Vectors & 3D GeometrymediumPYQ · JEE Main · 4 Apr 2026 · Shift 2 (Afternoon)Free

Express λ via Dot Products of A with û, v̂ | JEE Main 2026

JEE Maths question with a full step-by-step solution.

Question
Let u^\hat u and v^\hat v be unit vectors inclined at an acute angle such that ∣u^×v^∣=32|\hat u\times\hat v|=\dfrac{\sqrt3}{2}. If A⃗=λu^+v^+(u^×v^)\vec A=\lambda\hat u+\hat v+(\hat u\times\hat v), then λ\lambda is equal to
A43(A⃗⋅u^)−23(A⃗⋅v^)\dfrac43(\vec A\cdot\hat u)-\dfrac23(\vec A\cdot\hat v)correct
B23(A⃗⋅u^)−13(A⃗⋅v^)\dfrac23(\vec A\cdot\hat u)-\dfrac13(\vec A\cdot\hat v)
C43(A⃗⋅u^)+23(A⃗⋅v^)\dfrac43(\vec A\cdot\hat u)+\dfrac23(\vec A\cdot\hat v)
D(A⃗⋅u^)−12(A⃗⋅v^)(\vec A\cdot\hat u)-\dfrac12(\vec A\cdot\hat v)
Solution
Step 1: ∣u^×v^∣=sin⁡θ=32⇒θ=π3|\hat u\times\hat v|=\sin\theta=\dfrac{\sqrt3}{2}\Rightarrow\theta=\dfrac{\pi}{3}, so u^⋅v^=cos⁡π3=12\hat u\cdot\hat v=\cos\dfrac{\pi}{3}=\dfrac12. Step 2: Dot A⃗=λu^+v^+(u^×v^)\vec A=\lambda\hat u+\hat v+(\hat u\times\hat v) with u^\hat u (note u^⋅(u^×v^)=0\hat u\cdot(\hat u\times\hat v)=0):
A⃗⋅u^=λ+12 ⇒ 2A⃗⋅u^=2λ+1.(2)\vec A\cdot\hat u=\lambda+\frac12\ \Rightarrow\ 2\vec A\cdot\hat u=2\lambda+1.\quad(2)
 Step 3: Dot with v^\hat v:
A⃗⋅v^=λ⋅12+1 ⇒ A⃗⋅v^−λ2=1.(3)\vec A\cdot\hat v=\lambda\cdot\frac12+1\ \Rightarrow\ \vec A\cdot\hat v-\frac{\lambda}{2}=1.\quad(3)
 Step 4: Eliminate the constant: from (2), 2A⃗⋅u^−2λ=12\vec A\cdot\hat u-2\lambda=1; from (3), A⃗⋅v^−λ2=1\vec A\cdot\hat v-\dfrac{\lambda}{2}=1. Setting equal and solving,
λ=43(A⃗⋅u^)−23(A⃗⋅v^).\lambda=\frac43(\vec A\cdot\hat u)-\frac23(\vec A\cdot\hat v).
 Correct answer: (1)
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