JEE Main 2025 Jan 24 Shift 2, Mathematics Q14: Vector (or Cross) Product of Two Vectors

Compute $\vec{b}$:

$\vec{b}=\vec{a}\times (\hat{i}-2\hat{k})=\left|\begin{array}{c}\hat{i}\hat{j}\hat{k}\\ 3-12\\ 10-2\end{array}\right|=\hat{i}(2-0)-\hat{j}(-6-2)+\hat{k}(0+1)=2\hat{i}+8\hat{j}+\hat{k}$

Compute $\vec{c}$:

$\vec{c}=\vec{b}\times \hat{k}=(2\hat{i}+8\hat{j}+\hat{k})\times \hat{k}=2(\hat{i}\times \hat{k})+8(\hat{j}\times \hat{k})+(\hat{k}\times \hat{k})$

$=-2\hat{j}+8\hat{i}+\vec{0}=8\hat{i}-2\hat{j}$

Form the required vector:

$\vec{c}-2\hat{j}=8\hat{i}-4\hat{j}$

Project on $\vec{a}=(3,-1,2)$:

$\text{Projection}=\frac{(\vec{c}-2\hat{j})\cdot \vec{a}}{|\vec{a}|}=\frac{(8)(3)+(-4)(-1)+(0)(2)}{\sqrt{9+1+4}}=\frac{24+4}{\sqrt{14}}=\frac{28}{\sqrt{14}}=2\sqrt{14}$

Correct option: (3)