Q30 / 75 · JEE Main 2026 (24 January Shift 2) · Mathematics
Let $\vec{a}=2 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}$ and $\vec{b}=\hat{\mathrm{i}}-\hat{\mathrm{j}}+3 \hat{\mathrm{k}}$. If $\vec{c}$ is a vector such that $2(\vec{a} \times \vec{c})+3(\vec{b} \times \vec{c})=\overrightarrow{0}$ and $(\vec{a}-\vec{b}) \cdot \vec{c}=-97$, then $|\vec{c} \times \hat{\mathrm{k}}|^{2}$ is equal to
- A233
- Correct answerB218
- C193
- D205
Solution
$2(\vec{a} \times \vec{c}) + 3(\vec{b} \times \vec{c}) = \vec{0} \Rightarrow (2\vec{a} + 3\vec{b}) \times \vec{c} = \vec{0}$.
So $\vec{c} \parallel (2\vec{a} + 3\vec{b}) = (7, -13, 19)$. Let $\vec{c} = t(7, -13, 19)$.
$\vec{a} - \vec{b} = (1, -4, 2)$.
$(\vec{a} - \vec{b}) \cdot \vec{c} = t(7 + 52 + 38) = 97t = -97 \Rightarrow t = -1$.
$\vec{c} = (-7, 13, -19)$.
$\vec{c} \times \hat{k} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -7 & 13 & -19 \\ 0 & 0 & 1 \end{vmatrix} = (13, 7, 0)$.
$|\vec{c} \times \hat{k}|^2 = 169 + 49 = 218$.
So $\vec{c} \parallel (2\vec{a} + 3\vec{b}) = (7, -13, 19)$. Let $\vec{c} = t(7, -13, 19)$.
$\vec{a} - \vec{b} = (1, -4, 2)$.
$(\vec{a} - \vec{b}) \cdot \vec{c} = t(7 + 52 + 38) = 97t = -97 \Rightarrow t = -1$.
$\vec{c} = (-7, 13, -19)$.
$\vec{c} \times \hat{k} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -7 & 13 & -19 \\ 0 & 0 & 1 \end{vmatrix} = (13, 7, 0)$.
$|\vec{c} \times \hat{k}|^2 = 169 + 49 = 218$.
Hence, the correct answer is option: 'B'.