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MathMedMain 2026 • 4th Apr - Shift 2

Find Value of Lambda in Terms of Vector Dot Products

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=λu^+v^+(u^×v^)\vec{A} = \lambda \hat{u} + \hat{v} + (\hat{u} \times \hat{v}), then λ\lambda is equal to:

Options

A

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

Correct
B

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

C

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

D

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

Topics & Concepts

Vector AlgebraVector AdditionDot and Cross Product

Step-by-Step Solution

Given that u^\hat{u} and v^\hat{v} are unit vectors, we have: ∣u^∣=1and∣v^∣=1|\hat{u}| = 1 \quad \text{and} \quad |\hat{v}| = 1

Let θ\theta be the acute angle between u^\hat{u} and v^\hat{v}. We are given: ∣u^×v^∣=32|\hat{u} \times \hat{v}| = \frac{\sqrt{3}}{2}

Since ∣u^×v^∣=∣u^∣∣v^∣sin⁡θ=sin⁡θ|\hat{u} \times \hat{v}| = |\hat{u}| |\hat{v}| \sin\theta = \sin\theta, we get: sin⁡θ=32\sin\theta = \frac{\sqrt{3}}{2}

Because θ\theta is an acute angle, we can find cos⁡θ\cos\theta: cos⁡θ=1−sin⁡2θ=1−34=12\cos\theta = \sqrt{1 - \sin^2\theta} = \sqrt{1 - \frac{3}{4}} = \frac{1}{2}

Thus, the dot product of u^\hat{u} and v^\hat{v} is: u^⋅v^=∣u^∣∣v^∣cos⁡θ=1⋅1⋅12=12\hat{u} \cdot \hat{v} = |\hat{u}||\hat{v}|\cos\theta = 1 \cdot 1 \cdot \frac{1}{2} = \frac{1}{2}

We are given the vector expression: A⃗=λu^+v^+(u^×v^)\vec{A} = \lambda \hat{u} + \hat{v} + (\hat{u} \times \hat{v})

Taking the dot product of A⃗\vec{A} with u^\hat{u}: A⃗⋅u^=(λu^+v^+(u^×v^))⋅u^\vec{A} \cdot \hat{u} = (\lambda \hat{u} + \hat{v} + (\hat{u} \times \hat{v})) \cdot \hat{u}

Since (u^×v^)(\hat{u} \times \hat{v}) is perpendicular to u^\hat{u}, we have (u^×v^)⋅u^=0(\hat{u} \times \hat{v}) \cdot \hat{u} = 0. Thus: A⃗⋅u^=λ(u^⋅u^)+(v^⋅u^)=λ(1)+12=λ+12— (1)\vec{A} \cdot \hat{u} = \lambda(\hat{u} \cdot \hat{u}) + (\hat{v} \cdot \hat{u}) = \lambda(1) + \frac{1}{2} = \lambda + \frac{1}{2} \quad \text{--- (1)}

Next, taking the dot product of A⃗\vec{A} with v^\hat{v}: A⃗⋅v^=(λu^+v^+(u^×v^))⋅v^\vec{A} \cdot \hat{v} = (\lambda \hat{u} + \hat{v} + (\hat{u} \times \hat{v})) \cdot \hat{v}

Since (u^×v^)(\hat{u} \times \hat{v}) is perpendicular to v^\hat{v}, we have (u^×v^)⋅v^=0(\hat{u} \times \hat{v}) \cdot \hat{v} = 0. Thus: A⃗⋅v^=λ(u^⋅v^)+(v^⋅v^)=λ(12)+1=λ2+1— (2)\vec{A} \cdot \hat{v} = \lambda(\hat{u} \cdot \hat{v}) + (\hat{v} \cdot \hat{v}) = \lambda\left(\frac{1}{2}\right) + 1 = \frac{\lambda}{2} + 1 \quad \text{--- (2)}

Now, we want to express λ\lambda as a linear combination of (A⃗⋅u^)(\vec{A} \cdot \hat{u}) and (A⃗⋅v^)(\vec{A} \cdot \hat{v}): λ=c1(A⃗⋅u^)+c2(A⃗⋅v^)\lambda = c_1(\vec{A} \cdot \hat{u}) + c_2(\vec{A} \cdot \hat{v})

Substituting equations (1) and (2) into this expression: λ=c1(λ+12)+c2(λ2+1)\lambda = c_1\left(\lambda + \frac{1}{2}\right) + c_2\left(\frac{\lambda}{2} + 1\right) λ=(c1+c22)λ+(c12+c2)\lambda = \left(c_1 + \frac{c_2}{2}\right)\lambda + \left(\frac{c_1}{2} + c_2\right)

Equating the coefficients of λ\lambda and the constant terms on both sides:

  1. c1+c22=1c_1 + \frac{c_2}{2} = 1
  2. c12+c2=0  ⟹  c2=−c12\frac{c_1}{2} + c_2 = 0 \implies c_2 = -\frac{c_1}{2}

Substituting c2=−c12c_2 = -\frac{c_1}{2} into the first equation: c1+12(−c12)=1  ⟹  34c1=1  ⟹  c1=43c_1 + \frac{1}{2}\left(-\frac{c_1}{2}\right) = 1 \implies \frac{3}{4}c_1 = 1 \implies c_1 = \frac{4}{3}

Then: c2=−12(43)=−23c_2 = -\frac{1}{2}\left(\frac{4}{3}\right) = -\frac{2}{3}

Therefore, we have: λ=43(A⃗⋅u^)−23(A⃗⋅v^)\lambda = \frac{4}{3}(\vec{A} \cdot \hat{u}) - \frac{2}{3}(\vec{A} \cdot \hat{v})

This matches Option A.

Find Value of Lambda in Terms of Vector Dot Products | Mathematics PYQ Solution - JEE Challenger