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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

$\frac{\pi}{3}$

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 $\overline{c} = \overline{a} + 2\overline{b}$ and $\overline{d} = 5\overline{a} - 4\overline{b}$ are perpendicular,their dot product is zero:$\overline{c} \cdot \overline{d} = 0$$(\overline{a} + 2\overline{b}) \cdot (5\overline{a} - 4\overline{b}) = 0$$5(\overline{a} \cdot \overline{a}) - 4(\overline{a} \cdot \overline{b}) + 10(\overline{b} \cdot \overline{a}) - 8(\overline{b} \cdot \overline{b}) = 0$$5|\overline{a}|^2 + 6(\overline{a} \cdot \overline{b}) - 8|\overline{b}|^2 = 0$Since $|\overline{a}| = 1$ and $|\overline{b}| = 1$,and $\overline{a} \cdot \overline{b} = |\overline{a}||\overline{b}| \cos 	heta = \cos 	heta$:$5(1) + 6 \cos 	heta - 8(1) = 0$$6 \cos 	heta - 3 = 0$$6 \cos 	heta = 3$$\cos 	heta = \frac{3}{6} = \frac{1}{2}$$	heta = \cos^{-1}\left(\frac{1}{2}ight) = \frac{\pi}{3}$.

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

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