Let $\overrightarrow a$ and $\overrightarrow b$ be two non-zero vectors perpendicular to each other and $|\overrightarrow a | = |\overrightarrow b |$. If $|\overrightarrow a \times \overrightarrow b | = |\overrightarrow a |$, then the angle between the vectors $$\left( {\overrightarrow a + \overrightarrow b + \left( {\overrightarrow a \times \overrightarrow b } \right)} \right)$$ and ${\overrightarrow a }$ is equal to :

$\overrightarrow a$ is perpendicular to $\overrightarrow b$<br><br>$\therefore$ $\overrightarrow a$ . $\overrightarrow b$ = 0<br><br>Given, | $\overrightarrow a$ $\times$ $\overrightarrow b$ | = | $\overrightarrow a$ |<br><br>and | $\overrightarrow a$ | = | $\overrightarrow b$ |<br><br>$\therefore$ | $\overrightarrow a$ $\times$ $\overrightarrow b$ | = | $\overrightarrow a$ | = | $\overrightarrow b$ | = k(assume)<br><br>Now, angle between $\overrightarrow a$ and ($\overrightarrow a$ + $\overrightarrow b$ + ($\overrightarrow a$ $\times$ $\overrightarrow b$))<br><br>$$\cos \theta = {{\overrightarrow a ((\overrightarrow a + \overrightarrow b + (\overrightarrow a \times \overrightarrow b ))} \over {|\overrightarrow a |.|\overrightarrow a + \overrightarrow b + (\overrightarrow a \times \overrightarrow b )|}}$$<br><br>$$ = {{|\overrightarrow a {|^2} + \,\overrightarrow a .\,\overrightarrow b + \overrightarrow a (\overrightarrow a \times \overrightarrow b )} \over {|\overrightarrow a |.|\overrightarrow a + \overrightarrow b + (\overrightarrow a \times \overrightarrow b )|}}$$<br><br>[Note $$\overrightarrow a .(\overrightarrow a \times \overrightarrow b ) = [\overrightarrow a \overrightarrow a \overrightarrow b ] = 0$$]<br><br>$$ = {{|\overrightarrow a {|^2} + 0 + 0} \over {|\overrightarrow a |.|\overrightarrow a + \overrightarrow b + (\overrightarrow a \times \overrightarrow b )|}}$$<br><br>Now, $$|\overrightarrow a + \overrightarrow b + (\overrightarrow a \times \overrightarrow b ){|^2}$$<br><br>$$ = |\overrightarrow a {|^2} + |\overrightarrow a {|^2} + |\overrightarrow a \times \overrightarrow b {|^2} + $$<br><br>$$2\overrightarrow a .\overrightarrow b + 2\overrightarrow b .(\overrightarrow a \times \overrightarrow b ) + 2\overrightarrow a .(\overrightarrow a \times \overrightarrow b )$$<br><br>$= {k^2} + {k^2} + {k^2}$<br><br>$\therefore$ $$|\overrightarrow a + \overrightarrow b + (\overrightarrow a \times \overrightarrow b ){|^2} = 3{k^2}$$<br><br>$$ \Rightarrow |\overrightarrow a + \overrightarrow b + (\overrightarrow a \times \overrightarrow b )| = \sqrt 3 k$$<br><br>$\therefore$ $\cos \theta = {{{k^2}} \over {k(\sqrt 3 k)}} = {1 \over {\sqrt 3 }}$<br><br>$\Rightarrow \theta = {\cos ^{ - 1}}\left( {{1 \over {\sqrt 3 }}} \right)$