JEE Main 2025 Jan 22 Shift 1, Mathematics Q25: Vector (or Cross) Product of Two Vectors

$|\vec{a}{|}^2=1+4+4=9$, so $|\vec{a}|=3$. Dot product: $\vec{b}\cdot \vec{a}=\lambda +8$.

Projection: $\vec{c}=\frac{\vec{b}\cdot \vec{a}}{|\vec{a}{|}^2}\vec{a}=\frac{\lambda +8}{9}(\hat{i}+2\hat{j}+2\hat{k})$.

$\vec{a}+\vec{c}=\left(1+\frac{\lambda +8}{9}\right)(\hat{i}+2\hat{j}+2\hat{k})=\frac{\lambda +17}{9}(\hat{i}+2\hat{j}+2\hat{k})$.

$|\vec{a}+\vec{c}|=\frac{\lambda +17}{9}\cdot 3=\frac{\lambda +17}{3}=7\Rightarrow \lambda =4$.

So $\vec{b}=4\hat{i}+4\hat{k}$ and $\vec{c}=\frac{12}{9}(\hat{i}+2\hat{j}+2\hat{k})=\frac{4}{3}(\hat{i}+2\hat{j}+2\hat{k})$.

Area $=|\vec{b}\times \vec{c}|$. Computing: $\vec{b}\times \vec{c}=\left|\begin{array}{c}\hat{i}\hat{j}\hat{k}\\ 404\\ 4/38/38/3\end{array}\right|=\hat{i}\left(0\cdot \frac{8}{3}-4\cdot \frac{8}{3}\right)-\hat{j}\left(4\cdot \frac{8}{3}-4\cdot \frac{4}{3}\right)+\hat{k}\left(4\cdot \frac{8}{3}-0\right)$.

$=-\frac{32}{3}\hat{i}-\frac{16}{3}\hat{j}+\frac{32}{3}\hat{k}$. Magnitude $=\frac{1}{3}\sqrt{32^2+16^2+32^2}=\frac{1}{3}\sqrt{1024+256+1024}=\frac{1}{3}\sqrt{2304}=\frac{48}{3}=16$.