Three particles P, Q and R are moving along the vectors \overrightarrow A = \widehat i + \widehat j , \overrightarrow B = \widehat j + \widehat k and \overrightarrow C = - \widehat i + \widehat j respectively. They strike on a point and start to move in different directions. Now particle P is moving normal to the plane which contains vector \overrightarrow A and \overrightarrow B . Similarly particle Q is moving normal to the plane which contains vector \overrightarrow A and \overrightarrow C . The angle between the direction of motion of P and Q is \cos ^ - 1( 1 \over \sqrt x ) . Then the value of x is \underline\hspace2cm .

Vector Algebra PYQ — JEE Main 2021 · 22 Jul · Shift 2 · Q67 with Solution

View written solutionFree

Correct answer: 3

Given vectors

$\vec A = \hat i + \hat j = (1,1,0)$ $\vec B = \hat j + \hat k = (0,1,1)$ $\vec C = -\hat i + \hat j = (-1,1,0)$
Direction of motion of particle P

P moves normal to the plane containing $\vec A$ and $\vec B$ .

So its direction is along: $\vec n_P = \vec A \times \vec B$

Compute:
$\begin{vmatrix} \hat i & \hat j & \hat k \\ 1 & 1 & 0 \\ 0 & 1 & 1 \end{vmatrix}$$ $$= \hat i(1\cdot 1 - 0\cdot 1) - \hat j(1\cdot 1 - 0\cdot 0) + \hat k(1\cdot 1 - 1\cdot 0)$$ $$= \hat i - \hat j + \hat k$$ Hence, $$\vec n_P = (1,-1,1)$$$
Direction of motion of particle Q

Q moves normal to the plane containing $\vec A$ and $\vec C$ .

So its direction is along: $\vec n_Q = \vec A \times \vec C$

Compute:
$\begin{vmatrix} \hat i & \hat j & \hat k \\ 1 & 1 & 0 \\ -1 & 1 & 0 \end{vmatrix}$$ $$= \hat i(1\cdot 0 - 0\cdot 1) - \hat j(1\cdot 0 - 0\cdot (-1)) + \hat k(1\cdot 1 - 1\cdot (-1))$$ $$= 0\hat i - 0\hat j + 2\hat k$$ Hence, $$\vec n_Q = (0,0,2)$$$
Angle between directions of P and Q

Let the angle be $\theta$ .

$\cos \theta = \frac{\vec n_P \cdot \vec n_Q}{|\vec n_P|\,|\vec n_Q|}$

Dot product: $\vec n_P \cdot \vec n_Q = (1)(0)+(-1)(0)+(1)(2)=2$

Magnitudes: $|\vec n_P| = \sqrt{1^2+(-1)^2+1^2}=\sqrt{3}$ $|\vec n_Q| = \sqrt{0^2+0^2+2^2}=2$

Therefore, $\cos\theta = \frac{2}{\sqrt{3}\cdot 2} = \frac{1}{\sqrt{3}}$
Compare with given form

Given: $\theta = \cos^{-1}\left(\frac{1}{\sqrt{x}}\right)$

So, $\frac{1}{\sqrt{x}} = \frac{1}{\sqrt{3}}$

Hence, $x=3$
Comparison with stored answer

Derived answer = $3$

Stored correct answer = $3$

They agree.

More from Vector Algebra

View written solutionFree

Correct answer: 3

Given vectors

$\vec A = \hat i + \hat j = (1,1,0)$ $\vec B = \hat j + \hat k = (0,1,1)$ $\vec C = -\hat i + \hat j = (-1,1,0)$
Direction of motion of particle P

P moves normal to the plane containing $\vec A$ and $\vec B$ .

So its direction is along: $\vec n_P = \vec A \times \vec B$

Compute:
$\begin{vmatrix} \hat i & \hat j & \hat k \\ 1 & 1 & 0 \\ 0 & 1 & 1 \end{vmatrix}$$ $$= \hat i(1\cdot 1 - 0\cdot 1) - \hat j(1\cdot 1 - 0\cdot 0) + \hat k(1\cdot 1 - 1\cdot 0)$$ $$= \hat i - \hat j + \hat k$$ Hence, $$\vec n_P = (1,-1,1)$$$
Direction of motion of particle Q

Q moves normal to the plane containing $\vec A$ and $\vec C$ .

So its direction is along: $\vec n_Q = \vec A \times \vec C$

Compute:
$\begin{vmatrix} \hat i & \hat j & \hat k \\ 1 & 1 & 0 \\ -1 & 1 & 0 \end{vmatrix}$$ $$= \hat i(1\cdot 0 - 0\cdot 1) - \hat j(1\cdot 0 - 0\cdot (-1)) + \hat k(1\cdot 1 - 1\cdot (-1))$$ $$= 0\hat i - 0\hat j + 2\hat k$$ Hence, $$\vec n_Q = (0,0,2)$$$
Angle between directions of P and Q

Let the angle be $\theta$ .

$\cos \theta = \frac{\vec n_P \cdot \vec n_Q}{|\vec n_P|\,|\vec n_Q|}$

Dot product: $\vec n_P \cdot \vec n_Q = (1)(0)+(-1)(0)+(1)(2)=2$

Magnitudes: $|\vec n_P| = \sqrt{1^2+(-1)^2+1^2}=\sqrt{3}$ $|\vec n_Q| = \sqrt{0^2+0^2+2^2}=2$

Therefore, $\cos\theta = \frac{2}{\sqrt{3}\cdot 2} = \frac{1}{\sqrt{3}}$
Compare with given form

Given: $\theta = \cos^{-1}\left(\frac{1}{\sqrt{x}}\right)$

So, $\frac{1}{\sqrt{x}} = \frac{1}{\sqrt{3}}$

Hence, $x=3$
Comparison with stored answer

Derived answer = $3$

Stored correct answer = $3$

They agree.

Vector Algebra question

Vector Algebra question

More from Vector Algebra

Vector Algebra question

Vector Algebra question

More from Vector Algebra