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Vector Algebra question

2025 · 7 Apr · Shift 1 · Q44
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  5. /2025 · 7 Apr · Shift 1 · Q44

Vector Algebra question

2025 · 7 Apr · Shift 1 · Q44

JEE MainMathematicsVector AlgebraMCQ+4 / −1
Let the angle θ,0<θ<π2\theta, 0\lt \theta\lt \frac{\pi}{2}θ,0<θ<2π​ between two unit vectors a^\hat{a}a^ and b^\hat{b}b^ be sin⁡−1(659)\sin ^{-1}\left(\frac{\sqrt{65}}{9}\right)sin−1(965​​). If the vector c⃗=3a^+6b^+9(a^×b^)\vec{c}=3 \hat{a}+6 \hat{b}+9(\hat{a} \times \hat{b})c=3a^+6b^+9(a^×b^), then the value of 9(c⃗⋅a^)−3(c⃗⋅b^)9(\vec{c} \cdot \hat{a})-3(\vec{c} \cdot \hat{b})9(c⋅a^)−3(c⋅b^) is
  1. A
    31
  2. B
    29
  3. C
    24
  4. D
    27
View written solutionFree

Correct answer: D

  1. Given data
  • a^\hat aa^ and b^\hat bb^ are unit vectors.
  • Angle between them is θ=sin⁡−1(659),0<θ<π2\theta=\sin^{-1}\left(\frac{\sqrt{65}}{9}\right), \qquad 0<\theta<\frac{\pi}{2}θ=sin−1(965​​),0<θ<2π​
  • c⃗=3a^+6b^+9(a^×b^)\vec c=3\hat a+6\hat b+9(\hat a\times \hat b)c=3a^+6b^+9(a^×b^)

We need to find: 9(c⃗⋅a^)−3(c⃗⋅b^)9(\vec c\cdot \hat a)-3(\vec c\cdot \hat b)9(c⋅a^)−3(c⋅b^)


  1. Find cos⁡θ\cos\thetacosθ

Since sin⁡θ=659\sin\theta=\frac{\sqrt{65}}{9}sinθ=965​​ we get

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