Let the angle between two unit vectors a and b (0

Question

Let the angle between two unit vectors $\hat a$ and $\hat b$ ($0<	heta<\dfrac\pi2$) be $\sin^{-1}\left(\dfrac{\sqrt{65}}9ight)$. If $\vec c = 3\hat a+6\hat b+9(\hat a	imes\hat b)$, then $9(\vec c\cdot\hat a) - 3(\vec c\cdot\hat b)$ is equal to:

Accepted Answer

$\vec c\cdot\hat a = 3+6(\hat a\cdot\hat b)$, $\vec c\cdot\hat b = 3(\hat a\cdot\hat b)+6$.

$$9(\vec c\cdot\hat a)-3(\vec c\cdot\hat b) = \big[27+54(\hat a\cdot\hat b)\big] - \big[9(\hat a\cdot\hat b)+18\big] = 9+45(\hat a\cdot\hat b)$$

Since $\sin\theta = \dfrac{\sqrt{65}}9$, $\cos\theta = \hat a\cdot\hat b$, and $(\hat a\cdot\hat b)^2 = 1-\dfrac{65}{81} = \dfrac{16}{81} \Rightarrow \hat a\cdot\hat b = \dfrac49$ (positive since $\theta$ acute).

$$9+45\left(\frac49\right) = 9+20 = 29$$

Solutions · 1