Let $\mathbf{a},$ $\mathbf{b},$ $\mathbf{c}$ be three vectors with equal magnitudes, and such that any pair of them are orthogonal.  If $\theta$ is the angle between $\mathbf{a}$ and $\mathbf{a} + \mathbf{b} + \mathbf{c},$ then find $\cos \theta.$
I need to find $\cos \theta,$ which is the same as the dot product of $\mathbf{a}$ and $\mathbf{a} + \mathbf{b} + \mathbf{c}$ divided by the magnitude of $\mathbf{a}$ and the magnitude of $\mathbf{a} + \mathbf{b} + \mathbf{c}.$ Since any pair of the vectors are orthogonal, their dot product is zero. So $\mathbf{a} \cdot \mathbf{b} = \mathbf{a} \cdot \mathbf{c} = 0.$ The magnitude of $\mathbf{a}$ is the same as the magnitude of $\mathbf{b}$ and $\mathbf{c},$ so let's call it $m.$ Then $|\mathbf{a}|^2 = m^2.$ The magnitude of $\mathbf{a} + \mathbf{b} + \mathbf{c}$ is the square root of the sum of the squares of the components of the vector. So $|\mathbf{a} + \mathbf{b} + \mathbf{c}|^2 = (\mathbf{a} + \mathbf{b} + \mathbf{c}) \cdot (\mathbf{a} + \mathbf{b} + \mathbf{c}).$ Expanding the dot product, I get $|\mathbf{a} + \mathbf{b} + \mathbf{c}|^2 = \mathbf{a} \cdot \mathbf{a} + \mathbf{b} \cdot \mathbf{b} + \mathbf{c} \cdot \mathbf{c} + 2(\mathbf{a} \cdot \mathbf{b} + \mathbf{a} \cdot \mathbf{c} + \mathbf{b} \cdot \mathbf{c}).$ Again, since any pair of the vectors are orthogonal, the dot products are zero. So $|\mathbf{a} + \mathbf{b} + \mathbf{c}|^2 = \mathbf{a} \cdot \mathbf{a} + \mathbf{b} \cdot \mathbf{b} + \mathbf{c} \cdot \mathbf{c}.$ Each of these dot products is the square of the magnitude of the vector, which is $m.$ So $|\mathbf{a} + \mathbf{b} + \mathbf{c}|^2 = m^2 + m^2 + m^2 = 3m^2.$ Taking the square root of both sides, I get $|\mathbf{a} + \mathbf{b} + \mathbf{c}| = \sqrt{3m^2} = m\sqrt{3}.$ Now I can plug these values into the formula for $\cos \theta.$ I get $\cos \theta = \frac{\mathbf{a} \cdot \mathbf{a} + \mathbf{b} + \mathbf{c}}{|\mathbf{a}||\mathbf{a} + \mathbf{b} + \mathbf{c}|} = \frac{m^2}{m \cdot m\sqrt{3}} = \frac{1}{\sqrt{3}}.$

