Scalar dot product of two unit vectors

Given two unit vectors: $𝐀=(a,b)$ and: $𝐂=(c,d)$, we prove their dot product is: $ac+bd$.

Step 1: Squared Distance

The squared distance between: $𝐀$ and $𝐂$ is:

$${\parallel 𝐂-𝐀\parallel }^2={(a-c)}^2+{(b-d)}^2$$

Multiplying out:

$${\parallel 𝐂-𝐀\parallel }^2=a^2-2\cdot a\cdot c+c^2+b^2-2\cdot b\cdot d+d^2$$

Using the unit vector property: $a^2+b^2=1$ and $c^2+d^2=1$:

$${\parallel 𝐂-𝐀\parallel }^2=2-2(ac+bd)$$

Step 2: Distance and Dot Product

The squared distance between: $\mathbf{A}$ and: $\mathbf{C}$ can also be calculated using the dot product as a means of multiplying vectors:

$${\parallel \mathbf{C}-\mathbf{A}\parallel }^2=(\mathbf{C}-\mathbf{A})\cdot (\mathbf{C}-\mathbf{A})$$

Expanding the dot product:

$${\parallel 𝐂-𝐀\parallel }^2=\mathbf{C}\cdot \mathbf{C}-2\mathbf{A}\cdot \mathbf{C}+\mathbf{A}\cdot \mathbf{A}$$

Since $\mathbf{A}$ and $\mathbf{C}$ are unit vectors:

$${\parallel 𝐂-𝐀\parallel }^2=1-2\mathbf{A}\cdot \mathbf{C}+1=2-2\mathbf{A}\cdot \mathbf{C}$$

Final result

Equate to Step 1’s distance formula to Step 2’s:

$${\parallel 𝐂-𝐀\parallel }^2=2-2\mathbf{A}\cdot \mathbf{C}=2-2(ac+bd)$$ So:

$$\mathbf{A}\cdot \mathbf{C}=ac+bd$$ as was to be shown.