JEE Main 2023 (Online) 31st January Evening ShiftQuestion 1
Let $\vec a=\hat i+2\hat j+3\hat k$, $\vec b=\hat i-\hat j+2\hat k$ and $\vec c=5\hat i-3\hat j+3\hat k$ be three vectors. If $\vec r$ is a vector such that $\vec r\times\vec b=\vec c\times\vec b$ and $\vec r\cdot\vec a=0$, then $25|\vec r|^{2}$ is equal to
Vector Algebra
Solution
$(\vec r-\vec c)\times\vec b=0$, so $\vec r=\vec c+t\vec b$. $\vec r\cdot\vec a=8+5t=0$, $t=-\frac85$, $\vec r=(\frac{17}5,-\frac75,-\frac15)$ and $25|\vec r|^2=289+49+1=339$.