JEE Main 2024 (Online) 31st January Evening ShiftQuestion 1
Let $\vec{a}=3\hat{i}+2\hat{j}+\hat{k}$, $\vec{b}=2\hat{i}-\hat{j}+3\hat{k}$ and $\vec{c}$ be a vector such that $(\vec{a}+\vec{b})\times\vec{c}=2(\vec{a}\times\vec{b})+24\hat{j}-6\hat{k}$ and $(\vec{a}-\vec{b}+\hat{i})\cdot\vec{c}=-3$. Then $|\vec{c}|^{2}$ is equal to $\_\_\_\_\_\_$ .
Vector Algebra
Answer$38$
Solution
$(\vec a+\vec b)\times\vec c=(5,1,4)\times\vec c$. Also $\vec a\times\vec b=(7,-7,-7)$, so RHS $=(14,10,-20)$. And $\vec a-\vec b+\hat i=(2,3,-2)$. Solving $(5,1,4)\times\vec c=(14,10,-20)$ together with $2x+3y-2z=-3$ gives $\vec c=(5,-3,2)$. Check: $(5,1,4)\times(5,-3,2)=(2+12,\,20-10,\,-15-5)=(14,10,-20)$; $10-9-4=-3$. $|\vec c|^2=25+9+4=38$.