If $|\vec {A}.\vec {B}|=|\vec {A} \times \vec {B}|$, then find the resultant of $\vec {A}$ and $\vec {B}$ 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?
Assuming that $\vec A$ and $\vec B$ are non-zero:

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

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, $|\vec A + \vec B| = \sqrt{|\vec A|^2 + |\vec B|^2 - \sqrt2|\vec A||\vec B|}$.
Assuming that $\vec A$ and $\vec B$ are non-zero:

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

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