Let the angle , 0< < 2 between two unit vectors a and b be ^ -1 (… - Vidyadip

MCQ

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 $\vec{c}=3 \hat{a}+6 \hat{b}+9(\hat{a} \times \hat{b}), \quad$ then the value of $9(\vec{c} \cdot \hat{a})-3(\vec{c} \cdot \hat{b})$ is

A
31
B
27
C
29
D
24

✓

Answer

C. 29
$\overrightarrow{\mathrm{c}}=3 \vec{a}+6 \vec{b}+9(\vec{a} \times \vec{b})$
$\sin ^{-1}\left(\frac{\sqrt{65}}{9}\right) \Rightarrow \sin \theta=\frac{\sqrt{65}}{9} \Rightarrow \cos \theta=\frac{4}{9}$
$\vec{c} \cdot \vec{a}=3|\vec{a}|^{2}+6 \vec{a} \cdot \vec{b}=3+\frac{6.4}{9}=\frac{51}{9}$
$\vec{c} \cdot \vec{a}=3 \vec{a} \cdot \vec{b}+6|\vec{b}|^{2}=\frac{3.4}{9}+6=\frac{22}{3}$
$\therefore 9(\vec{c} \cdot \vec{a})-3(\vec{c} \cdot \vec{b})=51-22=29$

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