JEE Main 2018 (Online) 16th AprilQuestion 1
Let $\vec a=\hat i+\hat j+\hat k$, $\vec c=\hat j-\hat k$ and a vector $\vec b$ be such that $\vec a\times\vec b=\vec c$ and $\vec a\cdot\vec b=3$. Then $|\vec b|$ equals
Vector Algebra
Solution
$\vec a\times(\vec a\times\vec b)=3\vec a-3\vec b=\vec a\times\vec c=(-2,1,1)$, so $\vec b=\vec a-\frac13(-2,1,1)=(\frac53,\frac23,\frac23)$ and $|\vec b|^2=\frac{33}9=\frac{11}3$.