JEE Main 2025 Jan 24 Shift 1, Mathematics Q1: Scalar and Vector Triple Product

Any vector coplanar with $\vec{a}$ and $\vec{b}$ and perpendicular to $\vec{b}$ can be written as $\vec{c}=\lambda \vec{b}\times (\vec{a}\times \vec{b})$, since the BAC$-$CAB expansion guarantees the result lies in the $(\vec{a},\vec{b})$ plane and is automatically perpendicular to $\vec{b}$.

Using the BAC$-$CAB identity:

$\vec{c}=\lambda [(\vec{b}\cdot \vec{b})\vec{a}-(\vec{a}\cdot \vec{b})\vec{b}]$

Compute the dot products: $\vec{b}\cdot \vec{b}=9+1+1=11$ and $\vec{a}\cdot \vec{b}=3+2-3=2$.

$\vec{c}=\lambda (11\vec{a}-2\vec{b})=\lambda [(11-6)\hat{i}+(22-2)\hat{j}+(33+2)\hat{k}]=5\lambda (\hat{i}+4\hat{j}+7\hat{k})$

Apply $\vec{a}\cdot \vec{c}=5$:

$5\lambda (1+8+21)=5\Rightarrow 150\lambda =5\Rightarrow \lambda =\frac{1}{30}$

Hence $\vec{c}=\frac{1}{6}(\hat{i}+4\hat{j}+7\hat{k})$ and

$|\vec{c}|=\frac{\sqrt{1+16+49}}{6}=\frac{\sqrt{66}}{6}=\sqrt{\frac{11}{6}}$.

$\therefore$ Option (4).