How Do You Find the Angle Between Two Lines Using Vectors?

Use the direction vectors of the two lines and apply the scalar product formula. For direction vectors $\mathbf{b}$ and $\mathbf{d}$, the acute angle $\theta$ satisfies $\cos\theta=\dfrac{|\mathbf{b}\cdot\mathbf{d}|}{|\mathbf{b}||\mathbf{d}|}$.

The Method

In Mathematics: Analysis and Approaches HL, sub-topic AHL 3.14, a line is commonly written as

$$\mathbf{r}=\mathbf{a}+\lambda\mathbf{b},$$

where $\mathbf{a}$ is a position vector and $\mathbf{b}$ is its direction vector. Only direction vectors determine the angle between two lines; the position vectors determine where the lines are located and do not affect the angle.

For direction vectors $\mathbf{b}$ and $\mathbf{d}$:

1. Calculate $\mathbf{b}\cdot\mathbf{d}$.
2. Calculate $|\mathbf{b}|$ and $|\mathbf{d}|$.
3. Substitute into

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

The absolute value gives the acute angle between the lines. It also ensures that reversing either direction vector does not change the answer. If the scalar product is zero, the lines are perpendicular; if $|\mathbf{b}\cdot\mathbf{d}|=|\mathbf{b}||\mathbf{d}|$, they are parallel.

These special cases follow directly from $\mathbf{b}\cdot\mathbf{d}=|\mathbf{b}||\mathbf{d}|\cos\theta$. They can help you verify whether a calculated angle is reasonable before giving your final answer.

Lines in three-dimensional space may be skew, meaning they are non-parallel and do not intersect. This does not prevent an angle calculation: the angle between them is still defined as the angle between their direction vectors. This definition is independent of their location.

Worked Example

Let

$$\mathbf{b}=\begin{pmatrix}1\\2\\2\end{pmatrix},\qquad \mathbf{d}=\begin{pmatrix}2\\-1\\2\end{pmatrix}.$$

First, calculate the scalar product:

$$\mathbf{b}\cdot\mathbf{d}=1(2)+2(-1)+2(2)=4.$$

The magnitudes are

$$|\mathbf{b}|=\sqrt{1^2+2^2+2^2}=3, \qquad |\mathbf{d}|=\sqrt{2^2+(-1)^2+2^2}=3.$$

Therefore,

$$\cos\theta=\frac{|4|}{3(3)}=\frac{4}{9}, \qquad \theta=\cos^{-1}\left(\frac{4}{9}\right)\approx63.6^\circ.$$

The common misconception is using position vectors instead of direction vectors. Another error is omitting the absolute value, which can produce the supplementary obtuse angle rather than the acute angle between the lines.

Exam Technique

This AA HL method can appear on Paper 1 or Paper 2. Show the scalar product and both magnitudes before finding $\theta$. On a GDC paper, check the calculator is in the required angle mode and give a numerical answer to three significant figures unless instructed otherwise.