📂Mathematical Physics

The Relationship between the Dot Product of Two Vectors and the Angle Between Them

Theorem

Let the angle between two vectors $\mathbf{a} = (a_{1}, a_{2}, a_{3})$ and $\mathbf{b} = (b_{1}, b_{2}, b_{3})$ be $\theta$. Then, the following holds.

$$\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \cos \theta$$

Here, $\mathbf{a} \cdot \mathbf{b}$ is the dot product (inner product) of the two vectors.

Corollary

The necessary and sufficient condition for two non-zero vectors $\mathbf{a}$ and $\mathbf{b}$ to be orthogonal is as follows.

$$\mathbf{a} \cdot \mathbf{b} = 0$$

Proof

Consider the diagram below. Vectors $\mathbf{a}$ and $\mathbf{b}$, along with $\mathbf{a} - \mathbf{b}$, form a triangle.

Now, applying the law of cosines to this triangle, we obtain the following.

$$|\mathbf{a}|^2 + |\mathbf{b}|^2 - 2 |\mathbf{a}| |\mathbf{b}| \cos \theta = | \mathbf{a} - \mathbf{b} |^2$$

By the properties of the dot product $\mathbf{a} \cdot \mathbf{a} = | \mathbf{a} |^{2}$, the equation becomes:

$$\begin{align*} \mathbf{a} \cdot \mathbf{a} + \mathbf{b} \cdot \mathbf{b} - 2 |\mathbf{a}| |\mathbf{b}| \cos \theta &= (\mathbf{a} - \mathbf{b}) \cdot (\mathbf{a} - \mathbf{b}) \\ &= \mathbf{a} \cdot \mathbf{a} - 2 \mathbf{a} \cdot \mathbf{b} + \mathbf{b} \cdot \mathbf{b} \end{align*}$$

Eliminating the common terms yields:

$$\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \cos \theta$$

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Proof of the Corollary

$(\Longrightarrow)$

Assume $\mathbf{a}$ and $\mathbf{b}$ are orthogonal. Then,

$$\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \cos \frac{\pi}{2} = 0$$

Let $(\Longleftarrow)$ and $\mathbf{a} \cdot \mathbf{b} = 0$ be true. Then,

$$|\mathbf{a}| |\mathbf{b}| \cos \theta = 0$$

Since $\mathbf{a}$ and $\mathbf{b}$ are non-zero vectors, $|\mathbf{a}| \ne 0$ and $|\mathbf{b}| \ne 0$ hold. Therefore,

$$\cos\theta = 0 \implies \theta = \frac{\pi}{2}$$

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