If a and b are unit vectors such that (a + 2b | vedclass

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(B) Given that $a$ and $b$ are unit vectors,so $|a| = 1$ and $|b| = 1$.Since $(a + 2b)$ and $(5a - 4b)$ are perpendicular,their dot product is zero:$(

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$60$

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(B) Given that $a$ and $b$ are unit vectors,so $|a| = 1$ and $|b| = 1$.Since $(a + 2b)$ and $(5a - 4b)$ are perpendicular,their dot product is zero:$(a + 2b) \cdot (5a - 4b) = 0$Expanding the dot product:$5(a \cdot a) - 4(a \cdot b) + 10(b \cdot a) - 8(b \cdot b) = 0$$5|a|^2 + 6(a \cdot b) - 8|b|^2 = 0$Substituting $|a| = 1$ and $|b| = 1$:$5(1) + 6(a \cdot b) - 8(1) = 0$$6(a \cdot b) - 3 = 0$$6(a \cdot b) = 3$$a \cdot b = \frac{3}{6} = \frac{1}{2}$We know that $a \cdot b = |a||b| \cos 	heta$,where $	heta$ is the angle between $a$ and $b$.$(1)(1) \cos 	heta = \frac{1}{2}$$\cos 	heta = \frac{1}{2}$$	heta = 60^o$.

If $a$ and $b$ are unit vectors such that $(a + 2b)$ and $(5a - 4b)$ are perpendicular to each other,then the angle between $a$ and $b$ is ................... $^o$.

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