Q43 / 75 · JEE Main 2026 (21 January Shift 1) · Mathematics
Let $\overrightarrow{\mathrm{a}}=-\hat{i}+2 \hat{j}+2 \hat{k}, \overrightarrow{\mathrm{~b}}=8 \hat{i}+7 \hat{j}-3 \hat{k}$ and $\overrightarrow{\mathrm{c}}$ be vector such that $\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{b}}$. If $\vec{c} \cdot(\hat{i}+\hat{j}+\hat{k})=4$, then $|\vec{a}+\vec{c}|^{2}$ is equal to :
- Correct answerA27
- B33
- C35
- D30
Solution
Let $\vec{c} = x\hat{i} + y\hat{j} + z\hat{k}$.
From $\vec{a} \times \vec{c} = \vec{b}$: $z - y = 4$, $z + 2x = 7$, $y + 2x = 3$.
From $\vec{c} \cdot (\hat{i} + \hat{j} + \hat{k}) = 4$: $x + y + z = 4$.
Solving: $y = -1$, $z = 3$, $x = 2$.
$\vec{c} = 2\hat{i} - \hat{j} + 3\hat{k}$.
$\vec{a} + \vec{c} = (-1+2)\hat{i} + (2-1)\hat{j} + (2+3)\hat{k} = \hat{i} + \hat{j} + 5\hat{k}$.
$|\vec{a} + \vec{c}|^2 = 1 + 1 + 25 = 27$.
From $\vec{a} \times \vec{c} = \vec{b}$: $z - y = 4$, $z + 2x = 7$, $y + 2x = 3$.
From $\vec{c} \cdot (\hat{i} + \hat{j} + \hat{k}) = 4$: $x + y + z = 4$.
Solving: $y = -1$, $z = 3$, $x = 2$.
$\vec{c} = 2\hat{i} - \hat{j} + 3\hat{k}$.
$\vec{a} + \vec{c} = (-1+2)\hat{i} + (2-1)\hat{j} + (2+3)\hat{k} = \hat{i} + \hat{j} + 5\hat{k}$.
$|\vec{a} + \vec{c}|^2 = 1 + 1 + 25 = 27$.
Hence, the correct answer is option: 'A'.