JEE Main 2026 Apr 8 Shift 2, Mathematics Q14: Vector (or Cross) Product of Two Vectors

Rewrite: $2(\vec{a}\times \vec{b})-3(\vec{c}\times \vec{b})=\vec{0}⟹(2\vec{a}-3\vec{c})\times \vec{b}=\vec{0}$.

So $2\vec{a}-3\vec{c}=\lambda \vec{b}$, i.e. $\vec{c}=\frac{2\vec{a}-\lambda \vec{b}}{3}$.

Find $\lambda$: Use $\vec{a}\cdot \vec{c}=15$.

$|\vec{a}{|}^2=16+1+9=26$ and $\vec{a}\cdot \vec{b}=40-2-3=35$.

$\frac{2(26)-\lambda (35)}{3}=15⟹52-35\lambda =45⟹\lambda =\frac{7}{35}=\frac{1}{5}$.

Compute the dot product: Let $\vec{d}=\hat{i}+\hat{j}-3\hat{k}$.

$\vec{a}\cdot \vec{d}=4-1-9=-6$ and $\vec{b}\cdot \vec{d}=10+2+3=15$.

$\vec{c}\cdot \vec{d}=\frac{2(-6)-\frac{1}{5}(15)}{3}=\frac{-12-3}{3}=-5$.