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(B) Let $	heta_1$ be the angle between $(a + b)$ and $a$,and $	heta_2$ be the angle between $(a + b)$ and $b$.Since $(a + b)$ bisects the angle betw

Vedclass · Accepted Answer

$|a| = |b|$ or the angle between $a$ and $b$ is $0$

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(B) Let $	heta_1$ be the angle between $(a + b)$ and $a$,and $	heta_2$ be the angle between $(a + b)$ and $b$.Since $(a + b)$ bisects the angle between $a$ and $b$,we have $\cos 	heta_1 = \cos 	heta_2$.Using the definition of the dot product,$\cos 	heta_1 = \frac{(a + b) \cdot a}{|a + b||a|}$ and $\cos 	heta_2 = \frac{(a + b) \cdot b}{|a + b||b|}$.Equating these,we get $\frac{|a|^2 + a \cdot b}{|a + b||a|} = \frac{a \cdot b + |b|^2}{|a + b||b|}$.This simplifies to $\frac{|a|^2 + a \cdot b}{|a|} = \frac{a \cdot b + |b|^2}{|b|}$.Multiplying both sides by $|a||b|$,we get $|b|(|a|^2 + a \cdot b) = |a|(a \cdot b + |b|^2)$.$|a|^2|b| + |b|(a \cdot b) = |a|(a \cdot b) + |a||b|^2$.$|a|^2|b| - |a||b|^2 + |b|(a \cdot b) - |a|(a \cdot b) = 0$.$|a||b|(|a| - |b|) - (a \cdot b)(|a| - |b|) = 0$.$(|a| - |b|)(|a||b| - a \cdot b) = 0$.Since $a \cdot b = |a||b| \cos 	heta$,where $	heta$ is the angle between $a$ and $b$,we have $(|a| - |b|)(|a||b| - |a||b| \cos 	heta) = 0$.$|a||b|(|a| - |b|)(1 - \cos 	heta) = 0$.This implies $|a| = |b|$ or $\cos 	heta = 1$,which means $	heta = 0$.

The vector $a + b$ bisects the angle between the vectors $a$ and $b$,if

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