JEE Main 2026 — Mathematics Vectors & 3D Geometry

Given $|\hat{u}\times\hat{v}| = \dfrac{\sqrt{3}}{2}$∣u^×v^∣=23​​ and $\hat{u}, \hat{v}$u^,v^ are unit vectors.

$$\sin\theta = \dfrac{\sqrt{3}}{2}$$sinθ=23​​

Since the angle $\theta$θ is acute, $\theta = \dfrac{\pi}{3}$θ=3π​.

$$\hat{u}\cdot\hat{v} = \cos\dfrac{\pi}{3} = \dfrac{1}{2}$$u^⋅v^=cos3π​=21​

Given $\vec{A} = \lambda\hat{u} + \hat{v} + (\hat{u}\times\hat{v})$A=λu^+v^+(u^×v^).

Taking the dot product with $\hat{u}$u^:

$$\vec{A}\cdot\hat{u} = \lambda(\hat{u}\cdot\hat{u}) + \hat{v}\cdot\hat{u} + (\hat{u}\times\hat{v})\cdot\hat{u}$$A⋅u^=λ(u^⋅u^)+v^⋅u^+(u^×v^)⋅u^

$$\vec{A}\cdot\hat{u} = \lambda(1) + \dfrac{1}{2} + 0 = \lambda + \dfrac{1}{2}$$A⋅u^=λ(1)+21​+0=λ+21​

Taking the dot product with $\hat{v}$v^:

$$\vec{A}\cdot\hat{v} = \lambda(\hat{u}\cdot\hat{v}) + \hat{v}\cdot\hat{v} + (\hat{u}\times\hat{v})\cdot\hat{v}$$A⋅v^=λ(u^⋅v^)+v^⋅v^+(u^×v^)⋅v^

$$\vec{A}\cdot\hat{v} = \lambda\left(\dfrac{1}{2}\right) + 1 + 0 = \dfrac{\lambda}{2} + 1$$A⋅v^=λ(21​)+1+0=2λ​+1

From the first equation, we have $1 = 2(\vec{A}\cdot\hat{u}) - 2\lambda$1=2(A⋅u^)−2λ.

Substituting this into the second equation:

$$\vec{A}\cdot\hat{v} = \dfrac{\lambda}{2} + 2(\vec{A}\cdot\hat{u}) - 2\lambda$$A⋅v^=2λ​+2(A⋅u^)−2λ

$$\vec{A}\cdot\hat{v} = 2(\vec{A}\cdot\hat{u}) - \dfrac{3\lambda}{2}$$A⋅v^=2(A⋅u^)−23λ​

$$\dfrac{3\lambda}{2} = 2(\vec{A}\cdot\hat{u}) - (\vec{A}\cdot\hat{v})$$23λ​=2(A⋅u^)−(A⋅v^)

$$\lambda = \dfrac{4}{3}(\vec{A}\cdot\hat{u}) - \dfrac{2}{3}(\vec{A}\cdot\hat{v})$$λ=34​(A⋅u^)−32​(A⋅v^)

Answer: $\dfrac{4}{3}(\vec{A}\cdot\hat{u})-\dfrac{2}{3}(\vec{A}\cdot\hat{v})$34​(A⋅u^)−32​(A⋅v^)