Lesson Video: Projections | Nagwa Lesson Video: Projections | Nagwa

# Lesson Video: Projections Mathematics • Second Year of Preparatory School

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In this video, we will learn how to find the projection of a point, a line segment, a ray, or a line on another line and find the length of the projection.

15:13

### Video Transcript

In this video, we will learn how to find the projection of a point, a line segment, a ray, or a line on another line and find the length of the projection. But before we do that, let’s go ahead and review the definition of points, lines, line segments, and rays.

In math, a point is an exact location. It has no size, only position. However, as that would be very difficult to draw, we represent a point as a dot. A point has no dimensions. Moving onto a line, in geometry, a line is straight; it has no bends. It also has no thickness and extends in both directions without end. A line has one dimension. A ray is very similar to a line, except that a ray has an endpoint on one end. In other words, a ray extends only in one direction infinitely. We should also consider a line segment. A line segment has two endpoints. It does not extend infinitely in either direction.

We also want to consider two different relationships that lines, rays, and line segments can have with one another. Perpendicular means at a right angle to. If two lines are perpendicular, they intersect at a 90-degree angle. This, of course, could be true of a ray and a line, a line segment and a line, a line segment and a ray, or any combination of these types. We also want to consider what parallel means. Parallel lines are always the same distance apart, and they never intersect. And again, these same properties would apply to rays and line segments. Using these concepts, let’s explore projection.

First, we’ll explore the idea of projecting a point onto a line. Here we have some point, and we have a line some distance away from the point. To find a projection of this point onto the line, we can imagine a light source that is perpendicular to our target line. If we think back to the image on the opening screen, we can imagine that that light source is a laser. The key point here is that laser light beam needs to be perpendicular to our target line. Once we have a light source that is perpendicular to our target line, we want to imagine what kind of shadow doing this will create on our target line. If this laser perpendicular to the target line goes through our point, the shadow created would be another point. The shadow of this point onto this line further away is just a point. So where that shadow would fall is the projection.

If this is what the projection of a point onto a line looks like, what do you think the projection of a line segment onto a line will look like? Again, we’ll have a target line and some line segment that we want to project onto our target line. Just like when we were projecting a point onto a line, we’ll need to consider the light source being perpendicular to our target line. Doing this, we can see that our endpoint from our first line becomes part of the projection. However, when we’re dealing with a line segment, we should think about the light source being all the parallel lines that are perpendicular to the target line. Once we do that, we see that our projection has two endpoints, but that the full projection, the full shadow, would be another line segment. The projection of a line segment onto a line will be another line segment.

Notice that, at least in this case, it looks like our target line and our line segment are parallel. However, this will not always be the case. Let’s imagine this new line segment that we want to project onto the same target line. We can visualize this with a light source that is perpendicular to our target line. There’re more light sources that are all parallel to this perpendicular line. Doing this, we can see the endpoints of our projection, and the shadow would be the segment between these two endpoints. Again, we see that a projection of a line segment onto a line will be a line segment. In this case, the projection is a different length than the original line segment; it’s slightly shorter. But we’ll look at an example of how to calculate these lengths later.

Before we look at example problems, let’s consider one other case. We want to consider what happens if we project a ray onto a line. Again, we’ll have our target line, and then we have a ray that we want to project onto our target line. Remember, to do this visualization, we want the line of the light source to be perpendicular to our target line. And our light source can also include any lines parallel to the initial one, that is to say, all of the lines that are perpendicular to our target line.

So initially we see the projection of the endpoint. But we need to think carefully about this ray because a ray does not have a second endpoint. We see that this space would be part of the projection; it would be part of the shadow. Because a ray continues indefinitely on one side, we have to say the projection of that ray would continue indefinitely on one side. We could continue adding those light source lines all the way up. What we’re showing here is that a projection of a ray onto a line produces a ray.

Before we look at some examples, we need to make a quick note about target lines. In all of the examples, I’ve drawn the target line vertically, and that meant the line representing the light source. The line that has to be perpendicular to the target line was a horizontal line so that when we had a projection of the point, we got something like this. But a target line is not necessarily always vertical. If, for example, this is our target line, then our light source lines would look something like this, the key being that we find projections using perpendicular lines to the target line.

So if we wanted to project a line segment onto this target line, we get something like this. And if we had a horizontal target line which we wanted to project this ray onto, we could visualize that with a vertical light source. And we would get a protection that looks like this. Using all of this information, we’re ready to solve some example problems.

What is the projection of line segment 𝐴𝐷 onto line 𝐴𝐶?

First of all, we want to know what our target line is. This is the line on which our projection will fall. Since we want to know the projection of line segment 𝐴𝐷 onto line 𝐴𝐶, line 𝐴𝐶 is our target line. First, we identify line 𝐴𝐶, and then we need the segment that we want to make a projection of, which is 𝐴𝐷. To visualize a projection, we need to visualize a light source that is perpendicular to our target line. In this diagram, we already have a line segment that’s perpendicular to our target line. That’s the line segment 𝐷𝑀. We’ll use lines that are parallel to 𝐷𝑀 to find this projection.

If we want to know what kind of shadow this line segment 𝐴𝐷 would create, it would fall between 𝐴 and 𝑀 on our target line, with the line segment 𝐴𝑀 being its shadow. That shadow portion is the projection. And so we say that the projection of line segment 𝐴𝐷 onto line 𝐴𝐶 would be line segment 𝐴𝑀.

Let’s look at another example of a line segment being projected onto a line.

What is the projection of line segment 𝐴𝐷 onto line 𝐵𝐶?

When we’re working with these kind of projections, we always want to consider the target line first, the line on which the projection will fall. In this case, it’s the line 𝐵𝐶, so we’ll show that line 𝐵𝐶 is our target line. The segment we’re using that we want to project is 𝐴𝐷, this segment. We should be noticing that our line segment we want to make projection of is perpendicular to our target line. When we try to visualize a projection, we always use lines that are perpendicular to the target line. So how do we visualize line segment 𝐴𝐷?

Imagine that we have a laser light source where it’s shining a light perpendicular to our line 𝐵𝐶. What kind of shadow is it going to create? Well, when it hits point 𝐴, it’s just going to make a point as a projection. Its shadow will just be the point 𝐷. We might be able to visualize this a bit better if we take it out of the triangle. Once our light source hits point 𝐴, its shadow will just be a point. It will not be a line segment. And so the projection of line segment 𝐴𝐷 onto the line 𝐵𝐶 is the point 𝐷.

In our next example, we’ll use what we know about projections and what we know about right-angle triangles to find some missing lengths in a figure.

Given that 𝐴𝐵 equals 29, 𝐶𝐵 equals 20, and 𝐶𝐷 equals 35, calculate the length of the projection of line segment 𝐶𝐷 on line 𝐴𝐷.

Before we can calculate the length of this projection, we’ll need to see which line segment this projection is. Line 𝐴𝐷 is our target line. That’s the line upon which the projection will fall. And 𝐶𝐷 is the line segment that we’ll use to create our projection. But in order to create a projection, we’ll need perpendicular lines to our target line. Because we know that line 𝐵𝐶 and line 𝐴𝐷 are parallel, the angle 𝐵𝐶𝐴 is an alternate interior angle to the angle 𝐶𝐴𝐷, which means the line segment 𝐴𝐶 is perpendicular to the line segment 𝐴𝐷.

Once we imagine the light source as the set of all the perpendicular lines to our target line, we can find the endpoints of our projection. Starting at 𝐶, the projection of point 𝐶 onto 𝐴𝐷 would be the point 𝐴 and the projection of point 𝐷 onto 𝐴𝐷 would be itself. And that means the projection of 𝐶𝐷 is going to be equal to 𝐴𝐷. This is the value we want to calculate the length of. To do that, let’s clear out our light source lines and think about what we know about right triangles.

First of all, we know that 𝐴𝐵 has a measure of 29 and 𝐶𝐵 has a measure of 20 and 𝐶𝐷 has a measure of 35. What we have in our figure is two separate right triangles that make up this quadrilateral. And so we remember that we can find side lengths in right triangles using the Pythagorean theorem, where 𝑎 and 𝑏 represent the two smaller sides and 𝑐 represents the hypotenuse. 𝑎 squared plus 𝑏 squared equals 𝑐 squared. To use the Pythagorean theorem, you need at least two of the lengths. Since we don’t know the length of 𝐴𝐶, we can’t find the length of 𝐴𝐷. However, we can use the information about the smaller triangle 𝐴𝐵𝐶 to solve for the side length 𝐴𝐶 first.

When we plug in what we know, we get 20 squared plus 𝐴𝐶 squared equals 29 squared. 400 plus 𝐴𝐶 squared equals 841. So we’ll subtract 400 from both sides of the equation, and we’ll get 𝐴𝐶 squared equals 441. After that, we’ll take the square root of both sides. We’re only interested in the positive square root since we’re dealing with distance, and so we see that 𝐴𝐶 equals 21. Now that we know that 𝐴𝐶 is 21, we know two distances in our right triangle and we’ll be able to find the side length of our third distance 𝐴𝐷. But we’ll need to set up the Pythagorean theorem for a second time.

This time, we’ll have 21 squared plus 𝐴𝐷 squared equals 35 squared. 441 plus 80 squared equals 1225. And so we subtract 441 from both sides, and we’ll get 𝐴𝐷 squared equals 784. So we take the square root of both sides. Again, we’re only interested in the positive square root of 784, which is 28. The line segment 𝐴𝐷 will be equal to 28. The line segment 𝐴𝐷 is the projection of the line segment 𝐶𝐷 onto line 𝐴𝐷, and it has a measure of 28.

Before we finish with this video, let’s go over key points. To find the projection of a point, line segment, or ray onto a target line, we consider the set of all lines perpendicular to the target line. Treating the set of perpendicular lines as a light source allows us to see the projection as the shadow of the original image onto the target line. In this figure, we have an image, our target line, the set of perpendicular lines which have created this projection.

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