Given that overrightarrow A + overrightarrow | vedclass

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Question

Given, $A+B=C$

and $B=C-A$

Since, $C \perp A$, therefore

$B^{2}=C^{2}+A^{2}$

Also, $|C|=|A|,$ therefore

$B^{2}=2 A^{2} \Rightarrow B=\s

Vedclass · Accepted Answer

$\frac{{3\pi }}{4}radian$

Vedclass · Answer

Given, $A+B=C$

and $B=C-A$

Since, $C \perp A$, therefore

$B^{2}=C^{2}+A^{2}$

Also, $|C|=|A|,$ therefore

$B^{2}=2 A^{2} \Rightarrow B=\sqrt{2} A$

$\mathrm{Now}, A^{2}+B^{2}+2 A B \cos 	heta=C^{2}=A^{2}$

$	herefore \cos 	heta=\frac{1}{\sqrt{2}}$

This gives $	heta=\frac{3 \pi}{4}$ rad

Given that $\overrightarrow A + \overrightarrow B = \overrightarrow C $and that $\overrightarrow C $ is $ \bot $ to $\overrightarrow A $. Further if $|\overrightarrow A |\, = \,|\overrightarrow C |,$then what is the angle between $\overrightarrow A $ and $\overrightarrow B $

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