Q40 / 75 · JEE Main 2026 (04 April Shift 2) · Mathematics
Let $\hat{u}$ and $\hat{v}$ be unit vectors inclined at an acute angle such that $|\hat{u}\times\hat{v}|=\dfrac{\sqrt{3}}{2}$. If $\vec{A}=\lambda\hat{u}+\hat{v}+(\hat{u}\times\hat{v})$, then $\lambda$ is equal to:
- Correct answerA$\dfrac{4}{3}(\vec{A}\cdot\hat{u})-\dfrac{2}{3}(\vec{A}\cdot\hat{v})$
- B$\dfrac{2}{3}(\vec{A}\cdot\hat{u})-\dfrac{1}{3}(\vec{A}\cdot\hat{v})$
- C$\dfrac{4}{3}(\vec{A}\cdot\hat{u})+\dfrac{2}{3}(\vec{A}\cdot\hat{v})$
- D$(\vec{A}\cdot\hat{u})-\dfrac{1}{2}(\vec{A}\cdot\hat{v})$
Solution
Given $|\hat{u}\times\hat{v}| = \dfrac{\sqrt{3}}{2}$ and $\hat{u}, \hat{v}$ are unit vectors.
$\sin\theta = \dfrac{\sqrt{3}}{2}$
Since the angle $\theta$ is acute, $\theta = \dfrac{\pi}{3}$.
$\hat{u}\cdot\hat{v} = \cos\dfrac{\pi}{3} = \dfrac{1}{2}$
Given $\vec{A} = \lambda\hat{u} + \hat{v} + (\hat{u}\times\hat{v})$.
Taking the dot product with $\hat{u}$:
$\vec{A}\cdot\hat{u} = \lambda(\hat{u}\cdot\hat{u}) + \hat{v}\cdot\hat{u} + (\hat{u}\times\hat{v})\cdot\hat{u}$
$\vec{A}\cdot\hat{u} = \lambda(1) + \dfrac{1}{2} + 0 = \lambda + \dfrac{1}{2}$
Taking the dot product with $\hat{v}$:
$\vec{A}\cdot\hat{v} = \lambda(\hat{u}\cdot\hat{v}) + \hat{v}\cdot\hat{v} + (\hat{u}\times\hat{v})\cdot\hat{v}$
$\vec{A}\cdot\hat{v} = \lambda\left(\dfrac{1}{2}\right) + 1 + 0 = \dfrac{\lambda}{2} + 1$
From the first equation, we have $1 = 2(\vec{A}\cdot\hat{u}) - 2\lambda$.
Substituting this into the second equation:
$\vec{A}\cdot\hat{v} = \dfrac{\lambda}{2} + 2(\vec{A}\cdot\hat{u}) - 2\lambda$
$\vec{A}\cdot\hat{v} = 2(\vec{A}\cdot\hat{u}) - \dfrac{3\lambda}{2}$
$\dfrac{3\lambda}{2} = 2(\vec{A}\cdot\hat{u}) - (\vec{A}\cdot\hat{v})$
$\lambda = \dfrac{4}{3}(\vec{A}\cdot\hat{u}) - \dfrac{2}{3}(\vec{A}\cdot\hat{v})$
Answer: $\dfrac{4}{3}(\vec{A}\cdot\hat{u})-\dfrac{2}{3}(\vec{A}\cdot\hat{v})$
$\sin\theta = \dfrac{\sqrt{3}}{2}$
Since the angle $\theta$ is acute, $\theta = \dfrac{\pi}{3}$.
$\hat{u}\cdot\hat{v} = \cos\dfrac{\pi}{3} = \dfrac{1}{2}$
Given $\vec{A} = \lambda\hat{u} + \hat{v} + (\hat{u}\times\hat{v})$.
Taking the dot product with $\hat{u}$:
$\vec{A}\cdot\hat{u} = \lambda(\hat{u}\cdot\hat{u}) + \hat{v}\cdot\hat{u} + (\hat{u}\times\hat{v})\cdot\hat{u}$
$\vec{A}\cdot\hat{u} = \lambda(1) + \dfrac{1}{2} + 0 = \lambda + \dfrac{1}{2}$
Taking the dot product with $\hat{v}$:
$\vec{A}\cdot\hat{v} = \lambda(\hat{u}\cdot\hat{v}) + \hat{v}\cdot\hat{v} + (\hat{u}\times\hat{v})\cdot\hat{v}$
$\vec{A}\cdot\hat{v} = \lambda\left(\dfrac{1}{2}\right) + 1 + 0 = \dfrac{\lambda}{2} + 1$
From the first equation, we have $1 = 2(\vec{A}\cdot\hat{u}) - 2\lambda$.
Substituting this into the second equation:
$\vec{A}\cdot\hat{v} = \dfrac{\lambda}{2} + 2(\vec{A}\cdot\hat{u}) - 2\lambda$
$\vec{A}\cdot\hat{v} = 2(\vec{A}\cdot\hat{u}) - \dfrac{3\lambda}{2}$
$\dfrac{3\lambda}{2} = 2(\vec{A}\cdot\hat{u}) - (\vec{A}\cdot\hat{v})$
$\lambda = \dfrac{4}{3}(\vec{A}\cdot\hat{u}) - \dfrac{2}{3}(\vec{A}\cdot\hat{v})$
Answer: $\dfrac{4}{3}(\vec{A}\cdot\hat{u})-\dfrac{2}{3}(\vec{A}\cdot\hat{v})$
Hence, the correct answer is option: 'A'.