JEE Mains · Maths · STD 12 - 10. vector algebra
Let the angle \(\theta, 0 \lt \theta \lt \frac{\pi}{2}\) between two unit vectors \(\hat{\mathrm{a}}\) and \(\hat{\mathrm{b}}\) be \(\sin ^{-1}\left(\frac{\sqrt{65}}{9}\right)\). If the vector \(\overrightarrow{\mathrm{c}}=3 \hat{\mathrm{a}}+6 \hat{\mathrm{~b}}+9(\hat{\mathrm{a}} \times \hat{\mathrm{b}}),\) then the value of \(9(\overrightarrow{\mathrm{c}} \cdot \hat{\mathrm{a}})-3(\overrightarrow{\mathrm{c}} \cdot \hat{\mathrm{b}})\) is
- A 31
- B 27
- C 29
- D 24
Answer & Solution
Correct Answer
(C) 29
Step-by-step Solution
Detailed explanation
\begin{aligned} & \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…
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