Let \veca be a non-zero vector parallel to the line of intersection of the two planes described by \hati+\hatj, \hati+\hatk and \hati-\hatj, \hatj-\hatk . If \theta is the angle between the vector \veca and the vector \vecb=2 \hati-2 \hatj+\hatk and \veca \cdot \vecb=6 , then the ordered pair (\theta,|\veca \times \vecb|) is equal to :

Vector Algebra PYQ — JEE Main 2023 · 11 Apr · Shift 1 · Q37 with Solution

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Correct answer: D

Interpret the planes from their spanning vectors

A plane described by two non-parallel vectors has those vectors lying in the plane, so its normal vector is their cross product.

First plane contains vectors $\hat i+\hat j=(1,1,0)$ and $\hat i+\hat k=(1,0,1)$ .
Second plane contains vectors $\hat i-\hat j=(1,-1,0)$ and $\hat j-\hat k=(0,1,-1)$ .

The line of intersection of two planes is parallel to the cross product of their normals.

Find normal to the first plane

\vec n_1=(\hat i+\hat j)\times(\hat i+\hat k)

Using coordinates:

(1,1,0)\times(1,0,1) =\begin{vmatrix} \hat i & \hat j & \hat k\\ 1 & 1 & 0\\ 1 & 0 & 1 \end{vmatrix} =\hat i(1)-\hat j(1)+\hat k(-1)

So,

\vec n_1=\hat i-\hat j-\hat k=(1,-1,-1)

Find normal to the second plane

\vec n_2=(\hat i-\hat j)\times(\hat j-\hat k)

Using coordinates:

(1,-1,0)\times(0,1,-1) =\begin{vmatrix} \hat i & \hat j & \hat k\\ 1 & -1 & 0\\ 0 & 1 & -1 \end{vmatrix} =\hat i(1)-\hat j(-1)+\hat k(1)

Thus,

\vec n_2=\hat i+\hat j+\hat k=(1,1,1)

Direction of line of intersection

A vector parallel to the line of intersection is

\vec d=\vec n_1\times \vec n_2

So,

(1,-1,-1)\times(1,1,1) =\begin{vmatrix} \hat i & \hat j & \hat k\\ 1 & -1 & -1\\ 1 & 1 & 1 \end{vmatrix}

=\hat i((-1)(1)-(-1)(1))- \hat j((1)(1)-(-1)(1))+ \hat k((1)(1)-(-1)(1))

=\hat i(0)-\hat j(2)+\hat k(2) =(0,-2,2)

Hence the line of intersection is parallel to

(0,-1,1)= -\hat j+\hat k

Therefore, since $\vec a$ is non-zero and parallel to this line,

\vec a=\lambda(0,-1,1)

for some non-zero scalar $\lambda$ .

Use the condition $\vec a\cdot \vec b=6$

Given

\vec b=2\hat i-2\hat j+\hat k=(2,-2,1)

Now,

\vec a\cdot \vec b=\lambda(0,-1,1)\cdot(2,-2,1)

=\lambda\,(0+2+1)=3\lambda

Given $\vec a\cdot \vec b=6$ , so

3\lambda=6 \implies \lambda=2

Thus,

\vec a=(0,-2,2)

Find the angle $\theta$ between $\vec a$ and $\vec b$

First compute magnitudes:

|\vec a|=\sqrt{0^2+(-2)^2+2^2}=\sqrt{8}=2\sqrt2

|\vec b|=\sqrt{2^2+(-2)^2+1^2}=\sqrt9=3

Using dot product formula,

\cos\theta=\frac{\vec a\cdot\vec b}{|\vec a||\vec b|} =\frac{6}{(2\sqrt2)(3)}=\frac{1}{\sqrt2}

Hence,

\theta=\frac{\pi}{4}

Find $|\vec a\times \vec b|$

Using

|\vec a\times \vec b|=|\vec a||\vec b|\sin\theta

= (2\sqrt2)(3)\left(\frac{1}{\sqrt2}\right)=6

since for $\theta=\pi/4$ , $\sin\theta=1/\sqrt2$ .

So,

|\vec a\times \vec b|=6

Final ordered pair

(\theta,|\vec a\times \vec b|)=\left(\frac{\pi}{4},6\right)

This corresponds to Option D.

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View written solutionFree

Correct answer: D

Interpret the planes from their spanning vectors

A plane described by two non-parallel vectors has those vectors lying in the plane, so its normal vector is their cross product.

First plane contains vectors $\hat i+\hat j=(1,1,0)$ and $\hat i+\hat k=(1,0,1)$ .
Second plane contains vectors $\hat i-\hat j=(1,-1,0)$ and $\hat j-\hat k=(0,1,-1)$ .

The line of intersection of two planes is parallel to the cross product of their normals.

Find normal to the first plane

\vec n_1=(\hat i+\hat j)\times(\hat i+\hat k)

Using coordinates:

(1,1,0)\times(1,0,1) =\begin{vmatrix} \hat i & \hat j & \hat k\\ 1 & 1 & 0\\ 1 & 0 & 1 \end{vmatrix} =\hat i(1)-\hat j(1)+\hat k(-1)

So,

\vec n_1=\hat i-\hat j-\hat k=(1,-1,-1)

Find normal to the second plane

\vec n_2=(\hat i-\hat j)\times(\hat j-\hat k)

Using coordinates:

(1,-1,0)\times(0,1,-1) =\begin{vmatrix} \hat i & \hat j & \hat k\\ 1 & -1 & 0\\ 0 & 1 & -1 \end{vmatrix} =\hat i(1)-\hat j(-1)+\hat k(1)

Thus,

\vec n_2=\hat i+\hat j+\hat k=(1,1,1)

Direction of line of intersection

A vector parallel to the line of intersection is

\vec d=\vec n_1\times \vec n_2

So,

(1,-1,-1)\times(1,1,1) =\begin{vmatrix} \hat i & \hat j & \hat k\\ 1 & -1 & -1\\ 1 & 1 & 1 \end{vmatrix}

=\hat i((-1)(1)-(-1)(1))- \hat j((1)(1)-(-1)(1))+ \hat k((1)(1)-(-1)(1))

=\hat i(0)-\hat j(2)+\hat k(2) =(0,-2,2)

Hence the line of intersection is parallel to

(0,-1,1)= -\hat j+\hat k

Therefore, since $\vec a$ is non-zero and parallel to this line,

\vec a=\lambda(0,-1,1)

for some non-zero scalar $\lambda$ .

Use the condition $\vec a\cdot \vec b=6$

Given

\vec b=2\hat i-2\hat j+\hat k=(2,-2,1)

Now,

\vec a\cdot \vec b=\lambda(0,-1,1)\cdot(2,-2,1)

=\lambda\,(0+2+1)=3\lambda

Given $\vec a\cdot \vec b=6$ , so

3\lambda=6 \implies \lambda=2

Thus,

\vec a=(0,-2,2)

Find the angle $\theta$ between $\vec a$ and $\vec b$

First compute magnitudes:

|\vec a|=\sqrt{0^2+(-2)^2+2^2}=\sqrt{8}=2\sqrt2

|\vec b|=\sqrt{2^2+(-2)^2+1^2}=\sqrt9=3

Using dot product formula,

\cos\theta=\frac{\vec a\cdot\vec b}{|\vec a||\vec b|} =\frac{6}{(2\sqrt2)(3)}=\frac{1}{\sqrt2}

Hence,

\theta=\frac{\pi}{4}

Find $|\vec a\times \vec b|$

Using

|\vec a\times \vec b|=|\vec a||\vec b|\sin\theta

= (2\sqrt2)(3)\left(\frac{1}{\sqrt2}\right)=6

since for $\theta=\pi/4$ , $\sin\theta=1/\sqrt2$ .

So,

|\vec a\times \vec b|=6

Final ordered pair

(\theta,|\vec a\times \vec b|)=\left(\frac{\pi}{4},6\right)

This corresponds to Option D.

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