Lesson Video: Scalar Projection Mathematics • 12th Grade

In this video, we will learn how to find the scalar projection of a vector onto another vector.

15:12

Video Transcript

In this video, our topic is scalar projection. And as we might guess, this means something a bit different from what we’re seeing on the screen. This scalar projection we’ll talk about has to do with finding out how much of a given vector lies along another.

We’ll begin by talking about this idea of projection. Say that we’ve got a vector, we’ll call it vector 𝐀, that exists in three-dimensional space. In this case, we can say that the components of vector 𝐀 represent how much it lies along or overlaps the three axes 𝑥, 𝑦, and 𝑧. For example, if we were to drop vector 𝐀 down into the 𝑥- and 𝑧-plane, in other words, projecting it into this plane, then we would find that the magnitude of this vector’s 𝑥-value is 𝐀 sub 𝑥 and its 𝑧-value is 𝐀 sub 𝑧. We could say then that these components represent projections of our vector onto these coordinate axes.

But then there’s nothing that says we can only project a vector in these three directions 𝑥, 𝑦, and 𝑧. Really, we could choose a vector in an arbitrary direction and then find out how much of 𝐀 lies along it. Whatever answer we find, it will be a scalar rather than a vector quantity. This then is what we mean when we talk about scalar projection, how much of one vector overlaps another.

Let’s say that our vector in an arbitrary direction, we’ve called it vector 𝐁, has these three components. Knowing this, we can solve mathematically for the scalar projection of vector 𝐀 onto vector 𝐁 this way. We take the dot product of these two vectors, and we divide them by the magnitude of the vector being projected onto, in this case vector 𝐁. From this formulation of the scalar projection, we can see that we would need to know the components of our two vectors involved, in this case 𝐀 and 𝐁. But if we recall what the dot product of two vectors in general is equal to, we’ll see that this isn’t always the case.

Let’s recall that the dot product of two vectors, we can call them 𝐕 one and 𝐕 two, is equal to the product of the magnitudes of these vectors multiplied by the cosine of the angle between them. This relationship means we could replace 𝐀 dot 𝐁 in our scalar projection equation with the magnitude of 𝐀 times the magnitude of 𝐁 times the cosine of the angle between 𝐀 and 𝐁. Returning to our sketch, we can indicate that angle this way. And we see that we’ve called it 𝜃.

Looking at the second equivalent way of writing the scalar projection, we see that the magnitude of vector 𝐁 cancels from numerator and denominator. So we then have these two equivalent ways of writing the projection of a vector 𝐀 onto a vector 𝐁. And notice that, in the second way, we only need to know the magnitude of the first vector and the angle between the two.

Now, talking about scalar projection, it can be helpful to see what this really means visually. Let’s consider these two vectors 𝐂 and 𝐃, which are in two-dimensional space. If we were to project vector 𝐂 onto vector 𝐃, then we would be calculating this length here. It’s the distance along which these vectors overlap. And we’ve seen that, mathematically, this distance is given by the dot product of 𝐂 and 𝐃 divided by the magnitude of the vector being projected onto. So even when we’re working with three-dimensional vectors, we can keep this two-dimensional picture in mind. So we know the geometric meaning of what we’re doing when we compute a scalar projection.

Knowing all this, let’s get some practice with these ideas through an example exercise.

If the magnitude of 𝐀 equals five, the magnitude of 𝐁 equals 15, and the measure of the angle between them is 30 degrees, find the algebraic projection of 𝐁 in the direction of 𝐀.

Alright, so in this exercise, we have these two vectors 𝐀 and 𝐁. And we know that the magnitude of 𝐀 is one-third the magnitude of 𝐁 and also that they’re separated by an angle of 30 degrees. Knowing this, we want to solve for the algebraic projection, also known as the scalar projection, of vector 𝐁 in the direction of vector 𝐀. What we’re proposing then is effectively to lay vector 𝐁 down along this line we’ve drawn through vector 𝐀. Essentially, we’re solving for how much of vector 𝐁 is parallel to vector 𝐀.

We can figure out how to compute this scalar projection in one of two ways. The first way is to recall from memory that the scalar projection of one vector, we’ve called it 𝐕 one, onto another, 𝐕 two, is given by these two expressions here. So, for example, we could use the fact that the magnitude of the first vector multiplied by the cosine of the angle between the two vectors gives us this scalar projection to solve for the algebraic projection of 𝐁 onto 𝐀.

Notice though that we could also reach this conclusion based on our sketch of vectors 𝐀 and 𝐁. The vector 𝐁 essentially forms the hypotenuse of a right triangle. And so we see that it’s the magnitude of that hypotenuse, the magnitude of 𝐁, multiplied by the cosine of the angle between 𝐁 and the direction of 𝐀, which is 30 degrees, that gives us this algebraic projection. Either method we choose leads us to the same result.

We substitute in the magnitude of 𝐁, 15. And knowing that the cos of 30 degrees equals exactly the square root of three over two, we find our answer to be 15 times the square root of three over two. This is the algebraic or scalar projection of 𝐁 in the direction of 𝐀.

Now let’s look at an example where our scalar projection yields a negative value.

Given that the measure of the smaller angle between 𝐀 and 𝐁 is 150 degrees and the magnitude of 𝐁 equals 54, determine the component of vector 𝐁 along 𝐀.

Okay, in this exercise, we have these two vectors 𝐀 and 𝐁. And just to help us see how they relate to one another, let’s assume that they lie in the 𝑥𝑦-plane. We might then draw these two vectors like this. And we’re told that the smaller angle between these two vectors, that’s this one here, we can call it 𝜃, is 150 degrees. We’re asked to determine the component of vector 𝐁 along vector 𝐀.

Looking at our sketch though, we might wonder if the answer isn’t zero, because it looks as though none of vector 𝐁 lies along vector 𝐀. Here we have to be careful though because this phrase “along 𝐀” really means along the line that passes through vector 𝐀, in other words this dashed line, where the line is assumed to have a direction equal to the direction of vector 𝐀. In answering this question then, we’re going to be calculating this distance here. That’s the component of vector 𝐁 along 𝐀.

To get started solving for this, let’s recall that the scalar projection of one vector onto another is equal to the dot product of those vectors divided by the magnitude of the vector being projected onto. And this is also equal to the magnitude of the first vector, here we’ve called it 𝐕 one, multiplied by the cosine of the angle between the two vectors. It’s this form of the scalar projection equation that we can make use of in this particular exercise. After all, we know the magnitude of what we could call our first vector, the magnitude of 𝐁, and we also know the angle between our vectors.

What we want to calculate then is the magnitude of 𝐁 times the cos of 𝜃, or substituting in our given values 54 times the cos of 150 degrees. The cosine of this angle equals exactly negative the square root of three over two. So our scalar projection equals 54 times negative root three over two, or in simplified form negative 27 root three. This is the component of vector 𝐁 along vector 𝐀. And we can see from this result that, in general, a scalar projection can be negative. In this case, that came from the fact that the part of 𝐁 lying along the line through vector 𝐀 points in the opposite direction as vector 𝐀.

Let’s look now at an example where we work with vectors by their components.

Determine, to the nearest hundredth, the component of vector 𝐕 along 𝐀𝐁 given that 𝐕 equals negative seven, two, 10 and the coordinates of 𝐀 and 𝐁 are one, negative four, negative eight and three, two, zero, respectively.

Okay, in this exercise, we have a three-dimensional vector 𝐕 and points in three-dimensional space 𝐴 and 𝐵. Let’s say that point 𝐴 is located here and point 𝐵 is here. We want to solve for the component of this given vector 𝐕 along a vector 𝐀𝐁. This vector 𝐀𝐁 will go from point 𝐴 to point 𝐵 looking like this. And to calculate the component of vector 𝐕 along 𝐀𝐁, we’ll want to know the components of vector 𝐀𝐁.

To solve for those, we can subtract the coordinates of point 𝐴 from the coordinates of point 𝐵. In other words, we could write that vector 𝐀𝐁 equals 𝐁 minus 𝐀 in vector form. Substituting in the coordinates of 𝐵 and 𝐴, we find subtracting those of 𝐴 from those of 𝐵 gives us a vector with components of three minus one or two, two minus negative four or six, and zero minus negative eight or eight. So then we now have our vector 𝐀𝐁. And as we’ve seen, we want to solve for the component of vector 𝐕 that lies along 𝐀𝐁.

We can begin to do this by recalling that the scalar projection of one vector onto another is equal to the dot product of those two vectors divided by the magnitude of the vector being projected onto. In our example, as we calculate the component of vector 𝐕 along 𝐀𝐁, we’re computing the scalar projection of 𝐕 onto 𝐀𝐁. Therefore, we can say that the quantity we want to solve for is given by 𝐕 dot 𝐀𝐁 over the magnitude of vector 𝐀𝐁.

Remembering that the magnitude of a vector is equal to the square root of the sum of the squares of the components of that vector, we see that what we want to calculate is this dot product over this square root. Carrying out this dot product, we start by multiplying the respective components of these two vectors together. And then working in our denominator, we know that two squared is four, six squared is 36, and eight squared is 64. So our fraction simplifies to negative 14 plus 12 plus 80 divided by the square root of four plus 36 plus 64. This equals 78 over the square root of 104.

And we could leave this as our answer, except that we’re told to determine this overlap to the nearest hundredth. If we enter this fraction on our calculator then, to the nearest hundredth, it equals 7.65. That’s the component of vector 𝐕 along vector 𝐀𝐁.

Let’s look now at another scalar projection example.

𝐴𝐵𝐶 is a right triangle at 𝐵, where 𝐴𝐵 equals 17 centimeters, 𝐵𝐶 equals 11 centimeters, and 𝐷 is the midpoint of 𝐴𝐶. Find the algebraic projection of 𝐴𝐷 in the direction of 𝐶𝐵.

Okay, in this example, we have this right triangle that looks something like this. We’re told that the length of side 𝐴𝐵 is 17 centimeters, here we’ll leave off the units, while 𝐵𝐶 is 11 centimeters. And there’s a point called point 𝐷, which is midway between 𝐴 and 𝐶 on the hypotenuse of the triangle. Our question wants us to solve for the algebraic projection of a vector 𝐀𝐃 in the direction of another vector 𝐂𝐁.

First, let’s define 𝐀𝐃. This is a vector that starts at point 𝐴 and ends at point 𝐷. And likewise, 𝐂𝐁 is a vector that starts at point 𝐶 and ends at point 𝐵. So the idea is, we want to solve for the algebraic projection of this vector onto this one.

To begin figuring this out, let’s first solve for the components of these two vectors. Let’s say that point 𝐵 in our triangle is the origin of an 𝑥𝑦-coordinate frame. We see then that vector 𝐂𝐁 lies along the 𝑥-axis and line segment 𝐴𝐵 lies along the 𝑦. From this perspective, we can define the coordinates of the four points 𝐴, 𝐵, 𝐶, and 𝐷.

Point 𝐴 has an 𝑥-coordinate of zero and a 𝑦-coordinate of 17. Point 𝐵, because it exists at the origin, has coordinates zero, zero, while point 𝐶 has coordinates 11, zero. But now what about the coordinates of point 𝐷? Because point 𝐷 cuts in half the line segment 𝐴𝐶, that means its 𝑥- and 𝑦-coordinates are one-half of the side lengths of those two sides of our triangle. That is, the 𝑥-coordinate of 𝐷 is 11 divided by two or 5.5, while the 𝑦-coordinate is 17 divided by two or 8.5.

Knowing all this, we can now focus on solving for the components of our two vectors 𝐀𝐃 and 𝐂𝐁. Vector 𝐀𝐃 is equal to the coordinates of point 𝐷 minus those of point 𝐴 all in vector form. When we substitute in for the coordinates of points 𝐷 and 𝐴, we find that this subtraction gives us a result of 5.5 and negative 8.5. These then are the 𝑥- and 𝑦-components of vector 𝐀𝐃.

Next, let’s calculate the components of 𝐂𝐁. To do this, we’ll subtract the coordinates of point 𝐶 from those at point 𝐵, which means we’ll subtract the point 11, zero from the point zero, zero. This results in the vector negative 11, zero. Okay, great, so now we have our two vectors, and we want to solve for the algebraic projection of 𝐀𝐃 in the direction of 𝐂𝐁.

To solve for this, let’s remember that the scalar, also known as the algebraic, projection of one vector, 𝐕 one, onto another, 𝐕 two, is given by the dot product of these two vectors divided by the magnitude of the vector being projected onto. In our case then, what we want to calculate is 𝐀𝐃 dot 𝐂𝐁 divided by the magnitude of 𝐂𝐁. Substituting in for the components of these vectors and recalling that the magnitude of a vector equals the square root of the sum of the squares of its components, we next start to compute our dot product by multiplying together the respective components of our vectors.

Along with this, we notice that negative 11 squared is 121 and zero squared is zero, which means that this projection equals 5.5 times negative 11 divided by the square root of 121. But then 121 is 11 squared so that factors of 11 in numerator and denominator cancel out. And we’re left with negative 5.5. This is the algebraic projection of 𝐀𝐃 in the direction of 𝐂𝐁.

Let’s finish up now by summarizing a few key points about scalar projection. In this video, we saw that the component of one vector, call it vector 𝐀, in the direction of another vector, call it vector 𝐁, is called the scalar projection of 𝐀 onto 𝐁. Mathematically, this projection is equal to the dot product of 𝐀 and 𝐁 divided by the magnitude of 𝐁 or equivalently the magnitude of 𝐀 times the cosine of the angle between the two vectors. Geometrically, this picture shows us what this projection looks like in two-dimensional space. And lastly, we saw that a scalar projection can be positive, negative, or zero.

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