If a + b bisects the angle between a and b,th | vedclass

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(C) Let the angle between vectors $a$ and $b$ be $	heta$. The unit vectors along $a$ and $b$ are $\hat{a} = \frac{a}{|a|}$ and $\hat{b} = \frac{b}{|b

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Equal in magnitude

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(C) Let the angle between vectors $a$ and $b$ be $	heta$. The unit vectors along $a$ and $b$ are $\hat{a} = \frac{a}{|a|}$ and $\hat{b} = \frac{b}{|b|}$.The vector that bisects the angle between $a$ and $b$ is given by the sum of their unit vectors,which is $\hat{a} + \hat{b} = \frac{a}{|a|} + \frac{b}{|b|}$.Given that $a + b$ is the bisector,it must be parallel to $\hat{a} + \hat{b}$.Thus,$a + b = k \left( \frac{a}{|a|} + \frac{b}{|b|} ight)$ for some scalar $k$.Comparing the coefficients of $a$ and $b$,we get $1 = \frac{k}{|a|}$ and $1 = \frac{k}{|b|}$.This implies $|a| = k$ and $|b| = k$,therefore $|a| = |b|$.Hence,$a$ and $b$ are equal in magnitude.

If $a + b$ bisects the angle between $a$ and $b$,then $a$ and $b$ are

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