Angle between Vectors

Before we see how to find the angle between two intersecting lines in three dimensional space, it is essential that we first look at how to find the angle between two vectors.

I like to think of angle between two vectors as the measure of “how far away” the direction of one vector arrow is with respect to the direction of the other vector arrow, if that makes any sense. It measures how much ‘off direction’ one vector arrow is from the direction of the other vector arrow. 

If the vectors are in the same direction, angle between them is 0°. In the above picture, b⃗ is in the same direction as a⃗. 

c⃗ is just 10° off from the direction in which a⃗ is pointing, while the direction of d⃗ is 45° off from the direction of a⃗. 

f⃗ is pointing exactly opposite to the direction of a⃗. Vectors in opposite directions will make a 180° angle. The direction of e⃗ is 90° off from the direction of a⃗. It is a vector perpendicular to a⃗. 

With ‘angle between vectors’, we are basically asking by how much angle should we rotate one vector so that it is pointing in the direction of the other vector. This analogy is particularly useful when the vectors are in two separate locations in the coordinate system or when their tails are not together.

You can also think of it like this: 

A vector in general has no fixed location in space. Position is not a characteristic of a vector arrow. The  arrow is “movable” and can be positioned anywhere in 3D space as long as we are not changing its length and its direction. 

When we “move” one or both vector arrows and reposition them such that their tails are together, we say that the vectors are ‘tail-to-tail’. The smaller of the two angles formed at their tails is the angle between the two vectors. 

In the figure above, 30° is the least of the two angles between u⃗ and v⃗ tail-to-tail. The smallest angle between two vectors will always be in the range 

0° ≤ 𝜃 ≤ 180°. 

Finding Angle between Vectors

There are two methods to find the angle between two vectors :

Using Cosine Rule of Triangles

Using Dot Product

Using Dot Product Formula

Finding angle by taking the dot product of vectors is the easiest of the two ways provided that we remember the dot product formula(or formulae) well.

The main formula is : 

u⃗ · v⃗ =  | u⃗ | | v⃗ | cos 𝜃

where, 

𝜃 is the smallest angle between vector u⃗ & v⃗, tail to tail. 0° ≤ 𝜃 ≤ 180°.

u⃗ · v⃗ is the dot product of the vectors. More details about dot product here. 

| u⃗ | & | v⃗ | are the lengths of the two vector arrows or the magnitudes of the vectors.

There is another relation which we need here that relates the dot product with the three scalar components of vectors. Three scalar components are basically the size of the vector arrow along the three coordinate axes. Here’s an example. 

The scalar components of a vector are the coefficients of \(\hat{i}\), \(\hat{j}\) and \(\hat{k}\). The dot product of the vectors, in terms of scalar components, will be equal to

u⃗ · v⃗ = (1)(1) + (1)(-1) + (2)\(\sqrt{\frac{6}{5}}\)

                       = 2 \(\sqrt{\frac{6}{5}}\)

We are basically multiplying scalar components of one vector with scalar components of the other vector along each axis, and summing the three products up. Be careful to not exclude the negative sign! 

The dot product of vectors is commutative. That’s just another way of saying that u⃗ · v⃗ is the same as v⃗ · u⃗. 

Also, the magnitudes of the vectors can be found from their scalar components :

| u⃗ | = \(\sqrt{1^2 + 1^2 + 2^2}\)

       = \(\sqrt{1 + 1 + 4}\) = \(\sqrt{6}\)

| v⃗ | = \(\sqrt{1^2+ (-1)^2 + \sqrt{\frac{6}{5}}^2}\)     …..(2) 

      = \(\sqrt{1 + 1 + \frac{6}{5}}\) = \(\sqrt{\frac{16}{5}}\)

      = \(\frac{4}{\sqrt{5}}\)

Going back to the above relation between theta and the dot product,

2 \(\sqrt{\frac{6}{5}}\) = (\(\sqrt{6}\))(\(\frac{4}{\sqrt{5}}\)) cos 𝜃

So, cos 𝜃 = \(\frac{1}{2}\)

⇒ 𝜃 = \(cos^{-1}(\frac{1}{2})\)

                 = 60° 

              =   \(\frac{\pi}{3}\)(in radians)

In general, \(cos^{-1}\) of 1/2 is infinite solutions of the form 2n\(\pi\) ± \(\frac{\pi}{3}\) n ∈ Ζ, for different values of integer n. But we want the angle 𝜃 in the range 0° ≤ 𝜃 ≤ 180°. And hence 𝜃 is 60°.

So all in all, to find the angle between two vectors using the Dot Product we only require the three scalar components of each vector to be provided to us.