JEE Main 2025 (Online) 28th January Morning ShiftQuestion 1
Let $\vec{a}=\hat{i}+\hat{j}+\hat{k}$, $\vec{b}=2\hat{i}+2\hat{j}+\hat{k}$ and $\vec{d}=\vec{a}\times\vec{b}$. If $\vec{c}$ is a vector such that $\vec{a}\cdot\vec{c}=|\vec{c}|$, $|\vec{c}-2\vec{a}|^2=8$ and the angle between $\vec{d}$ and $\vec{c}$ is $\frac{\pi}{4}$, then $|10-3\vec{b}\cdot\vec{c}|+|\vec{d}\times\vec{c}|^2$ equals ____.
Vector Algebra
Answer$6$
Solution
$\vec d=(-1,1,0)$, $|\vec d|=\sqrt2$. $|\vec c-2\vec a|^2=|\vec c|^2-4|\vec c|+12=8\Rightarrow|\vec c|=2$, $\vec a\cdot\vec c=2$, $\vec d\cdot\vec c=2$. Then $|\vec d\times\vec c|^2=2\cdot4\cdot\frac12=4$ and $\vec c=(0,2,0)$ or $(-\frac23,\frac43,\frac43)$ give $\vec b\cdot\vec c=4$ or $\frac83$, so $|10-3\vec b\cdot\vec c|=2$ either way. Total $=6$.