Finding the Dot Product of Vectors

Introduction

If the two vectors are as follows

$$\overrightarrow{a} = (a_1, a_2, a_3, \cdots), \overrightarrow{b} = (b_1, b_2, b_3, \cdots)$$

and \(\theta\) is the angle between the two vectors

<method 1>

$$\overrightarrow{a} \cdot \overrightarrow{b} = a_1 b_1 + a_2 b_2 + a_3 b_3 + \cdots$$

<method 2>

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

Proof

Then, let me demonstrate why the two different methods yield the same outcome.

Let's consider the situation to be the same as above.

First, we will move the two vectors so that their initial points coincide. This will form a triangle, which can be represented as follows.

Now, based on the figure above, we will apply the law of cosines.

$$|\overrightarrow{c}|^2 = |\overrightarrow{a} - \overrightarrow{b}|^2 = |\overrightarrow{a}|^2 + |\overrightarrow{b}|^2 - 2|\overrightarrow{a}||\overrightarrow{b}|\cos\theta \cdots (1)$$

First, let's calculate \(|\overrightarrow{a} - \overrightarrow{b}|^2\).

• It can be expanded like this.

$$|\overrightarrow{a} - \overrightarrow{b}|^2 = (\overrightarrow{a} - \overrightarrow{b}) \cdot (\overrightarrow{a} - \overrightarrow{b})$$

• By the way, given that the commutative and distributive properties are valid, the right-hand side can be written as follows.

$$
|\overrightarrow{a} - \overrightarrow{b}|^2 = |\overrightarrow{a}|^2 + |\overrightarrow{b}|^2 - 2\overrightarrow{a} \cdot \overrightarrow{b}
\cdots (2)
$$

Applying equation (2) to equation (1) gives

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

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