JEE Main 2021 (Online) 27th July Morning ShiftQuestion 1
Let $\vec a=\hat i+\hat j+\hat k$, $\vec b$ and $\vec c=\hat j-\hat k$ be three vectors such that $\vec a\times\vec b=\vec c$ and $\vec a\cdot\vec b=1$. If the length of projection vector of the vector $\vec b$ on the vector $\vec a\times\vec c$ is $l$, then the value of $3l^2$ is equal to ________.
Vector Algebra
Answer$2$
Solution
$\vec a\times(\vec a\times\vec b)=\vec a(\vec a\cdot\vec b)-\vec b|\vec a|^2=\vec a-3\vec b=\vec a\times\vec c=(-2,1,1)$. So $\vec b=\frac{(1,1,1)-(-2,1,1)}{3}=(1,0,0)$ (check: $\vec a\times\vec b=(0,1,-1)=\vec c$). Projection length on $\vec v=\vec a\times\vec c=(-2,1,1)$: $l=\frac{|\vec b\cdot\vec v|}{|\vec v|}=\frac2{\sqrt6}$, $l^2=\frac23$, $3l^2=2$.