Q36 / 75 · JEE Main 2026 (24 January Shift 1) · Mathematics
Let $\vec{a}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}}, \vec{b}=\hat{\mathrm{i}}+\hat{\mathrm{j}}$ and $\vec{c}=\vec{a} \times \vec{b}$. Let $\vec{d}$ be a vector such that $|\vec{d}-\vec{a}|=\sqrt{11},|\vec{c} \times \vec{d}|=3$ and the angle between $\vec{c}$ and $\vec{d}$ is $\frac{\pi}{4}$. Then $\vec{a} \cdot \vec{d}$ is equal to
- A3
- B11
- Correct answerC0
- D1
Solution
$\vec{c} = \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -2 \\ 1 & 1 & 0 \end{vmatrix} = 2\hat{i} - 2\hat{j} + \hat{k}$
$|\vec{c}| = \sqrt{4+4+1} = 3$, $|\vec{a}| = \sqrt{4+1+4} = 3$
From $|\vec{c} \times \vec{d}| = 3$ and $\cos(\angle(\vec{c}, \vec{d})) = \frac{1}{\sqrt{2}}$:
$|\vec{c}||\vec{d}|\sin\frac{\pi}{4} = 3$ → $3|\vec{d}| \cdot \frac{1}{\sqrt{2}} = 3$ → $|\vec{d}| = \sqrt{2}$
From $|\vec{d} - \vec{a}|^2 = 11$:
$|\vec{d}|^2 - 2\vec{a} \cdot \vec{d} + |\vec{a}|^2 = 11$
$2 - 2\vec{a} \cdot \vec{d} + 9 = 11$
$\vec{a} \cdot \vec{d} = 0$
$|\vec{c}| = \sqrt{4+4+1} = 3$, $|\vec{a}| = \sqrt{4+1+4} = 3$
From $|\vec{c} \times \vec{d}| = 3$ and $\cos(\angle(\vec{c}, \vec{d})) = \frac{1}{\sqrt{2}}$:
$|\vec{c}||\vec{d}|\sin\frac{\pi}{4} = 3$ → $3|\vec{d}| \cdot \frac{1}{\sqrt{2}} = 3$ → $|\vec{d}| = \sqrt{2}$
From $|\vec{d} - \vec{a}|^2 = 11$:
$|\vec{d}|^2 - 2\vec{a} \cdot \vec{d} + |\vec{a}|^2 = 11$
$2 - 2\vec{a} \cdot \vec{d} + 9 = 11$
$\vec{a} \cdot \vec{d} = 0$
Hence, the correct answer is option: 'C'.