JEE Main 2023 (Online) 11th April Morning ShiftQuestion 1
Let $\vec{a}$ be a non-zero vector parallel to the line of intersection of the two planes described by $\hat{i}+\hat{j}$, $\hat{i}+\hat{k}$ and $\hat{i}-\hat{j}$, $\hat{j}-\hat{k}$. If $\theta$ is the angle between the vector $\vec{a}$ and the vector $\vec{b}=2\hat{i}-2\hat{j}+\hat{k}$ and $\vec{a}\cdot\vec{b}=6$, then the ordered pair $\left(\theta,|\vec{a}\times\vec{b}|\right)$ is equal to
Vector Algebra
Solution
Normals: $(\hat i+\hat j)\times(\hat i+\hat k)=(1,-1,-1)$ and $(\hat i-\hat j)\times(\hat j-\hat k)=(1,1,1)$. Line direction $=(1,-1,-1)\times(1,1,1)=(0,-2,2)$, so $\vec a=t(0,-1,1)$. $\vec a\cdot\vec b=t(2+1)=3t=6\Rightarrow t=2$, $\vec a=(0,-2,2)$, $|\vec a|=2\sqrt2$, $|\vec b|=3$. $\cos\theta=\frac{6}{6\sqrt2}\Rightarrow\theta=\frac\pi4$. $|\vec a\times\vec b|=|\vec a||\vec b|\sin\theta=2\sqrt2\cdot3\cdot\frac1{\sqrt2}=6$. Option (D).