Angle between two Intersecting Lines in 3D

In this post we will be looking at how to find the angle(s) between two intersecting lines.

Between any two intersecting lines in 2D or 3D, there are a total four angles formed at the point of intersection. All four are mutually related to one another. Each angle shares a simple relation with the other three. 

In the figure above, ∠a is vertically opposite to ∠c , and hence they are of equal measure. The other pair of vertically opposite angles is ∠b & ∠d. 

∠a = ∠c    &    ∠b = ∠d

Also, ∠a forms two linear pairs with its two adjacent angles, ∠b and ∠d. Linear pairs of angles are supplementary, meaning their sum equals 180°. 

 ∠a + ∠b = 180°

∠a + ∠d = 180°

So it all boils down to knowing the measure of just one out of four angles. The other three angles can then be found pretty easily. 

Now one method to find the measure of any one angle between two intersecting lines in 3D is from direction numbers of two lines. Direction numbers also go by the name of direction ratios. Three direction numbers of a line are the representative of the direction of that line in 3D space. For example, let’s say there is a line L1 in 3D space with direction numbers 1, 1, 2. It simply means that L1 is pointing in the direction of the vector arrow \(\hat{i} + 1\hat{j} + 2\hat{k}\). In simple terms, L1 is parallel to the vector arrow. Let’s name it \(\vec{u}\). 

In a little more accurate terms, one of the two opposite directions of line L1 is the same as the direction of \(\vec{u}\). 

Consider another line L2 intersecting L1 at point P. If 1, -1, \(\sqrt{\frac{6}{5}}\) is a set of direction numbers of L2, then it implies that line L2 is in the direction of the vector arrow \(\hat{i} - 1\hat{j} + \sqrt{\frac{6}{5}}\hat{k}\). Let’s name it \(\vec{v}\). \(\vec{u}\) & \(\vec{v}\) can be called direction vectors of lines L1 & L2.

Now the location or position is not a feature of a vector arrow. A vector arrow is “movable” and can be positioned or re-positioned anywhere in 3D space as long as we are not changing its length and/or direction, i.e., as long as we are not shrinking, extending or rotating it. 

So we can “move” vector arrow representing \(\vec{u}\), and put it on the line L1 such that the tail of the vector arrow is sitting on the point of intersection of the lines. Similarly, we can move vector arrow representing \(\vec{v}\), and put it on the line L2 such that its tail also sits on P. 

The two vector arrows are now positioned tail-to-tail(tail on tail). Let \(\theta\) be the angle between them. \(\theta\) also happens to be one of the four angles between the lines L1 & L2. So we have actually reduced the problem of finding an angle between two intersecting lines in 3D to finding the angle between two vectors. 

In my last post, i have already gone into some details explaining how to find the angle between two 3D vectors. You can check that out now if you want to. 

We can find the angle \(\theta\) between the two direction vectors by using the formula, 

\(cos \theta = \frac {\overrightarrow{u} \cdot \overrightarrow{v}}{|\overrightarrow{u}| |\overrightarrow{v}|}\)

                

 \(= \frac {ap + bq + cr }{\sqrt{a^2 + b^2 + c^2} \sqrt{p^2+ q^2 + r^2}}\)

……...where a, b & c are scalar components of \(\vec{u}\) and p, q & r are scalar 

                               components of \(\vec{v}\)

                                        0° ≤ \(\theta\) ≤ 180°

We will end up getting the measure of \(\theta\) as 60°. I won’t go into details on how we got this value because i have already done so in my previous post for the very same example of vectors.

So one of the angles between lines L1 & L2  measures 60°. So the measure of the other three angles will be 60°, (180° - 60°), (180° - 60°), i.e., 60°, 120°, 120°. 

To wrap it up, the formula for finding an angle between two lines in 3D is the same as the formula for finding the angle between two vectors. 

\(cos \theta = \frac {ap + bq + cr }{\sqrt{a^2 + b^2 + c^2} \sqrt{p^2+ q^2 + r^2}}\)

In the formula, a, b & c and p, q & r are scalar components of direction vectors of two lines. Or we can simply say that they are direction numbers of lines. 

You might also come across a slightly different version of this formula where the entire fraction on the right hand side will be put inside the modulus sign. 

\(cos \theta = = |\frac {ap + bq + cr }{\sqrt{a^2 + b^2 + c^2} \sqrt{p^2+ q^2 + r^2}}|\)

The formula remains the same for finding the angle between vectors, it is only for the angle between lines that you will see this subtle change. The reason for the Modulus sign is that generally, the angle between two intersecting lines is taken to be the smallest of the two distinct angles formed between the lines.

For any pair of intersecting lines, the smaller and the greater angles are a linear pair, like 60° and 120° for lines L1 and L2. And the cosine values of a supplementary pair of angles only differ in sign. 

For example, 

Cos 60° = \(\frac{1}{2}\)

Cos (180° - 60°) = cos 120° = - \(\frac{1}{2}\)

Cosine value of the bigger angle, which is obtuse, is always going to be negative. Suppose that instead of \(\vec{u}\) and \(\vec{v}\) we were given the direction vectors \(\vec{m}\) and \(\vec{n}\) for lines L1 and L2, as shown in the picture below. 

Using the dot product formula to find the angle between \(\vec{m}\) and \(\vec{n}\) will give us, cos \(\theta\) = - 1, a negative value which will then give us \(\theta\) = 120°, an obtuse angle or the greater of the two angles formed between the lines. 

But if we are agreeing to the textbook definition, we are looking for the smaller angle. One way to find it is simply subtract the larger angle from 180°; 180° - 120° = 60°. 

But we don’t need this additional step if we simply take the modulus of the cosine value in the dot product formula. By taking the Modulus, cos \(\theta\) = | -\(\frac{1}{2}\)| =\(\frac{1}{2}\), gives us \(\theta\) = 60°, the smaller of the two angles. 

I hope you get the point I am trying to make in the end. Though to be honest, I doubt that I have done a neat job clarifying it. Any thoughts or queries are welcomed in the comments.