Determine the unit vector of vector 𝐀 from the following graph.
In this question, we’re given a graphical representation of the vector 𝐀 and we need to determine the unit vector of our vector 𝐀. So the first thing we’re going to need to do is recall exactly what we mean by the unit vector of a vector 𝐀. This is usually represented 𝐀 hat. It’s a unit vector pointing in the same direction as vector 𝐀. And we have a few different ways we can find this. For example, we know a formula to find this for any nonzero vector 𝐀. 𝐀 hat will be equal to one over the magnitude of 𝐀 multiplied by vector 𝐀. And all this formula is really telling us is that we know vector 𝐀 points in the same direction as vector 𝐀. And if we multiply this by one over the magnitude of 𝐀, we’ll now have a unit vector. And we won’t change the direction our vector is pointing because the magnitude of 𝐀 is positive.
However, this isn’t the only way we can answer this question. And in fact, because we’re given a graphical representation of our vector 𝐀, we can try and do this graphically. We know that our vector 𝐀 hat needs to point in the same direction as vector 𝐀. So we can start our vector 𝐀 hat from the origin. And we know that it must point in exactly the same direction as vector 𝐀.
However, it’s important to realize that vector 𝐀 hat is going to be a unit vector. And remember, the magnitude of a vector represents the length of this vector graphically. So if we’re starting our vector 𝐀 hat at the origin and its magnitude is equal to one, then its end point must lie on the unit circle centered at the origin. And this is enough to find the vector 𝐀 hat. It points in the same direction as 𝐀 and has magnitude one.
However, this doesn’t yet give us the exact form of our vector 𝐀 hat. There’s a few different ways we could find this from our diagram. However, most of these are more complicated than the formula we already have. So we’ll just use this formula instead. We see, to use this formula, we need to find the magnitude of our vector 𝐀 and we need to multiply this by our vector 𝐀.
So let’s start by finding an expression for vector 𝐀. We can see we’re given a picture of vector 𝐀 graphically. And from this, we can find the horizontal and vertical components of vector 𝐀. We can see that 𝐀 starts at the origin and ends at a horizontal value of 12. So its horizontal component is 12. And 𝐀 starts at the origin and ends at a vertical position of five. So its vertical component is going to be five. Therefore, we can represent the vector 𝐀 as the vector 12, five.
Now, we need to find the magnitude of vector 𝐀. And there’s two different ways we could do this. We could use our formula for the magnitude. However, remember, graphically, the magnitude of a vector represents its length. And graphically, we can see our vector 𝐀 is the hypotenuse of a right-angled triangle with height five and width 12. Therefore, by using the Pythagorean theorem, we know the hypotenuse of this triangle or the magnitude of vector 𝐀 will be equal to the square root of 12 squared plus five squared. And in fact, this is a Pythagorean triple. If we calculate this, we see that this is equal to 13.
Now we’re ready to find an expression for our vector 𝐀 hat by using our formula. Our vector 𝐀 hat is going to be one over the magnitude of 𝐀 times vector 𝐀, which is one over 13 times the vector 12, five. And remember, to multiply a vector by a scalar, we multiply all of our components by our scalar. And doing this, we get the vector 12 over 13, five over 13.
And it’s worth pointing out here we can check this on our diagram. On our diagram, we’ve already found our vector 𝐀 hat. And we can check its horizontal and vertical components. On our diagram, it appears the horizontal component is approximately 0.9 and the vertical component is approximately 0.4. And if we calculate 12 over 13 and five over 13, we see this is very close. This justifies our answer.
Therefore, given a sketch of the vector 𝐀, we were able to determine the unit vector pointing in the same direction as 𝐀. We got the components of this vector were 12 over 13 and five over 13.