Let the angle θ, 0 < θ < π/2 between two unit vectors a and b sin^(-1) (√(65/9)).

Question

Let the angle \(\theta, 0<\theta<\frac{\pi}{2}\) between two unit vectors \(\hat{a}\) and \(\hat{b}\) 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 \(9(\vec{c} \cdot \hat{a})-3(\vec{c} \cdot \hat{b})\) is

(1) 31

(2) 27

(3) 29

(4) 24 

Answer 1

Correct option is: (3) 29 

\(\overrightarrow{\mathrm{c}}=3 \vec{a}+6 \vec{b}+9(\vec{a} \times \vec{b})\)

\(\sin ^{-1}\left(\frac{\sqrt{65}}{9}\right) \Rightarrow \sin \theta=\frac{\sqrt{65}}{9} \Rightarrow \cos \theta=\frac{4}{9}\)

\(\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{a}}=3|\overrightarrow{\mathrm{a}}|^{2}+6 \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}=3+\frac{6.4}{9}=\frac{51}{9}\)

\(\vec{c} \cdot \vec{a}=3 \vec{a} \cdot \vec{b}+6|\vec{b}|^{2}=\frac{3 \cdot 4}{9}+6=\frac{22}{3}\)

\(\therefore\ 9(\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{a}})-3(\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{b}})=51-22=29\)