Verified Solution • 2025–26 Session moderate

QJEEM23Madef6: JEE Main 2023 Physics - Class 11 JEEM23Madef6 Solution: Let be a non-zero vector parallel to the line of intersection…

Let a⃗\vec{a} be a non-zero vector parallel to the line of intersection of the two planes described by i^+j^,i^+k^\hat{i}+\hat{j}, \hat{i}+\hat{k} and i^−j^,j^−k^\hat{i}-\hat{j}, \hat{j}-\hat{k} . If θ\theta is the angle between the vector a⃗\vec{a} and the vector b⃗=2i^−2j^+k^\vec{b}=2 \hat{i}-2 \hat{j}+\hat{k} and a⃗⋅b⃗=6\vec{a} \cdot \vec{b}=6 , then the ordered pair (θ,∣a⃗×b⃗∣)(\theta,|\vec{a} \times \vec{b}|) is equal to :

Multiple Choice Options

  • A (π3,36)\left(\frac{\pi}{3}, 3 \sqrt{6}\right)
  • B (π3,6)\left(\frac{\pi}{3}, 6\right)
  • C (π4,36)\left(\frac{\pi}{4}, 3 \sqrt{6}\right)
  • D (π4,6)\left(\frac{\pi}{4}, 6\right)

Step-by-Step Verified Solution

We have, a⃗\vec{a} is non-zero vector parallel to the line of intersection of the two planes described by i^+j^,i^+k^\hat{\mathbf{i}}+\hat{\mathbf{j}}, \hat{\mathbf{i}}+\hat{\mathbf{k}} and i^−j^,j^−k^\hat{\mathbf{i}}-\hat{\mathbf{j}}, \hat{\mathbf{j}}-\hat{\mathbf{k}} .

Let n1\mathbf{n}_1 and n2\mathbf{n}_2 are the normal vector to the plane i^+j^,i^+k^\hat{\mathbf{i}}+\hat{\mathbf{j}}, \hat{\mathbf{i}}+\hat{\mathbf{k}} and i^−j^,j^−k^\hat{\mathbf{i}}-\hat{\mathbf{j}}, \hat{\mathbf{j}}-\hat{\mathbf{k}} , respectively.

amp;n1=∣i^amp;j^amp;k^1amp;1amp;01amp;0amp;1∣=i^−j^−k^amp;n2=∣i^amp;j^amp;k^1amp;−1amp;01amp;0amp;−1∣=i^+j^+k^amp;a⃗=λ∣n1×n2∣\begin{aligned}& n_1=\left|\begin{array}{ccc}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\1 & 1 & 0 \\1 & 0 & 1\end{array}\right|=\hat{\mathbf{i}}-\hat{\mathbf{j}}-\hat{\mathbf{k}} \\\\& n_2=\left|\begin{array}{ccc}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\1 & -1 & 0 \\1 & 0 & -1\end{array}\right|=\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}} \\\\& \vec{a}=\lambda\left|n_1 \times n_2\right|\end{aligned}

[ ∵a\because \mathbf{a} is parallel to line of intersection of both planes]

amp;=λ∣i^amp;j^amp;k^1amp;−1amp;−11amp;1amp;1∣amp;=λ(−2j^+2k^)\begin{aligned}& =\lambda\left|\begin{array}{ccc}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\1 & -1 & -1 \\1 & 1 & 1\end{array}\right| \\\\& =\lambda(-2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}})\end{aligned}

a⃗⋅b⃗=6[Given]\vec{a} \cdot \vec{b} =6 [Given]

λ(0+4+2)=6\lambda(0+4+2) =6

λ=1\lambda =1

∴a⃗amp;=−2j^+2k^cos⁡θamp;=a⃗⋅b⃗∣a⃗∣∣b⃗∣=64+44+4+1amp;=12\begin{aligned}\therefore \vec{a} & =-2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}} \\\\\cos \theta & =\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}=\frac{6}{\sqrt{4+4} \sqrt{4+4+1}} \\\\& =\frac{1}{\sqrt{2}}\end{aligned}

∴θ=π4\therefore \theta=\frac{\pi}{4}

a⃗×b⃗amp;=∣i^amp;j^amp;k^0amp;−2amp;22amp;−2amp;1∣amp;=2i^+4j^+4k^∣a⃗×b⃗∣amp;=4+16+16=6 Hence, (θ,∣a⃗×b⃗∣)amp;=(π4,6)\begin{aligned}\vec{a} \times \vec{b} & =\left|\begin{array}{ccc}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\0 & -2 & 2 \\2 & -2 & 1\end{array}\right| \\\\& =2 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}+4 \hat{\mathbf{k}} \\\\|\vec{a} \times \vec{b}| & =\sqrt{4+16+16}=6 \\\\\text { Hence, }(\theta,|\vec{a} \times \vec{b}|) & =\left(\frac{\pi}{4}, 6\right)\end{aligned}

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