If overline{a} and overline{b} are two unit v | vedclass

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Question

(A) 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

(A) Let $	heta$ be the angle between $\overline{a}$ and $\overline{b}$.Since $\overline{a}$ and $\overline{b}$ are unit vectors,$|\overline{a}| = 1$ and $|\overline{b}| = 1$.The vectors $5 \overline{a} + 4 \overline{b}$ and $\overline{a} - 2 \overline{b}$ are perpendicular,so their dot product is $0$.$(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 $|\overline{b}| = 1$,and $\overline{a} \cdot \overline{b} = |\overline{a}||\overline{b}| \cos 	heta = \cos 	heta$:$5(1)^2 - 6 \cos 	heta - 8(1)^2 = 0$$5 - 6 \cos 	heta - 8 = 0$$-6 \cos 	heta = 3$$\cos 	heta = -\frac{3}{6} = -\frac{1}{2}$$	heta = \cos^{-1}\left(-\frac{1}{2}ight) = \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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