JEE Main 2026 Apr 6 Shift 1, Mathematics Q23: Scalar and Vector Triple Product

Apply the BAC-CAB identity: $\vec{a}\times (\vec{a}\times \vec{c})=(\vec{a}\cdot \vec{c})\vec{a}-(\vec{a}\cdot \vec{a})\vec{c}=\vec{a}\times \vec{b}$.

With $\vec{a}\cdot \vec{c}=3$ and $\vec{a}\cdot \vec{a}=3$:

$3\vec{a}-3\vec{c}=\vec{a}\times \vec{b}$.

Compute $\vec{a}\times \vec{b}=\left|\begin{array}{c}\hat{i}\hat{j}\hat{k}\\ 111\\ 01-1\end{array}\right|=-2\hat{i}+\hat{j}+\hat{k}$.

So $\vec{c}=\vec{a}-\frac{1}{3}(-2\hat{i}+\hat{j}+\hat{k})=\frac{1}{3}(5\hat{i}+2\hat{j}+2\hat{k})$.

Now $\vec{a}-2\vec{b}=(\hat{i}+\hat{j}+\hat{k})-2(\hat{j}-\hat{k})=\hat{i}-\hat{j}+3\hat{k}$.

$\vec{c}\cdot (\vec{a}-2\vec{b})=\frac{1}{3}(5-2+6)=\frac{9}{3}=3$.