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(B) Let $	heta$ be the angle between $\overline{a}$ and $\overline{b}$.Since $\overline{a}$ and $\overline{b}$ are unit vectors,$|\overline{a}| = 1$

Vedclass · Accepted Answer

$\frac{2 \pi}{3}$

Vedclass · Answer

(B) Let $	heta$ be the angle between $\overline{a}$ and $\overline{b}$.Since $\overline{a}$ and $\overline{b}$ are unit vectors,$|\overline{a}| = 1$ and $|\overline{b}| = 1$.Given that $(5 \overline{a} + 4 \overline{b})$ and $(\overline{a} - 2 \overline{b})$ are perpendicular,their dot product is zero.$(5 \overline{a} + 4 \overline{b}) \cdot (\overline{a} - 2 \overline{b}) = 0$$5(\overline{a} \cdot \overline{a}) - 10(\overline{a} \cdot \overline{b}) + 4(\overline{b} \cdot \overline{a}) - 8(\overline{b} \cdot \overline{b}) = 0$$5|\overline{a}|^2 - 6(\overline{a} \cdot \overline{b}) - 8|\overline{b}|^2 = 0$Since $|\overline{a}| = 1$ and $|\overline{b}| = 1$,we have $5(1)^2 - 6(1)(1)\cos 	heta - 8(1)^2 = 0$.$5 - 6 \cos 	heta - 8 = 0$$-6 \cos 	heta = 3$$\cos 	heta = -\frac{3}{6} = -\frac{1}{2}$Since $\cos 	heta = -\frac{1}{2}$,the angle $	heta = \frac{2\pi}{3}$.

If $\overline{a}$ and $\overline{b}$ are two unit vectors such that $5 \overline{a} + 4 \overline{b}$ and $\overline{a} - 2 \overline{b}$ are perpendicular to each other,then the angle between $\overline{a}$ and $\overline{b}$ is

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