JEE Mathematics 2025 Question with Solution

Given: $\mathbf{a} = 3\hat{i} - \hat{j} + 2\hat{k}$, $\mathbf{b} = \mathbf{a} \times (\hat{i} - 2\hat{k})$ and $\mathbf{c} = \mathbf{b} \times \hat{k}$.

Find: The projection of $\mathbf{c} - 2\hat{j}$ on $\mathbf{a}$.

Use the scalar projection formula

$$\text{proj}_{\mathbf{a}}(\mathbf{v}) = \frac{\mathbf{v} \cdot \mathbf{a}}{\|\mathbf{a}\|}$$

so first compute $\mathbf{b}$ and $\mathbf{c}$.

For $\mathbf{u} = \hat{i} - 2\hat{k}$,

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

So,

$$\mathbf{b} = 2\hat{i} + 8\hat{j} + \hat{k}$$

Now,

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

Hence,

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

Now compute the dot product with $\mathbf{a}$:

$$(8\hat{i} - 4\hat{j}) \cdot (3\hat{i} - \hat{j} + 2\hat{k}) = 8 \times 3 + (-4) \times (-1) = 28$$

Also,

$$\|\mathbf{a}\| = \sqrt{3^2 + (-1)^2 + 2^2} = \sqrt{14}$$

Therefore, the projection is

$$\frac{28}{\sqrt{14}} = 2\sqrt{14}$$

Therefore, the correct option is A.

To find the projection of $\mathbf{c} - 2\hat{j}$ on $\mathbf{a}$, first compute the vectors $\mathbf{b}$ and $\mathbf{c}$ using the given cross products. Then use

$$\text{Proj}_{\mathbf{a}} \mathbf{v} = \frac{\mathbf{a} \cdot \mathbf{v}}{|\mathbf{a}|}$$

Substituting $\mathbf{v} = \mathbf{c} - 2\hat{j}$ gives the required value. From the working,

$$\mathbf{b} = 2\hat{i} + 8\hat{j} + \hat{k}, \quad \mathbf{c} = 8\hat{i} - 2\hat{j}$$

so

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

and hence the projection on $\mathbf{a}$ is

$$\frac{(8\hat{i} - 4\hat{j}) \cdot (3\hat{i} - \hat{j} + 2\hat{k})}{\sqrt{14}} = \frac{28}{\sqrt{14}} = 2\sqrt{14}$$

Thus, the answer is $2\sqrt{14}$.