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Lesson Explainer: Projections Mathematics

In this explainer, 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.

In general terms, we can think of a projection as the shadow casted by the object. This can be a useful visualization that has many real-world applications. For example, consider a ball that is thrown in an arc. We can project the ball onto the ground by considering its shadow with respect to the sun. Of course, the position of the sun will affect where the shadow is. We are only interested in vertical projection, which means the sun will be directly above in our example. We get the following.

It is useful to consider the projection of the ball on the ground in this case since we are often interested in the horizontal distance traveled by the ball rather than the total distance traveled. In this case, we can measure the distance traveled by its projection to the ground. In other words, the horizontal distance traveled by the ball is the same as the horizontal distance traveled by its shadow. We can formalize this definition for projecting a point onto a line as follows.

Definition: Projection of a Point onto a Straight Line

The projection of a point 𝐴 onto a straight line ⃖⃗𝐡𝐢 is the point π·βˆˆβƒ–οƒ©οƒ©οƒ©οƒ©βƒ—π΅πΆ such that βƒ–οƒ©οƒ©οƒ©οƒ©οƒ©βƒ—π΄π·βŸ‚βƒ–οƒ©οƒ©οƒ©οƒ©βƒ—π΅πΆ.

If π΄βˆˆβƒ–οƒ©οƒ©οƒ©οƒ©βƒ—π΅πΆ, then we say the projection of 𝐴 onto ⃖⃗𝐡𝐢 is just 𝐴.

With this definition in mind, we can think of finding the area of a triangle in another way. Recall that the area of a triangle is half the length of its base multiplied by its perpendicular height. If we consider a sketch where we choose 𝐡𝐢 as the base, we have the following.

We see that ⃖⃗𝐴𝐷 meets ⃖⃗𝐡𝐢 at right angles, so we can say that 𝐷 is the projection of 𝐴 onto ⃖⃗𝐡𝐢. Since 𝐴𝐷 is the perpendicular height of the triangle, we can also think about this as the length between 𝐴 and its projection onto the line through the base.

Let’s now see an example of finding the projection of a point onto a line.

Example 1: Finding the Projection of a Point on a Line

What is the projection of 𝐷 on ⃖⃗𝐴𝐢?

Answer

We recall that the projection of a point 𝐷 onto a straight line ⃖⃗𝐴𝐢 is the point 𝑃 on the line such that βƒ–οƒ©οƒ©οƒ©οƒ©βƒ—π‘ƒπ·βŸ‚βƒ–οƒ©οƒ©οƒ©οƒ©βƒ—π΄πΆ. We can see in the diagram that πΈβˆˆβƒ–οƒ©οƒ©οƒ©οƒ©βƒ—π΄πΆ and that βƒ–οƒ©οƒ©οƒ©οƒ©βƒ—πΈπ·βŸ‚βƒ–οƒ©οƒ©οƒ©οƒ©βƒ—π΄πΆ.

Hence, the projection of 𝐷 on ⃖⃗𝐴𝐢 is the point 𝐸.

We can extend the idea of the projection of a single point onto a line by considering how we could project a line segment onto a line. To go back to our ball example, we may want to know the horizontal distance traveled between two times. In this case, we would want to project both the start and endpoints of the ball’s trajectory onto the ground as shown.

In this scenario, we are interested in the line segment between the projections of the endpoints. This idea is how we project a line segment onto a straight line. We can describe this formally as follows.

Definition: Projection of a Line Segment onto a Straight Line

The projection of a line segment 𝐴𝐡 onto a straight line 𝐿 is the line segment 𝐴′𝐡′, where 𝐴′ and 𝐡′ are the projections of 𝐴 and 𝐡 onto 𝐿 respectively.

It is worth noting that if π΄π΅βŸ‚πΏ, then 𝐴′ and 𝐡′ will be the same point. In this case we say that the projection is the single point 𝐴′.

It is worth noting that, for any point 𝐢∈𝐴𝐡, if we name its projection onto 𝐿 as 𝐢′, then we must have πΆβ€²βˆˆπ΄β€²π΅β€².

Let’s now see an example of finding the projection of a line segment onto a given straight line.

Example 2: Finding the Projection of a Line Segment on a Line

What is the projection of 𝐴𝐷 on ⃖⃗𝐴𝐢?

Answer

We begin by recalling that the projection of a line segment 𝐴𝐷 onto a straight line is the line segment between the projections of the endpoints. Therefore, to project 𝐴𝐷 onto ⃖⃗𝐴𝐢 we first need to separately project 𝐴 and 𝐷 onto ⃖⃗𝐴𝐢.

We can then recall that the projection of a point 𝐷 onto a straight line ⃖⃗𝐴𝐢 is the point 𝑃 on the line such that βƒ–οƒ©οƒ©οƒ©οƒ©βƒ—π‘ƒπ·βŸ‚βƒ–οƒ©οƒ©οƒ©οƒ©βƒ—π΄πΆ.

We can see in the diagram that π‘€βˆˆβƒ–οƒ©οƒ©οƒ©οƒ©βƒ—π΄πΆ and that βƒ–οƒ©οƒ©οƒ©οƒ©οƒ©οƒ©βƒ—π·π‘€βŸ‚βƒ–οƒ©οƒ©οƒ©οƒ©βƒ—π΄πΆ. So, 𝑀 is the projection of 𝐷 onto ⃖⃗𝐴𝐢. Finally, we recall that, since π΄βˆˆβƒ–οƒ©οƒ©οƒ©οƒ©βƒ—π΄πΆ, its projection onto ⃖⃗𝐴𝐢 is itself.

Hence, the projection of 𝐴𝐷 on ⃖⃗𝐴𝐢 is 𝐴𝑀.

In our next example, we will consider the projection of a perpendicular line segment onto a straight line.

Example 3: Finding the Projection of a Line Segment Perpendicular to a Line

What is the projection of 𝐴𝐷 on ⃖⃗𝐡𝐢?

Answer

We begin by recalling that the projection of a line segment 𝐴𝐷 onto a straight line is the line segment between the projections of the endpoints. Therefore, to project 𝐴𝐷 onto ⃖⃗𝐡𝐢, we first need to separately project 𝐴 and 𝐷 onto ⃖⃗𝐡𝐢.

First, we can see that π·βˆˆβƒ–οƒ©οƒ©οƒ©οƒ©βƒ—π΅πΆ, hence its projection onto this line is itself. Next, we see in the diagram that π΄π·βŸ‚βƒ–οƒ©οƒ©οƒ©οƒ©βƒ—π΅πΆ, so the projection of 𝐴 onto this line is also 𝐷.

Therefore, since the line segment 𝐷𝐷 is just the point 𝐷, we have shown that projection of 𝐴𝐷 on ⃖⃗𝐡𝐢 is the point 𝐷.

In our next example, we will determine the length of the projection of a line segment.

Example 4: Finding the Length of the Projection of a Line Segment on a Line

Given that 𝐴𝐡=29, 𝐢𝐡=20, and 𝐢𝐷=35, calculate the length of the projection of 𝐢𝐷 on ⃖⃗𝐴𝐷.

Answer

We begin by recalling that the projection of a line segment 𝐢𝐷 onto a straight line is the line segment between the projections of the endpoints. Therefore, to project 𝐢𝐷 onto ⃖⃗𝐴𝐷, we first need to separately project 𝐢 and 𝐷 onto ⃖⃗𝐴𝐷.

We first note that π·βˆˆβƒ–οƒ©οƒ©οƒ©οƒ©οƒ©βƒ—π΄π·, so its projection onto this line is unchanged. To project 𝐢 onto ⃖⃗𝐴𝐷, we need to find a line perpendicular to ⃖⃗𝐴𝐷 that passes through 𝐢. We can do this by noting that ⃖⃗𝐴𝐷‖⃖⃗𝐡𝐢 and 𝐴𝐢 is a transversal of these parallel lines. Hence, angles ∠𝐡𝐢𝐴 and ∠𝐢𝐴𝐷 are alternate, so π‘šβˆ πΆπ΄π·=90.∘

Therefore, the projection of 𝐢 onto ⃖⃗𝐴𝐷 is 𝐴, so we need to determine 𝐴𝐷. This gives us the following.

We can determine this length by applying the Pythagorean theorem twice. First, on triangle △𝐴𝐢𝐡, we have 𝐴𝐡=𝐴𝐢+𝐢𝐡.

Substituting 𝐴𝐡=29 and 𝐢𝐡=20 into the equation gives us 29=𝐴𝐢+20.

We can then rearrange this to get 𝐴𝐢=29βˆ’20=441.

Taking the square root of both sides of the equation where we note that 𝐴𝐢 is a length and so is nonnegative gives us 𝐴𝐢=√441=21.

We can add this to the diagram.

We can now determine the length of 𝐴𝐷 by applying the Pythagorean theorem to △𝐢𝐴𝐷; we have 𝐢𝐷=𝐴𝐢+𝐴𝐷.

Substituting 𝐴𝐢=21 and 𝐢𝐷=35 into the equation gives us 35=21+𝐴𝐷.

We can then rearrange the equation and evaluate it to get 𝐴𝐷=35βˆ’21=784.

Finally, we take square roots of both sides of the equation to get 𝐴𝐷=√784=28.

A useful fact to note is that the length of the projection of a line segment is always less than or equal to the length of the original line segment. We can use this as a quick check for our answers. We can also prove this result by considering a projection.

We can show that 𝐴′𝐡′<𝐴𝐡 by translating line 𝐿 vertically to form a right triangle.

We can now see that 𝐴𝐡 is the hypotenuse of this right triangle, so 𝐴′𝐡′<𝐴𝐡. It is worth noting that we assume that one of 𝐴 or 𝐡 does not lie on 𝐿; otherwise, the projection leaves the line segment unchanged. This gives the following result.

Property: Projection Length of a Line Segment

The length of the projection of a line segment is always less than or equal to the length of the original line segment.

Let’s see an example of using this property after we find the length of the projection of a line segment to check our answer.

Example 5: Finding the Length of the Projection of a Line Segment on a Line

Find the length of the projection of 𝐴𝐸 on ⃖⃗𝐡𝐢.

Answer

We begin by recalling that the projection of a line segment 𝐴𝐸 onto a straight line is the line segment between the projections of the endpoints. Therefore, to project 𝐴𝐸 onto ⃖⃗𝐡𝐢, we first need to separately project 𝐴 and 𝐸 onto ⃖⃗𝐡𝐢.

To project these points onto the line, we need to find the lines through 𝐴 and 𝐸 that are perpendicular to ⃖⃗𝐡𝐢. We note that π΄π΅βŸ‚βƒ–οƒ©οƒ©οƒ©οƒ©βƒ—π΅πΆ and π΅βˆˆβƒ–οƒ©οƒ©οƒ©οƒ©βƒ—π΅πΆ, so 𝐡 is the projection of 𝐴 onto ⃖⃗𝐡𝐢. If we draw the perpendicular from ⃖⃗𝐡𝐢 to 𝐸, we note that this is the perpendicular height of △𝐷𝐢𝐸; we will call the point on the base 𝐹 as shown.

Hence, the projection of 𝐴𝐸 onto ⃖⃗𝐡𝐢 is 𝐡𝐹.

We can then note that △𝐷𝐢𝐸 is an isosceles triangle and so its perpendicular from vertex 𝐸 to the side 𝐢𝐷 will bisect 𝐢𝐷. Then, since 𝐡𝐢=𝐷𝐢, we must have that 𝐢𝐹=12𝐡𝐢.

We can determine the length of 𝐡𝐢 by applying the Pythagorean theorem to △𝐴𝐡𝐢. We have 𝐴𝐢=𝐴𝐡+𝐡𝐢.

Substituting 𝐴𝐡=40 and 𝐴𝐢=58 yields 58=40+𝐡𝐢.

We can then rearrange and evaluate to get 𝐡𝐢=58βˆ’40=1764.

Taking the square root of both sides of the equation where we note that 𝐡𝐢 is a length and so is nonnegative gives us 𝐡𝐢=√1764=42.cm

We can then determine the length of 𝐢𝐹 as half of 𝐡𝐢. We get 𝐢𝐹=12Γ—42=21.cm

Finally, 𝐡𝐹=𝐡𝐢+𝐢𝐹=42+21=63.cm

There is one final type of projection, which is the projection of a ray onto a line. We recall that a ray is a straight line with a given direction and start point but no endpoint. We can project a ray onto a line by projecting its start point and one other point on the ray and then taking the ray starting at the projected start point and going through the other projected point. We can write this formally as follows.

Definition: Projection of a Ray onto a Line Segment

The projection of a ray 𝐴𝐡 onto a straight line 𝐿 is the ray 𝐴′𝐡′ where 𝐴′ and 𝐡′ are the projections of 𝐴 and 𝐡 onto 𝐿 respectively.

It is worth noting that, if 𝐴𝐡 is perpendicular to 𝐿, then 𝐴′ and 𝐡′ will be the same point. In this case, we call the projection the single point 𝐴′.

Finally, it is worth noting that we can also project a line onto another line in the same way. In this case, if the two lines are not perpendicular, then the projection will be the second line. If the two lines are perpendicular, then the projection will just be the point of intersection.

Let’s see an example of using this definition.

Example 6: Recognizing the Projection of a Ray onto a Line

What is the result of projecting a ray onto a straight line, given that the two are not perpendicular?

Answer

We recall that the projection of a ray 𝐴𝐡 onto a straight line 𝐿 is the ray 𝐴′𝐡′ where 𝐴′ and 𝐡′ are the projections of 𝐴 and 𝐡 onto 𝐿 respectively. Since 𝐴𝐡 is not perpendicular to 𝐿, we know that 𝐴 and 𝐡 will have distinct projections onto 𝐿. If we call these projections 𝐴′ and 𝐡′, respectively, then we have the following.

Therefore, the projection of 𝐴𝐡 onto 𝐿 is the ray 𝐴′𝐡′.

Hence, the answer is a ray.

Let’s finish by recapping some of the important points from this explainer.

Key Points

  • The projection of a point 𝐴 onto a straight line ⃖⃗𝐡𝐢 is the point π·βˆˆβƒ–οƒ©οƒ©οƒ©οƒ©βƒ—π΅πΆ such that βƒ–οƒ©οƒ©οƒ©οƒ©οƒ©βƒ—π΄π·βŸ‚βƒ–οƒ©οƒ©οƒ©οƒ©βƒ—π΅πΆ. Where we note that, if π΄βˆˆβƒ–οƒ©οƒ©οƒ©οƒ©βƒ—π΅πΆ, then we say the projection of 𝐴 onto ⃖⃗𝐡𝐢 is just 𝐴.
  • The projection of a line segment 𝐴𝐡 onto a straight line 𝐿 is the line segment 𝐴′𝐡′, where 𝐴′ and 𝐡′ are the projections of 𝐴 and 𝐡 onto 𝐿 respectively. Where we note if the line segment and straight line are perpendicular, then both endpoints project to the same point and so we say the projection is a single point.
  • The length of the projection of a line segment is always less than or equal to the length of the original line segment.
  • The projection of a ray 𝐴𝐡 onto a straight line 𝐿 is the ray 𝐴′𝐡′ where 𝐴′ and 𝐡′ are the projections of 𝐴 and 𝐡 onto 𝐿 respectively. Where we note that, if 𝐴𝐡 is perpendicular to 𝐿, then 𝐴′ and 𝐡′ will be the same point. In this case, we call the projection the single point 𝐴′.

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