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Mathematics · Vector Algebra

JEE Main 2026 — 4 April, Evening Shift — Question 39

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}| = \frac{\sqrt{3}}{2}. If A⃗=λ(u^+v^)+(u^×v^)\vec{A} = \lambda(\hat{u}+\hat{v}) + (\hat{u}\times\hat{v}). Then λ\lambda is equal to:

  1. Option A:

    43A⃗⋅u^−23A⃗⋅v^\frac{4}{3}\vec{A}\cdot\hat{u} - \frac{2}{3}\vec{A}\cdot\hat{v}

    Correct
  2. Option B:

    23A⃗⋅u^−13A⃗⋅v^\frac{2}{3}\vec{A}\cdot\hat{u} - \frac{1}{3}\vec{A}\cdot\hat{v}

  3. Option C:

    43A⃗⋅u^+23A⃗⋅v^\frac{4}{3}\vec{A}\cdot\hat{u} + \frac{2}{3}\vec{A}\cdot\hat{v}

  4. Option D:

    12A⃗⋅u^−A⃗⋅v^\frac{1}{2}\vec{A}\cdot\hat{u} - \vec{A}\cdot\hat{v}

Answer: A

Step-by-step solution

∣u^×v^∣=32|\hat{\mathrm{u}} \times \hat{\mathrm{v}}|=\frac{\sqrt{3}}{2} ∣u^∣∣v^∣sin⁡θ=32|\hat{\mathrm{u}}||\hat{\mathrm{v}}| \sin \theta=\frac{\sqrt{3}}{2} ∴θ=π3\therefore \theta=\frac{\pi}{3} and u^⋅v^=∣u^∣∣v^∣cos⁡π3=12\hat{\mathrm{u}} \cdot \hat{\mathrm{v}}=|\hat{\mathrm{u}}||\hat{\mathrm{v}}| \cos \frac{\pi}{3}=\frac{1}{2}

\overrightarrow{\mathrm{A}}=\lambda \hat{\mathrm{u}}+\hat{\mathrm{v}}+\hat{\mathrm{u}} \times \hat{\mathrm{v}} \end{gathered}$$ Dot with û $\overrightarrow{\mathrm{A}} \cdot \hat{\mathrm{u}}=\lambda(1)+\hat{\mathrm{u}} . \hat{\mathrm{v}}+\hat{\mathrm{u}} \cdot(\hat{\mathrm{u}} \times \hat{\mathrm{v}})$ $\overrightarrow{\mathrm{A}} \cdot \hat{\mathrm{u}}=\lambda+\frac{1}{2}$ $$\begin{gathered} \Rightarrow 2 \overrightarrow{\mathrm{~A}} \cdot \hat{\mathrm{u}}=2 \lambda+1 \end{gathered}$$ Dot equation (1) with $\hat{\mathrm{v}}$ $\overrightarrow{\mathrm{A}} \cdot \hat{\mathrm{v}}=\lambda(\hat{\mathrm{u}} \cdot \hat{\mathrm{v}})+\hat{\mathrm{v}} \cdot \hat{\mathrm{v}}+\hat{\mathrm{v}} \cdot(\hat{\mathrm{u}} \times \hat{\mathrm{v}})$ $\overrightarrow{\mathrm{A}} \cdot \hat{\mathrm{v}}=\frac{\lambda}{2}+1$ $$\begin{gathered} \vec{A} \cdot \hat{v}-\frac{\lambda}{2}=1 \end{gathered}$$ From (2) and (3) $2 \overrightarrow{\mathrm{~A}} \cdot \hat{\mathrm{u}}-2 \lambda=\overrightarrow{\mathrm{A}} \cdot \hat{\mathrm{v}}-\frac{\lambda}{2}$ $\therefore \lambda=\frac{4}{3} \overrightarrow{\mathrm{~A}} \cdot \hat{\mathrm{u}}-\frac{2}{3} \overrightarrow{\mathrm{~A}} \cdot \hat{\mathrm{v}}$

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
4 April, Evening Shift
Subject
Mathematics
Chapter
Vector Algebra
Topic
Scalar or Dot Product of Two Vectors
Let hat u and hat v be unit vectors inclined at an acute angle such… | JEE Main 2026 PYQ with Solution · DhiX AI