$\begin{aligned} & \text { In option (a) } \\ & \text { (A. A) (B. Q) } \\ & \left(A^2\right)(B C \cos \theta) \\ & =A^2 B C \cos \theta \text { Scalar value } \\ \end{aligned}$

In option (b)

$\begin{aligned} (\mathbf{A} \times \mathbf{B})(\mathbf{B} \times \mathbf{Q} & =(A B \sin \theta \hat{\mathbf{n}}) \cdot(B C \sin \phi \hat{k}) \\ & =(A B \sin \theta)(B C \sin \phi)(\hat{\mathbf{n}} \cdot \hat{k}) \end{aligned}$

where, $\hat{\mathbf{n}}$ is unit vector perpendicular to both vector $\mathbf{A}$ and $\mathbf{B}$ and $\hat{k}$ is unit vector perpendicular to both vector $\mathbf{B}$ and $\mathbf{C}$.

$\therefore \hat{\mathbf{n}} . \hat{k}=$ scalar value

$\therefore(\mathbf{A} \times \mathbf{B}) \cdot(\mathbf{B} \times \mathbf{Q}$ is a scalar value In option (C)

$\begin{aligned} & \left(\mathbf{A} \times \mathbf{Q} \times\left(\mathbf{B} \times \mathbf{Q}=\left(A C \sin \theta_1\right) \hat{\mathbf{r}} \times\left(B C \sin \theta_2\right) \hat{\mathbf{r}}_2\right.\right. \\ & =\left(A C \sin \theta_1\right)\left(B C \sin \theta_2\right)\left(\hat{\mathbf{r}} \times \hat{\mathbf{r}}_2\right) \end{aligned}$

$\hat{\mathbf{r}}_1 \times \hat{\mathbf{r}}_2$ is a vector product, hence it gives a vector value.

Therefore, $\mathbf{A} \times \mathbf{Q} \times(\mathbf{B} \times \mathbf{Q}$ is a vector value.

In option (d) $\mathbf{A} \times(\mathbf{B} \times \mathbf{Q}$ is vector triple product which gives a vector value.

Given, applied force, F = 40 N

Horizontal component of force, $F_x=20\sqrt3$

and $\theta$ be the angle of force from horizontal X-axis.

According to given diagram,

The component of F along X-axis will be

$\begin{aligned} & F_x =F \cos \theta \\\\ \Rightarrow 20 \sqrt{3} & =40 \cos \theta \Rightarrow \sqrt{3} / 2=\cos \theta \\\\ \therefore & \quad \theta =\cos ^{-1}(\sqrt{3} / 2)=30^{\circ} \end{aligned}$

And the component of $F$ along $Y$-axis

$\begin{aligned} F_y & =F \sin \theta \\\\ & =F \sin 30^{\circ} \\\\ & =40 \times 1 / 2=20 \mathrm{~N} \end{aligned}$