JEE Main 2026 · 4 Apr · Shift 2 · Mathematics Q15

← JEE Main 2026 · 4 Apr · Shift 2

MathematicsSingle correct+4 / -1Question 15 of 75

Let $\hat u$ and $\hat v$ be unit vectors inclined at an acute angle such that $\left|\hat u\times\hat v\right|=\dfrac{\sqrt3}2$ . If $\vec A=\lambda\hat u+\hat v+\left(\hat u\times\hat v\right)$ , then $\lambda$ is equal to:

\frac43(\vec A\cdot\hat u)-\frac23(\vec A\cdot\hat v)

\frac23(\vec A\cdot\hat u)-\frac13(\vec A\cdot\hat v)

\frac43(\vec A\cdot\hat u)+\frac23(\vec A\cdot\hat v)

(\vec A\cdot\hat u)-\frac12(\vec A\cdot\hat v)

Answer and worked solution

Official answer: A

Dot products. $\hat u\cdot\hat v=\cos60^\circ=\frac12$ , so $\vec A\cdot\hat u=\lambda+\frac12$ and $\vec A\cdot\hat v=\frac\lambda2+1$ (the cross term drops out).

Eliminate. $\frac43\left(\lambda+\frac12\right)-\frac23\left(\frac\lambda2+1\right)=\lambda$ .

(A)

Topics:Dot product Cross product

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JEE Main 2026 · 4 Apr · Shift 2, Mathematics question 15