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Maths Question 15 – JEE-MAIN 2025

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Let the angle θ, 0<θ<π2 between two unit vectors a^ and b^ be sin−1⁡(659). If the vector c→=3a^+6b^+9(a^×b^), then the value of 9(c→⋅a^)−3(c→⋅b^) is

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

Step 1: Determine cos⁡θ and relevant dot products✦ Active

Given sin⁡θ=659 and 0<θ<π2, we find cos⁡θ=1−sin2⁡θ. Since a^ and b^ are unit vectors, |a^|=1 and |b^|=1. The dot products are calculated using their definitions and properties.

cos⁡θ=1−(659)2=1−6581=1681=49 a^⋅a^=1,b^⋅b^=1 a^⋅b^=b^⋅a^=|a^||b^|cos⁡θ=1⋅1⋅49=49 (a^×b^)⋅a^=0,(a^×b^)⋅b^=0
Step 2: Calculate c→⋅a^ and c→⋅b^○ Expand

Substitute the expression for c→ and use the dot product properties determined in Step 1.

c→⋅a^=(3a^+6b^+9(a^×b^))⋅a^=3(a^⋅a^)+6(b^⋅a^)+9((a^×b^)⋅a^) =3(1)+6(49)+9(0)=3+83=173 c→⋅b^=(3a^+6b^+9(a^×b^))⋅b^=3(a^⋅b^)+6(b^⋅b^)+9((a^×b^)⋅b^) =3(49)+6(1)+9(0)=43+6=223
Step 3: Compute the final expression○ Expand

Substitute the calculated values of c→⋅a^ and c→⋅b^ into the required expression.

9(c→⋅a^)−3(c→⋅b^)=9(173)−3(223) =3⋅17−1⋅22=51−22=29
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