JEE Main 2026 Jan 23 Shift 2, Mathematics Q10: Vector (or Cross) Product of Two Vectors

From $\vec{a}\times \vec{b}=2(\vec{a}\times \vec{c})$: $\vec{a}\times (\vec{b}-2\vec{c})=\vec{0}$, so $\vec{b}-2\vec{c}=\lambda \vec{a}$ for some scalar $\lambda$.

Taking magnitudes squared:

${\lambda }^2|\vec{a}{|}^2=|\vec{b}-2\vec{c}{|}^2=|\vec{b}{|}^2+4|\vec{c}{|}^2-4\vec{b}\cdot \vec{c}$

$\vec{b}\cdot \vec{c}=4\cdot 2\cdot \cos 60^{\circ }=4$.

${\lambda }^2=16+16-16=16\Rightarrow \lambda =\pm 4$.

Dotting $\vec{b}-2\vec{c}=\lambda \vec{a}$ with $\vec{c}$:

$\vec{b}\cdot \vec{c}-2|\vec{c}{|}^2=\lambda (\vec{a}\cdot \vec{c})$

$4-8=\pm 4(\vec{a}\cdot \vec{c})\Rightarrow |\vec{a}\cdot \vec{c}|=1$.