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JEE Main 20257 AprilMorning Shift
Exam
JEE Main
Year
2025
Date
7 April
Shift
Morning Shift
Subject
Mathematics
Chapter
Vector Algebra

JEE Main 2025 · 7 April · Morning Shift · Mathematics · Vector Algebra

Let the angle θ, 0 θ (π)/(2) between two unit vectors a and b be ^...

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

Options

  1. A31
  2. B27
  3. C29
  4. D24

Correct answer

C. 29

Step-by-step solution

<br/>c→=3a→+6 b→+9(a→×b→)sin⁡−1(659)⇒sin⁡θ=659⇒cos⁡θ=49c→⋅a→=3∣a→∣2+6a⃗⋅b⃗=3+6⋅49=519c→⋅a⃗=3a→⋅ b→+6∣ b→∣2=3⋅49+6=223∴9(c→⋅a→)−3(c→⋅b⃗)=51−22=29<br/>\begin{aligned}<br/>& \overrightarrow{\mathrm{c}}=3 \overrightarrow{\mathrm{a}}+6 \overrightarrow{\mathrm{~b}}+9(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}) \\ & \sin ^{-1}\left(\frac{\sqrt{65}}{9}\right) \Rightarrow \sin \theta=\frac{\sqrt{65}}{9} \Rightarrow \cos \theta=\frac{4}{9} \\ & \overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{a}}=3|\overrightarrow{\mathrm{a}}|^2+6 \vec{a} \cdot \vec{b}=3+\frac{6 \cdot 4}{9}=\frac{51}{9} \\ & \overrightarrow{\mathrm{c}} \cdot \vec{a}=3 \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{~b}}+6|\overrightarrow{\mathrm{~b}}|^2=\frac{3 \cdot 4}{9}+6=\frac{22}{3} \\ & \therefore 9(\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{a}})-3(\overrightarrow{\mathrm{c}} \cdot \vec{b})=51-22=29<br/>\end{aligned}
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