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JEE Mathematics 2025 Question with Solution

Let the angle θ, 0<θ<π2\theta, \ 0 < \theta < \frac{\pi}{2} between two unit vectors a^\hat{a} and b^\hat{b} be sin⁡−1(659)\sin^{-1} \left( \frac{\sqrt{65}}{9} \right). If the vector c⃗=3a^+6b^+9(a^×b^)\vec{c} = 3\hat{a} + 6\hat{b} + 9(\hat{a} \times \hat{b}), then the value of 9(c⃗⋅a^)−3(c⃗⋅b^)9(\vec{c} \cdot \hat{a}) - 3(\vec{c} \cdot \hat{b}) is:

  • A

    3131

  • B

    2727

  • C

    2929

  • D

    2424

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: The angle between unit vectors a^\hat{a} and b^\hat{b} is θ=sin⁡−1(659)\theta = \sin^{-1}\left(\frac{\sqrt{65}}{9}\right) and c⃗=3a^+6b^+9(a^×b^)\vec{c} = 3\hat{a} + 6\hat{b} + 9(\hat{a} \times \hat{b}).

Find: The value of 9(c⃗⋅a^)−3(c⃗⋅b^)9(\vec{c} \cdot \hat{a}) - 3(\vec{c} \cdot \hat{b}).

First determine cos⁡θ\cos\theta.

sin⁡θ=659\sin\theta = \frac{\sqrt{65}}{9}

Using sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1,

cos⁡θ=1−(659)2=1−6581=1681=49\cos\theta = \sqrt{1 - \left(\frac{\sqrt{65}}{9}\right)^2} = \sqrt{1 - \frac{65}{81}} = \sqrt{\frac{16}{81}} = \frac{4}{9}

So,

a^⋅b^=cos⁡θ=49\hat{a} \cdot \hat{b} = \cos\theta = \frac{4}{9}

Now use the perpendicularity of the cross product:

a^⋅(a^×b^)=0\hat{a} \cdot (\hat{a} \times \hat{b}) = 0 b^⋅(a^×b^)=0\hat{b} \cdot (\hat{a} \times \hat{b}) = 0

Compute c⃗⋅a^\vec{c} \cdot \hat{a}:

c⃗⋅a^=(3a^+6b^+9(a^×b^))⋅a^\vec{c} \cdot \hat{a} = (3\hat{a} + 6\hat{b} + 9(\hat{a} \times \hat{b})) \cdot \hat{a} =3+6(b^⋅a^)+0=3+6×49=3+249=173= 3 + 6(\hat{b} \cdot \hat{a}) + 0 = 3 + 6 \times \frac{4}{9} = 3 + \frac{24}{9} = \frac{17}{3}

Compute c⃗⋅b^\vec{c} \cdot \hat{b}:

c⃗⋅b^=(3a^+6b^+9(a^×b^))⋅b^\vec{c} \cdot \hat{b} = (3\hat{a} + 6\hat{b} + 9(\hat{a} \times \hat{b})) \cdot \hat{b} =3(a^⋅b^)+6+0=3×49+6=129+6=223= 3(\hat{a} \cdot \hat{b}) + 6 + 0 = 3 \times \frac{4}{9} + 6 = \frac{12}{9} + 6 = \frac{22}{3}

Substitute into the required expression:

9(c⃗⋅a^)−3(c⃗⋅b^)=9×173−3×223=51−22=299(\vec{c} \cdot \hat{a}) - 3(\vec{c} \cdot \hat{b}) = 9 \times \frac{17}{3} - 3 \times \frac{22}{3} = 51 - 22 = 29

Therefore, the correct option is C.

This matches option (3)\text{(3)}, whose value is 2929.

Direct Simplification

Given: c⃗=3a^+6b^+9(a^×b^)\vec{c} = 3\hat{a} + 6\hat{b} + 9(\hat{a} \times \hat{b}) and sin⁡θ=659\sin\theta = \frac{\sqrt{65}}{9}.

Find: The value of 9(c⃗⋅a^)−3(c⃗⋅b^)9(\vec{c} \cdot \hat{a}) - 3(\vec{c} \cdot \hat{b}).

Use the dot products directly:

c⃗⋅a^=3+6(a^⋅b^)\vec{c} \cdot \hat{a} = 3 + 6(\hat{a} \cdot \hat{b}) c⃗⋅b^=3(a^⋅b^)+6\vec{c} \cdot \hat{b} = 3(\hat{a} \cdot \hat{b}) + 6

because the cross product term is perpendicular to both a^\hat{a} and b^\hat{b}.

Then,

9(c⃗⋅a^)−3(c⃗⋅b^)=9(3+6a^⋅b^)−3(3a^⋅b^+6)9(\vec{c} \cdot \hat{a}) - 3(\vec{c} \cdot \hat{b}) = 9\big(3 + 6\hat{a} \cdot \hat{b}\big) - 3\big(3\hat{a} \cdot \hat{b} + 6\big) =27+54a^⋅b^−9a^⋅b^−18=9+45a^⋅b^= 27 + 54\hat{a} \cdot \hat{b} - 9\hat{a} \cdot \hat{b} - 18 = 9 + 45\hat{a} \cdot \hat{b}

Now,

a^⋅b^=cos⁡θ=81−659=169=49\hat{a} \cdot \hat{b} = \cos\theta = \frac{\sqrt{81 - 65}}{9} = \frac{\sqrt{16}}{9} = \frac{4}{9}

So,

9+45(49)=9+20=299 + 45\left(\frac{4}{9}\right) = 9 + 20 = 29

Therefore, the correct option is C.

Common mistakes

  • Using a^⋅b^=sin⁡θ\hat{a} \cdot \hat{b} = \sin\theta instead of a^⋅b^=cos⁡θ\hat{a} \cdot \hat{b} = \cos\theta is incorrect because the dot product of two vectors gives the cosine of the angle between them. First find cos⁡θ\cos\theta from the given sine value.

  • Treating (a^×b^)⋅a^(\hat{a} \times \hat{b}) \cdot \hat{a} or (a^×b^)⋅b^(\hat{a} \times \hat{b}) \cdot \hat{b} as non-zero is wrong because the cross product is perpendicular to both vectors. These dot products must be taken as 00.

  • Forgetting that a^\hat{a} and b^\hat{b} are unit vectors leads to missing the terms a^⋅a^=1\hat{a} \cdot \hat{a} = 1 and b^⋅b^=1\hat{b} \cdot \hat{b} = 1. Use the unit vector property before expanding the dot products.

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