How Do You Calculate the Scalar Product of Two Vectors?

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Methods for Calculating the Scalar Product

This is IB Mathematics: Analysis and Approaches HL, sub-topic AHL 3.13: Scalar (dot) product. Two equivalent formulas are used:

The component method is usually fastest when coordinates are given. Each product measures how much the vectors align in one coordinate direction, and the sum combines these contributions. In two dimensions, omit the third term: $\mathbf{a}\cdot\mathbf{b}=a_1b_1+a_2b_2$. The operation is commutative, so $\mathbf{a}\cdot\mathbf{b}=\mathbf{b}\cdot\mathbf{a}$.

For example, let

$\mathbf{a}=\begin{pmatrix}2\\-1\\3\end{pmatrix},\qquad \mathbf{b}=\begin{pmatrix}4\\5\\-2\end{pmatrix}.$

Multiply corresponding components, then add:

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

Therefore, $\mathbf{a}\cdot\mathbf{b}=-3$.

Another useful identity is the self-product, $\mathbf{a}\cdot\mathbf{a}=\lvert\mathbf{a}\rvert^2$. It follows because squaring and adding the components gives the square of the magnitude. This provides a quick check: the scalar product of a non-zero vector with itself must be positive. When finding an angle, ensure both vectors are non-zero before dividing by their magnitudes.

The geometric formula can be rearranged to find the angle between non-zero vectors:

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

For this example, $\lvert\mathbf{a}\rvert=\sqrt{14}$ and $\lvert\mathbf{b}\rvert=\sqrt{45}$. Hence $\cos\theta=-3/\sqrt{630}$, so $\theta=96.9^\circ$ to three significant figures. The negative result is consistent with an obtuse angle.

The scalar product also tests geometric relationships. For non-zero vectors, a zero scalar product proves perpendicularity. Parallel vectors satisfy $\lvert\mathbf{a}\cdot\mathbf{b}\rvert=\lvert\mathbf{a}\rvert\lvert\mathbf{b}\rvert$.

• $\mathbf{a}\cdot\mathbf{b}>0$: the angle is acute.
• $\mathbf{a}\cdot\mathbf{b}=0$: the vectors are perpendicular.
• $\mathbf{a}\cdot\mathbf{b}<0$: the angle is obtuse.

A common misconception is that the scalar product produces another vector. It produces one real number.

Exam Technique

On AA HL Paper 1 or Paper 2, show the component multiplication before giving the answer. When finding an angle, write the scalar-product equation first, substitute the magnitudes, and check the specified angle unit. Do not confuse the scalar product with the cross product, which produces a vector.