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Maths Question 15 – JEE-MAIN 2026

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Let u^ and v^ be unit vectors inclined at an acute angle such that |u^×v^|=32. If A→=λu^+v^+(u^×v^), then λ is equal to:

Understand the properties of unit vectors, dot products, and cross products, especially how they relate to the angle between vectors.

Step 1: Determine the angle between the unit vectors✦ Active

Given that u^ and v^ are unit vectors, we have |u^|=1 and |v^|=1. The magnitude of their cross product is given as |u^×v^|=32. Using the formula |u^×v^|=|u^||v^|sin⁡θ, we get:

1⋅1⋅sin⁡θ=32⟹sin⁡θ=32

Since θ is an acute angle, θ=π3 (60∘). Therefore, the dot product u^⋅v^ is:

u^⋅v^=|u^||v^|cos⁡θ=1⋅1⋅cos⁡(π3)=12
Step 2: Express A→⋅u^ and A→⋅v^ in terms of λ○ Expand

Given A→=λu^+v^+(u^×v^). Let's calculate the dot products with u^ and v^:

For A→⋅u^:

A→⋅u^=(λu^+v^+(u^×v^))⋅u^=λ(u^⋅u^)+(v^⋅u^)+((u^×v^)⋅u^)

Since u^⋅u^=|u^|2=1, v^⋅u^=12, and ((u^×v^)⋅u^)=0 (as u^×v^ is orthogonal to u^), we get:

A→⋅u^=λ(1)+12+0=λ+12

For A→⋅v^:

A→⋅v^=(λu^+v^+(u^×v^))⋅v^=λ(u^⋅v^)+(v^⋅v^)+((u^×v^)⋅v^)

Since u^⋅v^=12, v^⋅v^=|v^|2=1, and ((u^×v^)⋅v^)=0 (as u^×v^ is orthogonal to v^), we get:

A→⋅v^=λ(12)+1+0=λ2+1
💡 Teacher's Secret Hint

Remember that the scalar triple product of three vectors is zero if any two vectors are parallel or if the vectors are coplanar.

Step 3: Substitute expressions into options to find λ○ Expand

We have (A→⋅u^)=λ+12 and (A→⋅v^)=λ2+1. Let's test Option 1:

43(A→⋅u^)−23(A→⋅v^)=43(λ+12)−23(λ2+1)

Expand and simplify:

=43λ+46−26λ−23
=43λ+23−13λ−23
=(43−13)λ+(23−23)=33λ+0=λ

Since Option 1 simplifies to λ, it is the correct answer.

💡 Teacher's Secret Hint

Carefully perform the algebraic simplification after substitution. Only one option will yield λ.

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