Let u and v be unit vectors inclined at an acute angle such that | u × v |= 3 2 . If A =λ u + v +( u × v ) , then λ is equal to:

Correct answer: A. (4)/(3)( A · u )-(2)/(3)( A · v ) Given | u × v | = 3 2 and u , v are unit vectors. θ = 3 2 Since the angle θ is acute, θ = (π)/(3) . u · v = (π)/(3) = (1)/(2) Given A = λ u + v + ( u × v ) . Taking the dot product with u : A · u = λ( u · u ) + v · u + ( u × v )· u A · u = λ(1) + (1)/(2) + 0 = λ + (1)/(2) Taking the dot product with v : A · v = λ( u · v ) + v · v + ( u × v )· v A · v = λ((1)/(2)) + 1 + 0 = (λ)/(2) + 1 From the first equation, we have 1 = 2( A · u ) - 2λ . Substituting this into the second equation: A · v = (λ)/(2) + 2( A · u ) - 2λ A · v = 2( A · u ) - (3λ)/(2) (3λ)/(2) = 2( A · u ) - ( A · v ) λ = (4)/(3)( A · u ) - (2)/(3)( A · v ) Answer: (4)/(3)( A · u )-(2)/(3)( A · v )

JEE Main 20264 AprilEvening Shift

Exam: JEE Main
Year: 2026
Date: 4 April
Shift: Evening Shift
Subject: Mathematics
Chapter: Vector Algebra

JEE Main 2026 · 4 April · Evening Shift · Mathematics · Vector Algebra

Let u and v be unit vectors inclined at an acute angle such that |...

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:

Options

A $\dfrac{4}{3}(\vec{A}\cdot\hat{u})-\dfrac{2}{3}(\vec{A}\cdot\hat{v})$
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 $(\vec{A}\cdot\hat{u})-\dfrac{1}{2}(\vec{A}\cdot\hat{v})$

Correct answer

A. $\dfrac{4}{3}(\vec{A}\cdot\hat{u})-\dfrac{2}{3}(\vec{A}\cdot\hat{v})$

Step-by-step 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})

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