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?

  • $\begingroup$ You should have made clear that you are trying to find the magnitude of the resultant. $\endgroup$ – DHMO Apr 18 '17 at 1:38

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

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.