# Question Video: Find the Unit Vector in the Direction of a Vector Defined by Two Points

Find the unit vector in the direction of 𝐀𝐁, given 𝐀 = ⟨0, 1, −2⟩ and 𝐁 = ⟨1, 1, 2⟩.

03:51

### Video Transcript

Find the unit vector in the direction of the vector 𝐀𝐁, given that vector 𝐀 is zero, one, negative two and the vector 𝐁 is one, one, two.

In this question, we’re given two vectors: the vector 𝐀 and the vector 𝐁. We need to determine the unit vector which has the same direction as the vector of 𝐀𝐁. And to do this, let’s start by recalling what we mean by a unit vector. We say that any vector which has magnitude one is a unit vector, and we often represent this with a hat notation. For example, 𝐀 hat would be the vector in the same direction as vector 𝐀; however, its magnitude would be equal to one. So when the question asks us to find the unit vector in the direction of 𝐀𝐁, we could have also written this as 𝐀𝐁 hat. It’s the vector in the same direction as 𝐀𝐁. However, its magnitude will be equal to one.

We also know a formula for finding the unit directional vector pointing in the same direction as the vector 𝐯. 𝐯 hat will be equal to one over the magnitude of 𝐯 multiplied by the vector 𝐯. And in our case, we’re going to want to apply this to the vector 𝐀𝐁. So first, we’re going to need to find the vector 𝐀𝐁. First, recall the vector from 𝐀 to 𝐁 is going to be the vector 𝐁 minus the vector 𝐀. That’s the vector one, one, two minus the vector zero, one, negative two. And remember, to subtract two vectors with the same number of components, we need to subtract each of our corresponding components separately. This gives us the vector one minus zero, one minus one, and two minus negative two.

And now we can just calculate each of these components separately. We get the vector 𝐀𝐁 is the vector one, zero, four. Now we’re going to want to apply our formula to find the unit directional vector which points in the same direction as 𝐀𝐁. We get that 𝐀𝐁 hat is going to be equal to one over the magnitude of 𝐀𝐁 multiplied by our vector 𝐀𝐁. But we can’t evaluate this yet because we don’t know the magnitude of our vector 𝐀𝐁. So before we do this, we’re going to need to find the magnitude of vector 𝐀𝐁. And to do this, we’re going to need to recall how we find the magnitude of a vector. The magnitude of a vector is the square root of the sums of the squares of its components. So the magnitude of the vector 𝐚, 𝐛, 𝐜 will be the square root of 𝐚 squared plus 𝐛 squared plus 𝐜 squared.

We can then use this to find the magnitude of our vector 𝐀𝐁. Our value of 𝐚 is one, our value of 𝐛 is zero, and our value of 𝐜 is four. We need to take the square root of the sums of the squares of these components. This gives us the square root of one squared plus zero squared plus four squared. And if we calculate this, we see it’s equal to the square root of 17. Now, we can substitute both of these into our equation for 𝐀𝐁 hat. We get that 𝐀𝐁 hat is equal to one over root 17 multiplied by the vector one, zero, four. Now, we could leave our answer like this. However, we can simplify this because we can notice we’re multiplying a constant by a vector. So we can simplify this by using scalar multiplication of a vector.

We recall to multiply a scalar by a vector, we need to multiply all of our components by our scalar. In other words, 𝑘 multiplied by the vector 𝐚, 𝐛, 𝐜 is equal to the vector 𝑘𝐚, 𝑘𝐛, 𝑘𝐜. So we need to multiply all of the components of our vector by our scalar one over root 17. This gives us the vector one over root 17 times one, one over root 17 times zero, one over root 17 times four. And we can evaluate each of these components. We get one over root 17, zero, four over root 17. And we could just leave our answer like this. However, we can also simplify the components of our vector by rationalizing their denominators.

In our first and third component, we want to multiply by the vector root 17 divided by root 17. And then evaluating this and simplifying, we get our final answer which is the vector root 17 over 17, zero, four root 17 over 17. Therefore, we were able to show, given the vector 𝐀 is zero, one, negative two and the vector 𝐁 is one, one, two, we were able to find the vector 𝐀𝐁 by subtracting the vector 𝐀 from our vector of 𝐁. Then we were able to find the unit vector in the same direction as 𝐀𝐁 by multiplying 𝐀𝐁 by one over its magnitude. This gave us 𝐀𝐁 hat is equal to the vector root 17 over 17, zero, four root 17 over 17.