Reply To: Question 9 — JEE Main 23 Jan 2026 S Ii

Solution

$\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}=2 \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}} \Rightarrow \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}} \times(2 \overrightarrow{\mathrm{c}})=0$
$\Rightarrow \overrightarrow{\mathrm{a}} \times(\overrightarrow{\mathrm{b}}-2 \overrightarrow{\mathrm{c}})=0$
$\vec{a} \& \vec{b}-2 \vec{c}$ are parallel
$\overrightarrow{\mathrm{b}}-2 \overrightarrow{\mathrm{c}}=\lambda \overrightarrow{\mathrm{a}}$

Square $(\vec{b}-2 \vec{c})^{2}=(\lambda \vec{a})^{2}$
$\Rightarrow|\overrightarrow{\mathrm{b}}|^{2}+4|\overrightarrow{\mathrm{c}}|^{2}-4 \overrightarrow{\mathrm{~b}} \cdot \overrightarrow{\mathrm{c}}=\lambda^{2}|\overrightarrow{\mathrm{a}}|^{2}$
$\Rightarrow 16+4(4)-4|\vec{b} \| \vec{c}| \cos 60^{\circ}=\lambda^{2}(1)$
$\Rightarrow 32-4(4)(2)\left(\frac{1}{2}\right)=\lambda^{2} \Rightarrow \lambda^{2}=16 \Rightarrow \lambda= \pm 4$
$\because \overrightarrow{\mathrm{b}}-2 \overrightarrow{\mathrm{c}}=\lambda \overrightarrow{\mathrm{a}}$ dot with $\overrightarrow{\mathrm{c}}$
$\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}-2 \overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{c}}=\lambda \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}$
$\Rightarrow|\overrightarrow{\mathrm{b}} \| \overrightarrow{\mathrm{c}}| \cos 60^{\circ}-2|\overrightarrow{\mathrm{c}}|^{2}=\lambda(\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}})$
$\Rightarrow 4(2)\left(\frac{1}{2}\right)-2(4)=\lambda \overrightarrow{\mathrm{a}} . \overrightarrow{\mathrm{c}} \Rightarrow \lambda(\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}})=-4$
$\Rightarrow|\lambda||\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}|=4 \Rightarrow|\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}|=1$ Answer(