JEE Main 2026 Jan 24 Shift 1, Mathematics Q5: Vector (or Cross) Product of Two Vectors

Compute $\vec{c}$:

$\vec{c}=\vec{a}\times \vec{b}=\left|\begin{array}{c}\hat{i}\hat{j}\hat{k}\\ 21-2\\ 110\end{array}\right|=2\hat{i}-2\hat{j}+\hat{k}$.

$|\vec{c}|=\sqrt{4+4+1}=3$. Note $|\vec{a}{|}^2=4+1+4=9$.

Find $|\vec{d}|$: From $|\vec{c}\times \vec{d}|=|\vec{c}||\vec{d}|\sin (\pi /4)=3$:

$3\cdot |\vec{d}|\cdot \frac{1}{\sqrt{2}}=3\Rightarrow |\vec{d}|=\sqrt{2}$.

Use $|\vec{d}-\vec{a}{|}^2=11$:

$|\vec{a}{|}^2+|\vec{d}{|}^2-2\vec{a}\cdot \vec{d}=11$

$9+2-2\vec{a}\cdot \vec{d}=11\Rightarrow \vec{a}\cdot \vec{d}=0$.