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

We are given that $$\vec{a}$$ and $$\vec{b}$$ are non-zero, perpendicular, and equal in magnitude, so $$\vec{a} \cdot \vec{b} = 0$$ and $$|\vec{a}| = |\vec{b}|$$. Since the vectors are perpendicular, $$|\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}| = |\vec{a}|^2$$. The condition $$|\vec{a} \times \vec{b}| = |\vec{a}|$$ then gives $$|\vec{a}|^2 = |\vec{a}|$$, so $$|\vec{a}| = 1$$ and hence $$|\vec{b}| = 1$$.

Because $$\vec{a} \times \vec{b}$$ is perpendicular to both $$\vec{a}$$ and $$\vec{b}$$, and $$|\vec{a} \times \vec{b}| = 1$$, the three vectors $$\vec{a}$$, $$\vec{b}$$, and $$\vec{a} \times \vec{b}$$ are mutually orthogonal unit vectors. Let $$\vec{v} = \vec{a} + \vec{b} + (\vec{a} \times \vec{b})$$. Then $$|\vec{v}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + |\vec{a} \times \vec{b}|^2 = 1 + 1 + 1 = 3$$, so $$|\vec{v}| = \sqrt{3}$$.

The dot product $$\vec{v} \cdot \vec{a} = |\vec{a}|^2 + 0 + 0 = 1$$. Therefore the angle $$\theta$$ between $$\vec{v}$$ and $$\vec{a}$$ satisfies $$\cos\theta = \dfrac{\vec{v} \cdot \vec{a}}{|\vec{v}||\vec{a}|} = \dfrac{1}{\sqrt{3}}$$, giving $$\theta = \cos^{-1}\!\left(\dfrac{1}{\sqrt{3}}\right)$$.