Q32 / 75 · JEE Main 2026 (23 January Shift 1) · Mathematics
Let $\overrightarrow{\mathrm{a}}=-\hat{i}+\hat{j}+2 \hat{k}, \overrightarrow{\mathrm{~b}}=\hat{i}-\hat{j}-3 \hat{k}, \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}$ and $\overrightarrow{\mathrm{d}}=\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}}$. Then $(\vec{a}-\vec{b}) \cdot \vec{d}$ is equal to :
- A-4
- B4
- C2
- Correct answerD-2
Solution
$\vec{c} = \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 1 & 2 \\ 1 & -1 & -3 \end{vmatrix} = (-1, -1, 0)$.
$\vec{d} = \vec{c} \times \vec{a} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & -1 & 0 \\ -1 & 1 & 2 \end{vmatrix} = (-2, 2, -2)$.
$\vec{a} - \vec{b} = (-2, 2, 5)$.
$(\vec{a}-\vec{b}) \cdot \vec{d} = (-2)(-2) + (2)(2) + (5)(-2) = 4 + 4 - 10 = -2$.
$\vec{d} = \vec{c} \times \vec{a} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & -1 & 0 \\ -1 & 1 & 2 \end{vmatrix} = (-2, 2, -2)$.
$\vec{a} - \vec{b} = (-2, 2, 5)$.
$(\vec{a}-\vec{b}) \cdot \vec{d} = (-2)(-2) + (2)(2) + (5)(-2) = 4 + 4 - 10 = -2$.
Hence, the correct answer is option: 'D'.