Lesson Video: Graphical Operations on Vectors Mathematics • 12th Grade

Video Transcript

In this video, we’ll learn how to perform operations on vectors graphically using triangle and parallelogram rules. Let’s begin by recalling what we know about vectors. A vector quantity has both a size, we call this the magnitude, and a direction. We represent these either using columns or these angled parentheses, where the vector two, five represents a movement two units right and five units up. And the vector negative one, negative four represents a movement one unit left and four units down.

Very occasionally, we’ll use 𝑖 and 𝑗 notation, where 𝑖 is a unit vector in the horizontal direction and 𝑗 is a unit vector in the vertical direction. We represent a vector using a line segment labeled with an arrow. And we say that the vector that joins 𝐀 to 𝐁 is the vector 𝐀𝐁, with an arrow above the 𝐀 and 𝐁 as shown.

Vectors can be added, subtracted, and multiplied by a scalar. That’s a value which is purely magnitude. And we can use them to solve geometrical problems. Let’s look at a couple of examples to remind ourselves how we interpret vectors plotted on a coordinate grid.

Consider the vector in the given diagram. What are the coordinates of its terminal point? What are the coordinates of its initial point? What are the components of the vector?

We say that a vector has an initial point, that’s where it starts, and a terminal point, that’s where it ends. So to find the coordinates of the terminal point of the vector in our diagram, that’s the vector 𝐯, we need to find the point at which the line segment that represents that vector ends. Remember, the arrow represents the direction of the vector. So in this case, we move from left to right. This means the vector ends here. We can therefore say its terminal point has coordinates at two, one.

Next, we’re looking to find the coordinates of its initial point. And remember, we said that that’s the start of the line segment. That’s here. The coordinates of this point are negative one, three. So that’s the coordinates of the initial point of our vector.

The third and final part of this question asks us to find the components of the vector. We split two-dimensional vectors into components that represent the horizontal and vertical motion separately. In order to see what these are, we’re going to add a right-angled triangle onto our vector 𝐯. This will split it into its horizontal and vertical components. The right-angled triangle we add is as shown.

Let’s look at the motion in the horizontal direction. We start at 𝑥 equals negative one, and we move one, two, three spaces to the right. Next, we consider the vertical motion. We start at a 𝑦-coordinate of three, and we move one, two units down. We can therefore write the vector 𝐯 as shown. We can use a column or these angled brackets. And the vector is three, negative two.

In our second example, we’ll consider what we mean by equivalent vectors and how the geometry of a hexagon can help us to identify these.

In the given figure, 𝐴𝐵𝐶𝐷𝐸𝐹 is a regular hexagon with center 𝑚. Complete the following. The vector 𝐀𝐁 is equivalent to what? Is it (A) the vector 𝐌𝐄, (B) the vector 𝐅𝐌, (C) the vector 𝐁𝐌, (D) the vector 𝐃𝐂, or (E) the vector 𝐃𝐌?

We say that two vectors are equal or equivalent if they have the same magnitude and direction. And this is regardless of where they lie. For example, let vector 𝐚 be equal to one, three and 𝐛 be equal to one, three. These both represent a movement of one unit right and three units up. They’re equivalent vectors. We say vector 𝐚 is equal to vector 𝐛.

Now, we’re looking to find a vector which is equivalent to the vector 𝐀𝐁. We don’t really know officially the magnitude and direction of vector 𝐀𝐁. But we can infer its equivalent vectors from our diagram using a bit of geometrical reasoning. The vector 𝐀𝐁 moves from 𝐀 to 𝐁, as shown. And we know in a regular hexagon, opposite sides are parallel and equal in length. We also know that we can split a regular hexagon into six equilateral triangles about the center. And in doing so, we know that these sides are parallel to these sides.

Since the triangles are equilateral, this means each of these sides must also be equal in length. And actually, this means the vector 𝐀𝐁 has several equivalents. We move from left to right, and we see that it’s equivalent to the vector 𝐅𝐌. Similarly, it’s equivalent, it’s equal to the vector 𝐌𝐂. And it has one further equivalent vector. The vector 𝐄𝐃 is equal in magnitude and direction. Of those vectors in our list, we can see the correct answer is (B). It’s the vector 𝐅𝐌.

Note that had the vector say 𝐌𝐅 been listed, we could not have deduced that 𝐀𝐁 and 𝐌𝐅 are equal. We could, however, say that the vector 𝐀𝐁 is equal to the negative vector 𝐌𝐅, since changing the sign changes the direction in which we move.

Now, there are actually a bunch of other equivalent vectors in our regular hexagon. Take the vector 𝐀𝐅 for example. It’s equal in direction and magnitude to the vector 𝐁𝐌 for the same reasons. It’s equivalent to the vector 𝐌𝐄. And each of these is also equivalent to the vector 𝐂𝐃.

In this example, we’ll discover how to calculate the value of the scalar used to multiply a vector.

Given the information in the diagram below, if the vector 𝐁𝐂 is equal to 𝑘 times the vector 𝐄𝐃, find 𝑘.

In our diagram, we’ve actually been given two similar triangles. That is, one is an enlargement of the other. Triangle 𝐀𝐃𝐄 has been enlarged onto 𝐀𝐁𝐂. Now, we know this since angle 𝐴 here is a shared angle. We see that angle 𝐴𝐸𝐷 and 𝐴𝐶𝐵 are equal as are angles 𝐴𝐷𝐸 and 𝐴𝐵𝐶. And this is because corresponding angles are equal and we know sides 𝐸𝐷 and 𝐶𝐵 are parallel.

Since the triangles have equal angles, they must be similar. And this means we’re able to calculate a scale factor of enlargement. This is found by dividing the length of the enlarged triangle by the corresponding length of the original. We sometimes write this as new length divided by old length.

If we take the enlarged triangle to be 𝐴𝐵𝐶 and the original triangle to be 𝐴𝐷𝐸, we see that we can find the scale factor by dividing the length 𝐴𝐵 by the length 𝐴𝐷. The length of 𝐴𝐵 is actually the sum of the two dimensions given. It’s 7.8 plus 5.2, which is 13 centimeters. And so the scale factor for enlargement here is 13 divided by 5.2, which is five over two.

Now, the scale factor is simply a multiplier. We know that to enlarge triangle 𝐴𝐷𝐸 onto 𝐴𝐵𝐶, we’d multiply any of its lengths by the scale factor of five over two. Now, we’re trying to find a relationship between the vectors 𝐁𝐂 and 𝐄𝐃. Well, we know that the dimension 𝐶𝐵 must be five over two times the dimension 𝐸𝐷. But since these lines are also parallel, we know that they have the same direction. And this means we can say that the vector 𝐂𝐁 must be five over two times the vector 𝐄𝐃.

The problem is, we want to work out the vector 𝐁𝐂 in terms of the vector 𝐄𝐃. Currently, we have the vector 𝐂𝐁 in terms of 𝐄𝐃. And so we’re traveling in the opposite direction. And we remember that to do this with vectors, we change the sign so that the vector 𝐁𝐂 is equal to the negative vector 𝐂𝐁. We just showed that the vector 𝐂𝐁 is five over two times the vector 𝐄𝐃. So this must mean that the vector 𝐁𝐂 is negative five over two times the vector 𝐄𝐃. Comparing this to the original form in the question, and we see then that 𝑘 is equal to negative five over two.

In our next example, we’ll look at how to find the resultant of two vectors and what we mean by the triangle rule for addition of vectors.

The blue vector represents the complex number 𝐳 one. The green vector represents the complex number 𝐳 two. What does the red vector represent?

Now, try not to worry too much that the axis labels are not what you’re used to. We’ll consider this to be much like an 𝑥𝑦-plane. We’ve got two vectors, given by 𝐳 one and 𝐳 two. Then, if we look at the red line, we see it travels from the initial point of 𝐳 one to the terminal point — that’s the end point — of 𝐳 two.

And when looking at vectors, I like to think about them a little bit like a metro or a tube map. Sometimes we want to travel from one station to another but can’t get directly there. And instead, we travel to an intermediate station, change trains, and travel on to the final destination. Our final destination is the same. We just had to go about it in a different way. In this case, the ultimate destination is from the point zero, zero to the point seven, one. Rather than going straight there though, we traveled from zero, zero to two, three along the vector 𝐳 one. And then we traveled from two, three to seven, one along the vector 𝐳 two.

In vector form, we say that our journey is 𝐳 one plus 𝐳 two. Now, we can check this by looking at the components of each vector. 𝐳 one is given by the vector two, three. To travel from the initial to the terminal point of our second vector, we travel five right and two down. So its components are five, negative two.

We said that we thought the red vector was the sum of these. Two, three plus five, negative two. Well, we find the sum of these vectors by adding their component parts. We add two and five to give us seven. Then we add three and negative two to get one. So 𝐳 one plus 𝐳 two is seven, one. And if we compare that to the red vector, we see that is also seven, one. The red vector represents 𝐳 one plus 𝐳 two.

Now, the red vector actually has a formal name. It’s called the resultant of the vectors 𝐳 one and 𝐳 two. And because of the shape it makes, we define the triangle rule for addition of vectors as follows. We say that if two vectors are represented as two sides of a triangle, then the third side of that triangle represents the magnitude and direction of their resultant. This is shown in the general diagram. The vector 𝐀𝐂 is the resultant of the vector 𝐀𝐁 and 𝐁𝐂. And we say that 𝐀𝐂 is 𝐀𝐁 plus 𝐁𝐂.

In our final example, we’ll look at how to find the resultant of two vectors using the parallelogram method.

The figure shows two vectors, 𝐯 and 𝐮, where the magnitude of 𝐯 is equal to five and the magnitude of 𝐮 is equal to seven. Use the parallelogram method to find the magnitude of the resultant of these two vectors. Give your answer to two decimal places.

The parallelogram method is so called because of the shape that it creates, as shown in the figure. This says that if two vectors acting simultaneously at a point can be represented both in magnitude and direction by the adjacent sides of a parallelogram drawn from a point. Then the resultant vector is represented both in magnitude and direction by the diagonal of the parallelogram passing through that point.

So as the diagram shows, the resultant of 𝐯 and 𝐮 — that’s their sum — is given by the diagonal of the parallelogram that’s drawn in pink. But how does that help? Well, we’re given some information about the magnitude of the vectors 𝐮 and 𝐯. Remember, the magnitude is the size or the length of the vector. So we can label 𝐮 as seven units and 𝐯 as five units. We can also add in some missing angles. We know that cointerior angles sum to 180 degrees. And we know that these sides are parallel. It’s a parallelogram. So this angle here is given by 180 minus 125, which is equal to 55 degrees.

Now, in fact, we can label the lengths of two further sides in our parallelogram. We know that opposite sides in a parallelogram are equal in length. So we can label these sides as five and seven. And now we’re going to split our parallelogram into two triangles. We’re trying to find the magnitude of 𝐯 plus 𝐮, so the length of that diagonal line.

Let’s call that length 𝑥 units. We now see we have a non-right-angled triangle, for which we know the lengths of two of its sides and the angle between them. This means we can use the cosine rule to find the missing length 𝑥. The cosine rule says that 𝑎 squared is equal to 𝑏 squared plus 𝑐 squared minus two 𝑏𝑐 cos 𝐴. Since the angle we have is 55 degrees, we label this vertex 𝐴. Then the side opposite is lowercase 𝑎. We can label the other two sides in any order. Let’s label five as 𝑏 and seven as 𝑐.

We can substitute everything we know about our triangle into this formula. And we get 𝑥 squared equals five squared plus seven squared minus two times five times seven times cos of 55 degrees. Evaluating this expression on the right-hand side gives us 33.84 and so on. Now, we’ll solve this equation for 𝑥 by taking the square root of both sides. That gives us 5.818 and so on, which, correct to two decimal places, is 5.82. So we found the length of 𝑥 to be 5.82 units, which means the magnitude of the resultant of 𝐮 and 𝐯 is 5.82.

We’ll now summarize the key points from this lesson. In this video, we learned that vectors can be described by their initial point, that’s their start point, their terminal point, that’s their end point, and using their individual components. And these components represent the horizontal and vertical bits of the movement. We saw that we can use geometrical properties of shapes to solve problems regarding the equivalents of two vectors and finding scalar multiples.

We learned about the triangle rule. And this said if two vectors are represented as two sides of the triangle, then the third side of that triangle represents the magnitude and direction of the resultant vector. If our two vectors are 𝐚 and 𝐛, the resultant is 𝐚 plus 𝐛. And of course, we mark that with an arrow as shown.

Finally, we learned about the parallelogram method. This says that if two vectors act simultaneously at a point and they’re represented both in magnitude and direction by the adjacent sides of a parallelogram drawn from a point. Then the resultant vector is represented both in magnitude and direction by the diagonal of the parallelogram passing through that point, as shown.