How Do You Find the Angle Between Two Vectors Using the Scalar Product?

Use the scalar product formula $\mathbf{a}\cdot\mathbf{b}=|\mathbf{a}||\mathbf{b}|\cos\theta$, then rearrange and apply inverse cosine. For two non-zero vectors,

$$\theta=\cos^{-1}\left(\frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}||\mathbf{b}|}\right).$$

This is AHL 3.13: Scalar and vector products, so it is HL-only content. This explanation assumes an HL short-response question; a GDC is available on every Mathematics: Applications and Interpretation paper.

The Method

For vectors $\mathbf{a}=(a_1,a_2,a_3)$ and :

1. Calculate the scalar product:

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

The sign of the scalar product also indicates the angle type. If $\mathbf{a}\cdot\mathbf{b}>0$, the angle is acute; if it equals zero, the vectors are perpendicular; if it is negative, the angle is obtuse.

Worked Example

Let $\mathbf{a}=(2,1,-2)$ and $\mathbf{b}=(1,3,2)$, with the angle requested in degrees.

$$\mathbf{a}\cdot\mathbf{b}=2(1)+1(3)+(-2)(2)=1$$ $$|\mathbf{a}|=\sqrt{2^2+1^2+(-2)^2}=3,\qquad |\mathbf{b}|=\sqrt{1^2+3^2+2^2}=\sqrt{14}$$

Therefore,

$$\theta=\cos^{-1}\left(\frac{1}{3\sqrt{14}}\right)=84.9^\circ$$

to three significant figures. The positive scalar product is consistent with this acute angle.

Before evaluating, check that neither vector is the zero vector. If either magnitude is zero, the denominator is zero and the angle is undefined. Also note that the formula returns the smaller angle between the vectors, from $0^\circ$ to $180^\circ$. In a geometric or modelling context, state what that angle represents rather than giving only a number.

As a quick check, compare the angle with the sign of the dot product. A mismatch usually reveals an arithmetic or GDC mode error.

A common misconception is to use the cross product. The scalar product produces a scalar involving $\cos\theta$, so it gives the angle directly. The cross product instead produces a vector perpendicular to both original vectors.

Exam Technique

Show the scalar product, both magnitudes, and the substituted inverse-cosine expression before writing the GDC value. Keep full precision until the final answer. At HL, use radians unless the question includes a degree symbol or explicitly requests degrees; a wrong GDC angle mode produces an incorrect answer.