# Lesson Video: Angle between Two Vectors in Space Mathematics • 12th Grade

In this video, we will learn how to find the angle between two vectors in space using their dot product.

17:58

### Video Transcript

Angle between Two Vectors in Space

In this video, we’re going to learn how we can find the angle between any two vectors in space by using the dot product. And we’ll see how to do this in a few situations, for example, given the component forms of a vector or given a graphical representation of the vectors.

To do this, we’re first going to need to recall a couple of facts about vectors. First, we know how to find the dot product of two vectors of equal dimensions. If 𝐮 is the vector with components 𝐮 one, 𝐮 two, up to 𝐮 𝑛 and 𝐯 is the vector with components 𝐯 one, 𝐯 two, up to 𝐯 𝑛, then the dot product between 𝐮 and 𝐯 is equal to the sum of the products of the corresponding components. 𝐮 dot 𝐯 is 𝐮 one 𝐯 one plus 𝐮 two 𝐯 two and we sum all the way up to 𝐮 𝑛 times 𝐯 𝑛. And we’ve seen a few different ways we can apply the dot product.

For example, if 𝜃 is the angle between vectors 𝐮 and 𝐯, then we know the cos of 𝜃 will be equal to the dot product between 𝐮 and 𝐯 divided by the magnitude of 𝐮 times the magnitude of 𝐯. And it’s worth pointing out there’s a second way of viewing this formula. If we let 𝐮 hat be the unit vector pointing in the same direction as vector 𝐮 and 𝐯 hat be the unit vector pointing in the same direction as vector 𝐯 — so 𝐮 hat is 𝐮 divided by the magnitude of 𝐮 and 𝐯 hat is 𝐯 divided by the magnitude of 𝐯 — then the cos of 𝜃 will also be equal to the dot product between 𝐮 hat and 𝐯 hat. This gives us a nice geometric interpretation of the dot product.

This formula is what we’re going to use to find the angle between our two vectors. We’ll calculate this expression and then take the inverse cos of both sides of the equation. However, there is one thing worth pointing out here about our value of 𝜃. We recall if we’re working in degrees, then the inverse cosine function will have a range between zero and 180 degrees inclusive. So if we only take the inverse cos of this expression, our answer will always be between zero and 180 degrees inclusive.

And this has a useful result geometrically. If we sketch the vectors 𝐮 and 𝐯 starting at the same point, then by using this formula to find the value of 𝜃, we will always get the smaller angle. And of course, we can find the other angle directly from the sketch. These two angles sum to give us 360. So its angle will be 360 degrees minus 𝜃. And an alternative method of seeing why this is true is to think about what happens when we take the inverse cos of both sides of the equation.

We know there’s multiple solutions for this. And we also know that if 𝜃 is a solution to this, then 360 minus 𝜃 is also a solution because the cos of 𝜃 is equal to the cos of 360 minus 𝜃. The last thing we’re going to point out is that all of what we have just discussed is true if instead we were working in radians. However, our values of 𝜃 would range between zero and 𝜋 instead. Let’s now see some examples of how we’re going to apply this to find the angle between two vectors.

Given that the modulus of vector 𝐀 is 35 and the modulus of vector 𝐁 is 23 and the dot product between 𝐀 and 𝐁 is equal to negative 805 root two divided by two, determine the measure of the smaller angle between the two vectors.

In this question, we’re given some information about vectors 𝐀 and 𝐁. And we’re asked to determine the smaller angle between these two vectors. Sometimes in these questions we like to sketch a picture of what’s happening. However, the information we’re given about our vectors won’t allow us to sketch a picture. We don’t know the components of vectors 𝐀 and 𝐁. Instead, we only know their modulus and their dot product.

So we’re going to need to rely entirely on our formula. Remember, this tells us if 𝜃 is the angle between two vectors 𝐀 and 𝐁, then the cos of 𝜃 will be equal to the dot product between 𝐀 and 𝐁 divided by the modulus of 𝐀 times the modulus of 𝐁. And we would find the value of 𝜃 by taking the inverse cosine of both sides of this equation. And this gives us a useful result because the inverse cosine function has a range between zero and 180 degrees.

Therefore, it doesn’t really matter how we draw our vectors 𝐀 and 𝐁. If the value of 𝜃 is between zero and 180, it will always give us the smaller angle between these two vectors. The only possible caveat to this would be if of our vectors point in exactly opposite directions. Then the angle measured in both directions will be equal to 180 degrees. However, as we’ll see, that’s not what’s happening in this question.

Let’s now find the smaller angle between our two vectors 𝐀 and 𝐁. It solves the equation the cos of 𝜃 will be equal to the dot product between 𝐀 and 𝐁 divided by the modulus of 𝐀 times the modulus of 𝐁. In the question, we’re told the dot product between 𝐀 and 𝐁 is equal to negative 805 root two over two, the modulus of 𝐀 is equal to 35, and the modulus of 𝐁 is equal to 23. So we can substitute these values directly into our formula, giving us the cos of 𝜃 is negative 805 root two over two all divided by 35 times 23.

We can simplify this. Remember, dividing by a number is the same as multiplying by the reciprocal of that number, giving us the cos of 𝜃 is negative 805 root two divided by two times 35 times 23. And if we were to evaluate 35 times 23, we would see it’s exactly equal to 805. So we can cancel these, leaving us with the cos of 𝜃 is equal to negative root two over two.

And finally, we can solve for our value of 𝜃 by taking the inverse cos of both sides of the equation. Remember, we know this will give us the smaller angle between our two vectors. This gives us 𝜃 is the inverse cos of negative root two over two, which we can calculate is 135 degrees.

Let’s now see an example of how we would calculate the angle between two vectors given their component forms.

Find the angle 𝜃 between the vectors 𝐕 four, two, negative one and 𝐖 eight, four, negative two.

In this question, we’re given two vectors in component form. And we’re asked to find the angle 𝜃 between these two vectors. To do this, we know a formula for finding the angle between any two vectors. We recall if 𝜃 is the angle between vectors 𝐕 and 𝐖, then 𝜃 satisfies the equation the cos of 𝜃 is equal to the dot product between 𝐕 and 𝐖 divided by the modulus of 𝐕 times the modulus of 𝐖.

And since we’re given 𝐕 and 𝐖 in component form, we can calculate all of these values. So we can find our value of 𝜃. Let’s start by calculating the dot product between 𝐕 and 𝐖. So we want to find the dot product between the vectors four, two, negative one and eight, four, negative two. Remember, to find a dot product between two vectors, we need to multiply the corresponding components together and then add all of these together.

Multiplying the first components of each vector together, we get four times eight. Multiplying the second components, we get two times four. And multiplying the third components, we get negative one times negative two. So the dot product of these two vectors will be the sum of these three products. And we can evaluate this expression. We get the dot product between 𝐕 and 𝐖 is 42.

However, this is not the only thing we need to calculate. We also need to find the modulus of 𝐕 and the modulus of 𝐖. To do this, we’re first going to need to recall how we find the modulus of a vector. Remember, the modulus of a vector is the square root of the sum of the squares of its components. In other words, the modulus of the vector 𝐚, 𝐛, 𝐜 is the square root of 𝐚 squared plus 𝐛 squared plus 𝐜 squared. And we know the components of 𝐕 are four, two, and negative one. So the modulus of 𝐕 is the square root of four squared plus two squared plus negative one squared.

And if we evaluate this expression, we see it’s equal to the square root of 21. We can then do exactly the same thing to find the modulus of 𝐖. It’s equal to the square root of eight squared plus four squared plus negative two all squared. And if we were to evaluate and simplify this expression, we would see that the modulus of 𝐖 is two root 21. Now that we’ve found the dot product between 𝐕 and 𝐖 and the modulus of 𝐕 and the modulus of 𝐖, we can substitute these into our equation involving 𝜃.

We showed the dot product between 𝐕 and 𝐖 is 42, the modulus of 𝐕 is root 21, and the modulus of 𝐖 is two root 21. Therefore, the cos of 𝜃 is 42 over root 21 times two root 21. However, if we start evaluating this expression, we see something interesting. In the denominator of this expression, root 21 multiplied by two root 21 simplifies to give us 42. And 42 over 42 is one. So our entire equation simplifies to give us the cos of 𝜃 is equal to one, and we can solve for 𝜃. We take the inverse cosine of both sides of the equation, giving us 𝜃 is the inverse cos of one, which we know is zero degrees.

And we could stop here; however, this gives us a useful piece of information. If the angle between 𝐕 and 𝐖 is zero degrees, then 𝐕 and 𝐖 point in the same direction. In other words, we’ve also shown that the vectors 𝐕 and 𝐖 are parallel. And in fact, we could directly prove this. We would see that our vector 𝐖 is just two times the vector 𝐕. And there’s a useful result we can get from this. Because our scalar is positive, the angle between these two vectors will be zero. However, if this scalar was negative, then the angle between them would be 180 degrees because then our vectors would point in exactly opposite directions. Either way, we were able to show the angle 𝜃 between the vectors 𝐕 and 𝐖 given to us in the question was zero degrees.

Let’s see another example of finding the angle between two vectors.

Find the angle 𝜃 between the vectors 𝐯 is equal to 𝐢 and 𝐰 is equal to three 𝐢 plus two 𝐣 plus four 𝐤. Give your answer correct to two decimal places.

In this question, we’re given two vectors 𝐯 and 𝐰 in terms of the unit directional vectors 𝐢, 𝐣, and 𝐤. And we’re asked to find the angle 𝜃 between these two vectors. And we need to give our answer to two decimal places.

To do this, we can start by recalling we have a formula to find the angle between two vectors. Since 𝜃 is the angle between vectors 𝐯 and 𝐰, the cos of 𝜃 must be equal to the dot product between 𝐯 and 𝐰 divided by the magnitude of 𝐯 times the magnitude of 𝐰. So to find the value of 𝜃, we need to find the dot product between 𝐯 and 𝐰, the magnitude of 𝐯, and the magnitude of 𝐰. Then all we’ll need to do is take the inverse cosine of both sides of the equation.

There’s several different ways of doing this. For example, we could work directly with the unit directional vector notation for 𝐯 and 𝐰. However, we could also write these vectors component-wise by taking the coefficients of the unit directional vectors. Either method will work; it’s personal preference which we’ll use.

We’ll write 𝐯 and 𝐰 component-wise. 𝐯 is the vector one, zero, zero and 𝐰 is the vector three, two, four. Let’s now start finding the values of our equation. Let’s start with the dot product between 𝐯 and 𝐰. Remember, to find the dot product of two vectors, we need to find the product of the corresponding components and then add the results together. In this case, the first component of 𝐯 times the first component of 𝐰 is one times three, the second component of 𝐯 times the second component of 𝐰 is zero times two, and the third component of 𝐯 times the third component of 𝐰 is zero times four.

So the dot product is the sum of these, one times two plus zero times two plus zero times four. And we can just calculate this expression. The second and third terms are zero. So this just gives us three. We now want to find the magnitude of our two vectors. Let’s start with the magnitude of 𝐯. Remember, 𝐯 is the unit directional vector 𝐢. And remember, 𝐢 is a unit directional vector. It has magnitude one.

Now to find the magnitude of 𝐰 is more difficult. So we’ll write this in component notation. And we’ll recall to find the magnitude of a vector, we need to find the square root of the sum of the squares of the components. So the magnitude of the vector 𝐚, 𝐛, 𝐜 will be the square root of 𝐚 squared plus 𝐛 squared plus 𝐜 squared. So for vector 𝐰, this will be the square root of three squared plus two squared plus four squared, which we can calculate is equal to the square root of 29.

Now that we found these values, we can substitute them into our equation for 𝜃. We showed the dot product between 𝐯 and 𝐰 is three, the magnitude of 𝐯 is one, and the magnitude of 𝐰 is root 29. So we must have the cos of 𝜃 is three divided by one times the square root of 29. And now we can solve the 𝜃 by taking the inverse cos of both sides of our equation. This gives us that 𝜃 is the inverse cos of three divided by root 29. And if we calculate this and round our answer to two decimal places, we see that 𝜃 is 56.15 degrees.

Let’s now see an example of how we would find the angle between two vectors given in a diagram.

Find the angle between the vectors shown in the following diagram.

In this question, we need to find the angle between two vectors. And we’re given these vectors on a diagram. And we can also see the angle between the two vectors given on our diagram. We have a few different options of how we could calculate this. For example, we could just do this by using trigonometry. However, we also have a formula for finding the angle between two vectors. We recall if 𝜃 is the angle between two vectors 𝐮 and 𝐯, then the cos of 𝜃 will be equal to the dot product of 𝐮 and 𝐯 divided by the magnitude of 𝐮 times the magnitude of 𝐯. We can then use this to solve for the value of 𝜃 by taking the inverse cosine of both sides of the equation.

So to answer this question, we’re going to need to find the dot product between our vectors 𝐮 and 𝐯 and the magnitude of 𝐮 and the magnitude of 𝐯. And to do this, we’re going to want to write our vectors in component form. We’ll do this by using the diagram. Let’s start with our vector 𝐮. We could see on our diagram it starts at the origin and then at the endpoint it has an 𝑥-coordinate of negative two root three. So the change in 𝑥 is negative two root three. Similarly, we can see its 𝑦-coordinate starts at zero and ends at two and its change in 𝑦 is two. So we can represent 𝐮 as the vector with horizontal component negative two root three and vertical component two.

We can do exactly the same for vector 𝐯. We can see it starts at the origin and then ends at an 𝑥-coordinate of negative two and it starts at the origin and ends at a 𝑦-coordinate of negative two. So the change in 𝑥 is negative two, and the change in y is negative two. So 𝐯 is the vector negative two, negative two. Now we need to find the dot product of these two vectors and their magnitudes.

Let’s start by calculating the dot product between 𝐮 and 𝐯. Remember, to find the dot product of two vectors, we need to find the products of corresponding components and then add all of these together. So we multiply the first components of 𝐮 and 𝐯 together to give us negative two root three times negative two. And then we add the products of their second components. That’s two multiplied by negative two. And if we calculate this expression, we see it’s equal to four root three minus four.

But we’re not done yet. We still need to find the magnitude of 𝐮 and the magnitude of 𝐯. Let’s start by finding the magnitude of 𝐮. Remember, we can find this by taking the square root of the sums of the squares of the components. So the magnitude of 𝐮 is the square root of negative two root three all squared plus two squared, which, we can calculate, gives us the square root of 12 plus four, which is root 16, which is of course just equal to four. We can then do exactly the same to find the magnitude of 𝐯. We square each component of 𝐯, add these together, and take the square root. The magnitude of 𝐯 is the square root of negative two squared plus negative two squared, which of course simplifies to give us the square root of four plus four, which is equal to root eight.

And now that we found all of these values, we’re ready to substitute them into our equation for 𝜃. Substituting in 𝐮 dot 𝐯 is four root three minus four, the magnitude of 𝐮 is four, and the magnitude of 𝐯 is root eight, we get the cos of 𝜃 is four root three minus four all divided by four root eight. And it’s worth pointing out here we can simplify this expression to give us root six minus root two all divided by four. However, it’s not necessary because all we need to do now is take the inverse cosine of both sides of the equation.

This gives us that 𝜃 is the inverse cos of root six minus root two all divided by four, which, we can calculate, gives us 75 degrees. And this is our final answer because if we look in our diagram, there are two possible angles between vectors 𝐯 and 𝐮. There’s the angle shown in our diagram and there’s the angle we could take in the opposite direction. However, this secondary angle shown in green is bigger than 75 degrees, so it can’t possibly be 75 degrees. In fact, it would be 360 minus 75 degrees. Therefore, we were able to show the angle between the two vectors in our diagram 𝐮 and 𝐯 is given by 75 degrees.

Let’s now go through one last example of how we can use our formula to find information about vectors.

The angle between vector 𝐀 and vector 𝐁 is 22 degrees. If the magnitude of vector 𝐀 is equal to three times the magnitude of vector 𝐁 is equal to 25.2, find the dot product between 𝐀 and 𝐁 to the nearest hundredth.

In this question, we’re given some information about two vectors 𝐀 and 𝐁. First, we’re told the angle between these two vectors is equal to 22 degrees. Next, we’re also told information about their magnitudes. We know the magnitude of 𝐀 is equal to 25.2, and we know that three times the magnitude of 𝐁 is also equal to 25.2. So the magnitude of 𝐀 is three times bigger than the magnitude of 𝐁. We need to use this to find the dot product of 𝐀 and 𝐁. And we need to give our answer to the nearest hundredth.

To answer this question, we need to notice that we know a formula which connects the angle between two vectors with their dot product. We recall if 𝜃 is the angle between two vectors 𝐀 and 𝐁, then we know that the cos of 𝜃 must be equal to the dot product between 𝐀 and 𝐁 divided by the magnitude of 𝐀 times the magnitude of 𝐁. And in this question, we already know some of these values. For example, we’re told the angle between our two vectors is 22 degrees. Next, we’re also told the magnitude of 𝐀 is equal to 25.2.

And we could also find the magnitude of 𝐁 using the information given to us in the question. One way of doing this is to notice that three times the magnitude of 𝐁 is equal to 25.2. We can then solve this to find the magnitude of 𝐁 by dividing both sides of our equation through by three. And calculating this, we get the magnitude of 𝐁 is 8.4. So, in fact, the only unknown in this equation is the dot product between 𝐀 and 𝐁. And that’s exactly what we’re asked to calculate.

So we’ll substitute the angle of 𝜃 equal to 22 degrees, the magnitude of 𝐀 equal to 25.2, and the magnitude of 𝐁 equal to 8.4 into our equation. This gives us the cos of 22 degrees should be equal to the dot product between 𝐀 and 𝐁 divided by 25.2 times 8.4. And now we can just rearrange this equation for the dot product between 𝐀 and 𝐁. We multiply through by 25.2 times 8.4. This gives us 𝐀 dot 𝐁 is 25.2 times 8.4 times the cos of 22 degrees. And we can calculate this to the nearest hundredth or to two decimal places. It’s equal to 196.27.

Let’s now go over the key points of this video. First, we know if 𝜃 is the angle between two vectors 𝐮 and 𝐯, then the cos of 𝜃 will be equal to the dot product of 𝐮 and 𝐯 divided by the magnitude of 𝐮 times the magnitude of 𝐯. And this works so long as neither vector 𝐮 or 𝐯 is equal to zero. And to take the dot product of 𝐮 and 𝐯, we need them to have the same dimension. We can also use this formula to find the angle between two vectors by taking the inverse cosine of both sides of this equation. But we do need to be careful because the inverse cosine function has a range between zero and 180 degrees inclusive, or if we’re working in radians, zero to 𝜋 inclusive. So this method will give us the smaller of the two angles between our vectors 𝐮 and 𝐯.