Let the angle theta, 0

Question

(C) Given $\vec{c} = 3\hat{a} + 6\hat{b} + 9(\hat{a} 	imes \hat{b})$.Since $\sin 	heta = \frac{\sqrt{65}}{9}$,we have $\cos^2 	heta = 1 - \sin^2

Vedclass · Accepted Answer

$29$

Vedclass · Answer

(C) Given $\vec{c} = 3\hat{a} + 6\hat{b} + 9(\hat{a} 	imes \hat{b})$.Since $\sin 	heta = \frac{\sqrt{65}}{9}$,we have $\cos^2 	heta = 1 - \sin^2 	heta = 1 - \frac{65}{81} = \frac{16}{81}$,so $\cos 	heta = \frac{4}{9}$ (as $0 Now,calculate $\vec{c} \cdot \hat{a} = (3\hat{a} + 6\hat{b} + 9(\hat{a} 	imes \hat{b})) \cdot \hat{a} = 3(\hat{a} \cdot \hat{a}) + 6(\hat{b} \cdot \hat{a}) + 9((\hat{a} 	imes \hat{b}) \cdot \hat{a})$.Since $\hat{a} \cdot \hat{a} = 1$,$\hat{b} \cdot \hat{a} = \cos 	heta = \frac{4}{9}$,and $(\hat{a} 	imes \hat{b}) \cdot \hat{a} = 0$,we get $\vec{c} \cdot \hat{a} = 3(1) + 6(\frac{4}{9}) + 0 = 3 + \frac{8}{3} = \frac{17}{3}$.Next,calculate $\vec{c} \cdot \hat{b} = (3\hat{a} + 6\hat{b} + 9(\hat{a} 	imes \hat{b})) \cdot \hat{b} = 3(\hat{a} \cdot \hat{b}) + 6(\hat{b} \cdot \hat{b}) + 9((\hat{a} 	imes \hat{b}) \cdot \hat{b})$.Since $\hat{a} \cdot \hat{b} = \frac{4}{9}$,$\hat{b} \cdot \hat{b} = 1$,and $(\hat{a} 	imes \hat{b}) \cdot \hat{b} = 0$,we get $\vec{c} \cdot \hat{b} = 3(\frac{4}{9}) + 6(1) + 0 = \frac{4}{3} + 6 = \frac{22}{3}$.Finally,$9(\vec{c} \cdot \hat{a}) - 3(\vec{c} \cdot \hat{b}) = 9(\frac{17}{3}) - 3(\frac{22}{3}) = 3(17) - 22 = 51 - 22 = 29$.

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

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