JEE Main · 2023 · 11 Apr 2023 · Shift 1 · Three Dimensional Geometry

Each plane contains the two given vectors, so its normal is their cross product: ${\vec{n}}_1=(\hat{i}+\hat{j})\times (\hat{i}+\hat{k})=\hat{i}-\hat{j}-\hat{k}$ and ${\vec{n}}_2=(\hat{i}-\hat{j})\times (\hat{j}-\hat{k})=\hat{i}+\hat{j}+\hat{k}$. The line of intersection is parallel to ${\vec{n}}_1\times {\vec{n}}_2=-2\hat{j}+2\hat{k}$, so $\vec{a}=\lambda (\hat{j}-\hat{k})$. From $\vec{a}\cdot \vec{b}=6$ with $\vec{b}=2\hat{i}-2\hat{j}+\hat{k}$, $-3\lambda =6$, giving $\lambda =-2$ and $\vec{a}=-2\hat{j}+2\hat{k}$. Then $|\vec{a}|=2\sqrt{2}$ and $|\vec{b}|=3$, so $\cos \theta =\frac{\vec{a}\cdot \vec{b}}{|\vec{a}||\vec{b}|}=\frac{6}{2\sqrt{2}\cdot 3}=\frac{1}{\sqrt{2}}$, which gives $\theta =\frac{\pi }{4}$. Also $|\vec{a}\times \vec{b}|=|\vec{a}||\vec{b}|\sin \theta =2\sqrt{2}\cdot 3\cdot \frac{1}{\sqrt{2}}=6$. The required ordered pair is $\left(\frac{\pi }{4},6\right)$.