## Problem on dot products and vector length

Prove that $\bfx \cdot \bfx = \left| \bfx \right|^2$ in two ways:
1. Directly (in the case of 3d vectors)
2. By the dot product angle formula
• ## Solution

#### Part (a) - Direct proof

To prove this formula directly, we appeal to the definitions of dot product and length in terms of the components of $\bfx$.
Let $\bfx = \langle x_1, x_2, x_3 \rangle.$
Recall that
Appealing to this formula for $\bfx \cdot \bfx$, we get $$\langle x_1, x_2, x_3 \rangle \cdot \langle x_1, x_2, x_3 \rangle = x_1^2 + x_2^2 + x_3^2.$$
Recall that
Hence, we see that $\bfx \cdot \bfx = \left | \bfx \right |^2.$

#### Proof by the dot product angle formula

Recall the dot product angle formula:
Applying the dot product angle formula for $\bfy = \bfx$, we get $$\bfx \cdot \bfx = \left| \bfx \right| ^2.$$