Vectors – Cross Product – JEE Main 24 January 2025 Shift 2

Solution:

$$ \vec{b} = \vec{a} \times (\hat{i}-2\hat{k}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -1 & 2 \\ 1 & 0 & -2 \end{vmatrix} = 2\hat{i} + 8\hat{j} + \hat{k} $$

$$ \vec{c} = \vec{b} \times \hat{k} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 8 & 1 \\ 0 & 0 & 1 \end{vmatrix} = 8\hat{i} – 2\hat{j} $$

$$ \vec{c} – 2\hat{j} = (8\hat{i} – 2\hat{j}) – 2\hat{j} = 8\hat{i} – 4\hat{j} $$

Projection of $(\vec{c} – 2\hat{j})$ on $\vec{a}$:
$$ \frac{(\vec{c}-2\hat{j}) \cdot \vec{a}}{|\vec{a}|} = \frac{(8\hat{i}-4\hat{j}) \cdot (3\hat{i}-\hat{j}+2\hat{k})}{\sqrt{3^2+(-1)^2+2^2}} $$

$$ = \frac{24+4+0}{\sqrt{14}} = \frac{28}{\sqrt{14}} = 2\sqrt{14} $$

Ans. (3)