JEE Main 2026 — 24 January, Evening Shift — Question 4

Given: $\vec{a} \times \vec{b} = 2(\vec{a} \times \vec{c})$ $\Rightarrow \vec{a} \times (\vec{b} - 2\vec{c}) = 0$ $\Rightarrow \vec{b} - 2\vec{c}$ is parallel to $\vec{a}$, so $\vec{b} - 2\vec{c} = \lambda \vec{a}$ for some scalar $\lambda$ Take magnitude squared: $|\vec{b} - 2\vec{c}|^2 = \lambda^2 |\vec{a}|^2$ $\Rightarrow |\vec{b}|^2 + 4|\vec{c}|^2 - 4\vec{b}\cdot\vec{c} = \lambda^2$ Given $|\vec{b}|=4, |\vec{c}|=2, \vec{b}\cdot\vec{c}=4\cdot2\cdot\cos60^\circ=4$ $\Rightarrow 16 + 16 - 16 = \lambda^2 \Rightarrow \lambda^2 = 16 \Rightarrow \lambda = \pm 4$ Thus $\vec{b} - 2\vec{c} = \pm 4\vec{a}$ Dot both sides with $\vec{c}$: $\vec{b}\cdot\vec{c} - 2|\vec{c}|^2 = \pm 4(\vec{a}\cdot\vec{c})$ $\Rightarrow 4 - 8 = \pm 4(\vec{a}\cdot\vec{c}) \Rightarrow -4 = \pm 4(\vec{a}\cdot\vec{c})$ $\Rightarrow \vec{a}\cdot\vec{c} = \mp 1$ Hence $|\vec{a}\cdot\vec{c}| = 1$