How do I prove this equality? 
A vector x $\in \mathbb{R}^3$ makes angles $a, b, c$ with the
  three axes. Using the dot product with standard basis vectors, show:
$$\cos^2a + \cos^2b + \cos^2c = 1$$

I wasn't sure how to start this one. Conceptually, I can see what was going on, but I would think to create a number of simultaneous equations using $a + b = \frac{\pi}{2}$ etc. I wasn't sure what to do with standard basis vectors.
 A: HINT 1:
Recall that if $\vec{x} = x_a \hat{e}_a + x_b \hat{e}_b + x_c \hat{e}_c$ and $\vec{y} = y_a \hat{e}_a + y_b \hat{e}_b + y_c \hat{e}_c$, then $$\vec{x} \cdot \vec{y} = x_a y_a + x_b y_b + x_c y_c = \Vert x \Vert_2 \Vert y \Vert_2 \cos(\theta_{xy})$$ where $\theta_{xy}$ is the angle between the vectors $\vec{x}$ and $\vec{y}$.
HINT 2:
Look at the norm of the vector, which make angles $a,b,c$ with the three axes.
Move the cursor over the gray area for a complete answer. 

 Let the vector be $\vec{x} = x_a \hat{e}_a + x_b \hat{e}_b + x_c \hat{e}_c$. Note that $$x_a = \vec{x} \cdot \hat{e}_a = \Vert x \Vert_2 \cos(a)$$ $$x_b = \vec{x} \cdot \hat{e}_b = \Vert x \Vert_2 \cos(b)$$ $$x_c = \vec{x} \cdot \hat{e}_c = \Vert x \Vert_2 \cos(c)$$ Further, $$\Vert x \Vert_2^2 = \vert x_a \vert^2 + \vert x_b \vert^2 + \vert x_c \vert^2.$$ Hence, $$\Vert x \Vert_2^2 = \Vert x \Vert_2^2 \cos^2(a) + \Vert x \Vert_2^2 \cos^2(b) + \Vert x \Vert_2^2 \cos^2(c)$$ Canceling off $\Vert x \Vert_2^2$, we get $$\cos^2(a) + \cos^2(b) + \cos^2(c) = 1$$

A: Let be the position vector $\vec{D}$ of the point D(A, B, C) and the angles a,b, c
with the three axes. We have that:
$\cos^2 a = \frac{{OA}^{2}}{{OD}^{2}};\cos^2 b = \frac{{OB}^{2}}{{OD}^{2}};\cos^2 c = \frac{{OC}^{2}}{{OD}^{2}} $.
$$\cos^2a + \cos^2b + \cos^2c = \frac{{OA}^{2}+{OB}^{2}+{OC}^{2}}{{OD}^{2}}=\frac{{OD}^{2}}{{OD}^{2}}=1.$$
Q.E.D.
