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If $|\vec {A}.\vec {B}|=|\vec {A} \times \vec {B}|$, then find the resultant of $\vec {A}$ and $\vec {B}$

My Attempt: $$|\vec {A}.\vec {B}|=|\vec {A} \times \vec {B}|$$ $$|\vec {A}|.|\vec {B}|.\cos \theta=|\vec {A}|.|\vec {B}|\sin \theta n^{\cap}$$ $$\cos \theta=\sin \theta n^{\cap}$$

How do I proceed further?

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  • $\begingroup$ You should have made clear that you are trying to find the magnitude of the resultant. $\endgroup$ – DHMO Apr 18 '17 at 1:38
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Assuming that $\vec A$ and $\vec B$ are non-zero:

$$\begin{array}{rcl} |\vec A \cdot \vec B| &=& |\vec A \times \vec B| \\ |\vec A| |\vec B| \cos \theta &=& |\vec A| |\vec B| \sin \theta \\ \cos \theta &=& \sin \theta \\ \theta &=& 45^\circ \end{array}$$

Therefore, $|\vec A + \vec B|^2 = |\vec A|^2 + |\vec B|^2 - 2|\vec A||\vec B|\cos45^\circ = |\vec A|^2 + |\vec B|^2 - \sqrt2|\vec A||\vec B|$.

Therefore, $|\vec A + \vec B| = \sqrt{|\vec A|^2 + |\vec B|^2 - \sqrt2|\vec A||\vec B|}$.

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  • $\begingroup$ Do we not need $n$ for calculation? $\endgroup$ – pi-π Apr 18 '17 at 1:34
  • $\begingroup$ @Ramanujan $\vec A \times \vec B = |\vec A||\vec B| (\sin \theta) \hat n$, where $|\hat n| = 1$. Then take magnitude of both sides. $\endgroup$ – DHMO Apr 18 '17 at 1:35
  • $\begingroup$ and one more thing, How did you get $-$ sign in your answer? The book's answer is $\sqrt {A^2+B^2+\sqrt {2} AB}$. $\endgroup$ – pi-π Apr 18 '17 at 1:38
  • $\begingroup$ @Ramanujan That would be when $\theta = 135^\circ$. $\endgroup$ – DHMO Apr 18 '17 at 1:40
  • $\begingroup$ Why is $|\vec {A}+\vec {B}|^{2}=|\vec {A}|^2+|\vec {B}|^2-2|\vec {A}|.|\vec {B}|.\cos \theta$? $\endgroup$ – pi-π Apr 18 '17 at 1:43

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