[shahkb4]

shahkb4

3 June, 2026 at 8:15 pm #1038

Solution

$\overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{a}} \times \hat{\mathrm{b}}=\left|\begin{array}{ccc}\hat{\lambda} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 2 & -1 & 1 \\ 0 & \lambda & 2\end{array}\right|=(-2-\lambda) \hat{\mathrm{i}}-4 \hat{\mathrm{j}}+2 \lambda \hat{\mathrm{k}}$
$|\overrightarrow{\mathrm{c}}|=\sqrt{53}=\sqrt{(-2-\lambda)^{2}+16+4 \lambda^{2}}$
$\Rightarrow 53=4+\lambda^{2}+4 \lambda+16+4 \lambda^{2}$
$\Rightarrow 5 \lambda^{2}+4 \lambda-33=0 \Rightarrow(5 \lambda-11)(\lambda+3)=0$
$\lambda=11 / 5$ (reject) $\lambda=-3$ (accept); $\lambda \in \mathrm{Z}$
$\overrightarrow{\mathrm{c}}=\hat{\mathrm{i}}-4 \hat{\mathrm{j}}-6 \hat{\mathrm{k}}$
let $\vec{d}=x \hat{i}+y \hat{j}+z \hat{k}$
$\overrightarrow{\mathrm{d}}$ lies in yz plane, $\mathrm{x}=0$
$\overrightarrow{\mathrm{d}}=y \hat{\mathrm{j}}+z \hat{\mathrm{k}}$
$|\overrightarrow{\mathrm{d}}|=2=\sqrt{\mathrm{y}^{2}+\mathrm{z}^{2}} \Rightarrow \mathrm{y}^{2}+\mathrm{z}^{2}=4$
$\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{d}}=|\overrightarrow{\mathrm{c}} \| \overrightarrow{\mathrm{d}}| \cos \theta$
$\Rightarrow(\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{d}})^{2}=|\overrightarrow{\mathrm{c}}|^{2}|\overrightarrow{\mathrm{~d}}|^{2} \cos ^{2} \theta \leq|\overrightarrow{\mathrm{c}}|^{2}|\overrightarrow{\mathrm{~d}}|^{2}$
$\Rightarrow(\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{d}})^{2} \leq(53)(4)=212$
Clearly $\overrightarrow{\mathrm{c}}$ \& $\overrightarrow{\mathrm{d}}$ are not parallel
It means $(\overrightarrow{\mathrm{c}} . \overrightarrow{\mathrm{d}})^{2}<212$
$\vec{c}=\hat{i}-4 \hat{j}-6 \hat{k} ; \vec{d}=y \hat{j}+z \hat{k}$
$\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{d}}=-4 \mathrm{y}-6 \mathrm{z}$
$\Rightarrow(4 y+6 z)^{2} \leq\left(\sqrt{4^{2}+6^{2}} \sqrt{y^{2}+z^{2}}\right)^{2}$
$\Rightarrow(\overrightarrow{\mathrm{c}} . \overrightarrow{\mathrm{d}})^{2} \leq(52)(4)=208$ Answer(3)