JEE Main 2026 Apr 4 Shift 2, Mathematics Q15: Scalar (or Dot) Product of Two Vectors

$|\hat{u}\times \hat{v}|=\sin \theta =\sqrt{3}/2$, with $\theta$ acute, so $\theta =\pi /3$ and $\hat{u}\cdot \hat{v}=\cos \theta =1/2$.

Dot $\vec{A}$ with $\hat{u}$: $\vec{A}\cdot \hat{u}=\lambda +\hat{u}\cdot \hat{v}+0=\lambda +\frac{1}{2}$, since $\hat{u}\cdot (\hat{u}\times \hat{v})=0$.

So $2\vec{A}\cdot \hat{u}=2\lambda +1$  (*)

Dot $\vec{A}$ with $\hat{v}$: $\vec{A}\cdot \hat{v}=\lambda (\hat{u}\cdot \hat{v})+1+0=\frac{\lambda }{2}+1$.

So $\vec{A}\cdot \hat{v}-1=\frac{\lambda }{2}$  (**)

From (*) and (**): $2\vec{A}\cdot \hat{u}-2\lambda =\vec{A}\cdot \hat{v}-\frac{\lambda }{2}$, giving $\lambda =\frac{4}{3}(\vec{A}\cdot \hat{u})-\frac{2}{3}(\vec{A}\cdot \hat{v})$.