JEE Main 2025 (Online) 3rd April Evening ShiftQuestion 1
Let $\vec{a}=\hat{i}+2\hat{j}+\hat{k}$, $\vec{b}=3\hat{i}-3\hat{j}+3\hat{k}$, $\vec{c}=2\hat{i}-\hat{j}+2\hat{k}$ and $\vec{d}$ be a vector such that $\vec{b}\times\vec{d}=\vec{c}\times\vec{d}$ and $\vec{a}\cdot\vec{d}=4$. Then $\left|\left(\vec{a}\times\vec{d}\right)\right|^{2}$ is equal to $\_\_\_\_\_\_$ .
Vector Algebra
Answer$128$
Solution
$(\vec b-\vec c)\times\vec d=0\Rightarrow\vec d=t(1,-2,1)$. $\vec a\cdot\vec d=t(1-4+1)=-2t=4\Rightarrow t=-2$, $\vec d=(-2,4,-2)$, $|\vec d|^2=24$. $|\vec a\times\vec d|^2=|\vec a|^2|\vec d|^2-(\vec a\cdot\vec d)^2=6\cdot24-16=128$.