How Do You Calculate the Scalar (Dot) Product of Two Vectors?

Multiply corresponding components of the two vectors, then add the results. The answer is a single number called the scalar product or dot product, not another vector.

Calculating From Components

For vectors $\mathbf{a}=(a_1,a_2,a_3)$ and $\mathbf{b}=(b_1,b_2,b_3)$,

$$\mathbf{a}\cdot\mathbf{b}=a_1b_1+a_2b_2+a_3b_3.$$

The same rule applies in two dimensions, using only two component products. The vectors must have the same number of components.

Calculate it in three steps: pair components in the same positions, multiply each pair, and sum all products. Do not multiply magnitudes unless you are using the angle formula. Because each product is a number and the final operation is addition, the result has no direction.

For example, let

$$\mathbf{a}=(2,-1,3), \qquad \mathbf{b}=(4,5,-2).$$

Then

$$\begin{aligned} \mathbf{a}\cdot\mathbf{b} &=(2)(4)+(-1)(5)+(3)(-2)\\ &=8-5-6\\ &=-3. \end{aligned}$$

Therefore, $\mathbf{a}\cdot\mathbf{b}=-3$. Notice that negative components must keep their signs during multiplication. This component method is normally the quickest route and provides clear working for an examiner.

Why the Dot Product Matters

The dot product is connected to the angle between two vectors by

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

Therefore,

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

For the example, $|\mathbf{a}|=\sqrt{14}$ and $|\mathbf{b}|=\sqrt{45}$. Hence,

$$\theta=\cos^{-1}\left(\frac{-3}{\sqrt{14}\sqrt{45}}\right)=96.9^\circ.$$

This is consistent with the negative dot product. Since both magnitudes are positive, the sign of the dot product matches the sign of $\cos\theta$.

A common misconception is confusing the dot product with the cross product. The dot product produces a scalar; the cross product produces a vector perpendicular to both original vectors.

IB Exam Technique

This is HL-only content from AHL 3.13 Scalar and vector products; a Paper 1 short-response style is assumed. Show the component multiplication before the result and retain full GDC precision until the final answer. When finding an angle, write the substitution before using inverse cosine. If asked whether vectors are perpendicular, calculate the dot product and explicitly conclude that $\mathbf{a}\cdot\mathbf{b}=0$ implies perpendicularity.