Problem on dot products and vector length
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Prove that $\bfx \cdot \bfx = \left| \bfx \right|^2$ in two ways:
Directly (in the case of 3d vectors)
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
Direct computation of dot product
For any two vectors $\langle x_1, x_2, x_3 \rangle$ and $\langle y_1, y_2, y_3\rangle$, $$\langle x_1, x_2, x_3\rangle \cdot \langle y_1, y_2, y_3 \rangle = x_1 y_1 + x_2 y_2 + x_3 y_3.$$
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
Definition of vector length
The length of the vector $\bfx = \langle x_1, x_2, x_3 \rangle$ is given by $$\left | \bfx \right| = \sqrt{x_1^2 + x_2^2 + x_3^2}.$$
Hence, we see that $\bfx \cdot \bfx = \left | \bfx \right |^2.$
Proof by the dot product angle formula
Recall the dot product angle formula:
Dot product angle formula
If the angle between the vectors $\bfx$ and $\bfy$ is $\theta$, then $$\bfx \cdot \bfy = \left| \bfx \right| \left| \bfy \right| \cos \theta.$$
The angle between two vectors is always between zero and pi
The angle between the vector $\bfx$ and itself is $\theta=0$. Hence $\cos\theta=1$.
Applying the dot product angle formula for $\bfy = \bfx$, we get $$ \bfx \cdot \bfx = \left| \bfx \right| ^2.$$
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