If the vector a + b makes equal angles with v | vedclass

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(B) Let the angle between $(a + b)$ and $a$ be $	heta_1$,and the angle between $(a + b)$ and $b$ be $	heta_2$. Given $	heta_1 = 	heta_2$,we have $

Vedclass · Accepted Answer

$|a| = |b|$ or the angle between $a$ and $b$ is zero.

Vedclass · Answer

(B) Let the angle between $(a + b)$ and $a$ be $	heta_1$,and the angle between $(a + b)$ and $b$ be $	heta_2$. Given $	heta_1 = 	heta_2$,we have $\cos 	heta_1 = \cos 	heta_2$.$\frac{(a + b) \cdot a}{|a + b| |a|} = \frac{(a + b) \cdot b}{|a + b| |b|}$$\Rightarrow \frac{|a|^2 + b \cdot a}{|a + b| |a|} = \frac{a \cdot b + |b|^2}{|a + b| |b|}$$\Rightarrow \frac{|a|^2}{|a + b| |a|} + \frac{b \cdot a}{|a + b| |a|} = \frac{a \cdot b}{|a + b| |b|} + \frac{|b|^2}{|a + b| |b|}$$\Rightarrow \frac{|a|}{|a + b|} + \frac{a \cdot b}{|a + b| |a|} = \frac{a \cdot b}{|a + b| |b|} + \frac{|b|}{|a + b|}$$\Rightarrow \frac{|a| - |b|}{|a + b|} + \frac{a \cdot b}{|a + b|} \left( \frac{1}{|a|} - \frac{1}{|b|} ight) = 0$$\Rightarrow \frac{|a| - |b|}{|a + b|} + \frac{a \cdot b}{|a + b|} \left( \frac{|b| - |a|}{|a| |b|} ight) = 0$$\Rightarrow (|a| - |b|) \left( \frac{1}{|a + b|} - \frac{a \cdot b}{|a + b| |a| |b|} ight) = 0$$\Rightarrow (|a| - |b|) \left( 1 - \frac{a \cdot b}{|a| |b|} ight) = 0$This implies $|a| = |b|$ or $\frac{a \cdot b}{|a| |b|} = 1$,which means $\cos 	heta = 1$,so the angle between $a$ and $b$ is $0$.

If the vector $a + b$ makes equal angles with vectors $a$ and $b$,then:

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