Q41 / 75 · JEE Main 2026 (22 January Shift 2) · Mathematics
Let $\vec{a}=2 \hat{i}-\hat{j}+\hat{k}$ and $\vec{b}=\lambda \hat{j}+2 \hat{k}, \lambda \in \boldsymbol{Z}$ be two vectors. Let $\vec{c}=\vec{a} \times \vec{b}$ and $\vec{d}$ be a vector of magnitude 2 in $y z$-plane. If $|\overrightarrow{\mathrm{c}}|=\sqrt{53}$, then the maximum possible value of $(\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{d}})^{2}$ is equal to :
- A26
- B52
- Correct answerC208
- D104
Solution
$\vec{a} = 2\hat{i} - \hat{j} + \hat{k}$
$\vec{b} = \lambda\hat{j} + 2\hat{k}$ ; $\lambda \in Z$
$\vec{c} = \vec{a} \times \vec{b} = (-2 - \lambda)\hat{i} - 4\hat{j} + 2\lambda\hat{k}$
$|\vec{c}| = \sqrt{53}$
$\Rightarrow 5\lambda^2 + 4\lambda - 33 = 0$
$\lambda = 2.2$ or $-3$
$\Rightarrow \boxed{\lambda = -3}$
$\vec{c} = \hat{i} - 4\hat{j} - 6\hat{k}$
let $\vec{d} = y\hat{j} + z\hat{k}$
$|\vec{d}| = 2$
$\Rightarrow y^2 + z^2 = 4$
$(\vec{c} \cdot \vec{d})^2 = (4y + 6z)^2 \leq \left(\sqrt{4^2 + 6^2} \times \sqrt{y^2 + z^2}\right)^2 \leq 208$
$\vec{b} = \lambda\hat{j} + 2\hat{k}$ ; $\lambda \in Z$
$\vec{c} = \vec{a} \times \vec{b} = (-2 - \lambda)\hat{i} - 4\hat{j} + 2\lambda\hat{k}$
$|\vec{c}| = \sqrt{53}$
$\Rightarrow 5\lambda^2 + 4\lambda - 33 = 0$
$\lambda = 2.2$ or $-3$
$\Rightarrow \boxed{\lambda = -3}$
$\vec{c} = \hat{i} - 4\hat{j} - 6\hat{k}$
let $\vec{d} = y\hat{j} + z\hat{k}$
$|\vec{d}| = 2$
$\Rightarrow y^2 + z^2 = 4$
$(\vec{c} \cdot \vec{d})^2 = (4y + 6z)^2 \leq \left(\sqrt{4^2 + 6^2} \times \sqrt{y^2 + z^2}\right)^2 \leq 208$
Hence, the correct answer is option: 'C'.