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Hi there.

My name's Ms. Lambel.

You've made a superb choice deciding to join me today to do some maths.

Let's get going.

Welcome to today's lesson.

The title of today's lesson is Parallel Vectors.

And, no surprises, that's within the unit, Vectors.

By the end of this lesson, you'll be able to identify column vectors which are parallel.

Quick recap of some words that we are going to be using in today's lesson, all of which you should be familiar with.

But I always think it's worth doing a quick double check, make sure that we're happy with those words before we start using them in the lesson.

Two lines are parallel if they are straight lines that are always the same, non-zero, distance apart.

Think about that.

It's like the sides of your table or the sides of your exercise book.

The gradient is a measure of how steep a line is.

It is calculated by finding the rate of change in the y direction with respect to the positive x direction.

Displacement is the distance from the starting point when measured in a straight line.

Effectively, it's the shortest distance between two points.

Today's learning I've split into two separate cycles.

In the first one, we will look at identifying parallel vectors using the gradient, and in the second learning cycle we will look at how we can use other ways to identify parallel vectors.

We're gonna get started then on identifying parallel vectors using the gradient.

A vector has a magnitude, size, and direction.

The length of the line represents the magnitude of the vector, and the arrowhead points in the direction that vector is moving.

Here are four equilateral triangles.

I've highlighted line number 1 with a purple arrow.

Which other vectors are the same as Vector 1? They are moving in exactly the same direction, and they are exactly the same length.

Which ones? And hopefully you decided that Vector 2 and also Vector 7.

Now let's take a look at Vector 3.

Which other vectors are the same as Vector 3, the same length and exactly the same direction? This time, you should have decided Vector 5 and Vector 8.

We can see there I've highlighted those on the diagram.

And, finally, Vector 4.

Which other vectors are the same as Vector 4? And those vectors are Vector 6 and Vector 9.

Hopefully, you were able to identify those.

We know that vectors have a magnitude and direction but not an exact position.

This means that all parallel lines with the same length and direction are indeed the same vector.

If vectors are represented on a grid, it is easy to see if they are parallel.

Are 3, 6 and 5, 10 parallel? Let's draw the vector 3, 6.

I'm going to have a starting point, and I'm going to move three to the right and six up.

That's the vector 3, 6.

I've now drawn the vector 5, 10, and we can see that they have the same gradient.

The distance between the two lines remains unchanged.

They are parallel.

If we look at the gradient, we can see for the first one, as I move one horizontally, I move two up.

And then if we look at the second vector, as I move one horizontally, I move two up.

They have the same gradient, they have the gradient of two, therefore they are parallel.

We've now gotta show that question.

We need to show that negative 9, 3 and 6, negative 2 are parallel.

We're gonna draw them negative 9, 3 and 6, negative 2.

Are they parallel? Let's work out the gradient.

Remember, when you're working out the gradient, make sure that you choose points along your line that you know the exact coordinates of the start and end points if it was on a coordinate grid.

Here, I can see that I've moved three places to the right and one place down.

If I look at the next one, I've also moved three places to the right and one place down.

Both vectors have a gradient of one third, therefore, they are parallel.

Now I'd like you please to have a go at this check for understanding.

It's a show that question.

I need you please to use the grid to show that the vectors negative 9, negative 6, and 6, 4 are parallel.

Remember, if a question says show that they are parallel, then they are parallel.

You need to just prove that with your diagrams and your gradients.

Pause the video now, and then when you come back we'll check those answers for you.

And how did you get on with that? Great.

Here is the vector negative 9, negative 6, and that's the vector 6, 4.

Let's take two points on our line.

We can see here that the horizontal displacement is 3 and the vertical displacement is 2.

Let's take a look now at the other vector.

The horizontal displacement is 3 and the vertical displacement is 2.

Both vectors have a gradient of two thirds, therefore, they are parallel.

We've shown that they are parallel.

What happens if we take the grid away? Are 6, 3 and negative 4, negative 2 parallel? Let's start with the first vector, 6, 3.

The change in x is the horizontal displacement.

That's 6.

And the change in y is the vertical displacement.

And for the first column vector, that's 3.

Remember, when we're looking at the gradient, we need to know the change in y given the change in x is 1.

Let's have a look at our ratio table.

To get from 6 to 1, I'm dividing by 6, therefore, I need to do the same to the y.

3 divided by 6 is 0.

Now we will check what the gradient is of the second column vector, negative 4, negative 2.

The horizontal displacement represents the change in x.

That's negative 4.

And the change in y is the vertical displacement, which is negative 2.

Again, I need to know the change in y.

for change in x of 1.

What do I divide negative 4 by to get 1? And that's negative 4.

Negative 2 divided by negative 2 is 0.

Here we can see that both vectors have a gradient of 0.

5, therefore, they are parallel.

Let's take a look at another one.

Again, we'll start with the first column vector, which is 5, 11.

The change in x is 5 and the change in y is 11.

For the gradient, we need to know what the change of y is for a change of x of 1.

What do I do to 5 to get to 1? I divide by 5, or you may say multiply by 0.

I do the same then to 11.

11 divided by 5 is 2.

Now let's consider the second column vector.

The change in x is negative 3 and the change in y is negative 7.

We need to work out the change when the change in x is 1.

What do I do to get from negative 3 to 1? I divide by negative 3.

Negative 7 divided by negative 3 is 7 over 3.

Here we can see that the vectors have different gradients, so therefore they are not parallel.

And one more.

This time, the question says show that they are parallel.

So we know they're parallel.

So in this type of question, you've almost got a bit of a headstart.

Because if you get to the end and they're not parallel, you know you've made a mistake.

Let's take a look.

The first vector, negative 10, 2.

Change in x, negative 10, change in y is 2.

I divide by negative 10 to give me the change of x of 1, and that gives me negative 0.

Now let's have a look at the second column vector, negative 15, 1, and then change in y is 3.

So this time I'm dividing by negative 15.

3 divided by negative 15 is negative 0.

We can see that both vectors have a gradient of negative 0.

2, meaning that they therefore are parallel.

We'll do one more together, and then you can have a go at one independently.

The change in x for the first column vector is negative 6, and the change in y is negative 10.

We need to work out the change in y when the change of x is 1.

So I'm going to divide by negative 6.

Negative 10 divided by negative 6 is five thirds.

Notice there I've written it as its exact fractional value rather than having to worry about recurring decimals.

Now we're gonna consider the vector 4, 7.

This time, I've dividing by 4 to get 1.

7 divided by 4 is seven quarters.

Here, both vectors have different gradients, so that means they are, that's right, not parallel.

Now it's your turn.

Are the vectors 12, 8 and negative 6, negative 4 parallel? Remember, you need to show me whether they are or not, and you need to use the gradient to show me that.

Pause the video and then come back when you're done.

So were they parallel? Let's take a look.

The change in x is 12, change in y is 8.

This is for the first column vector.

So I'm dividing by 12.

8 divided by 12 is two thirds.

If I look at the second column vector, the change in x is negative 6, the change in y is negative 4.

So this time I'm dividing by negative 6.

Negative 4 divided by negative 6 is two thirds.

Both vectors have the same gradient, therefore, they are parallel.

Did you end up with them being parallel? Of course you did, well done.

Now you're ready for Task A.

Which of the following pairs of vectors are parallel? And I'd like you please to find the gradients to support your answers.

So just like those last few examples we've been through together.

Pause the video and come back when you are done.

Well done.

And then two more, parts C and D.

The same thing.

Can you find the gradients to support your answers as to whether these pairs are parallel or not? Great work.

Let's check our answers.

So, A, they were parallel.

We can see both of those there have a gradient of 1.

And if you need to pause the video just to take a look at those tables in more detail, then obviously you can.

B, they were also parallel.

We can see that both of them had a gradient of negative three eighths.

C, they were parallel.

We can see again they both have the same gradient, this time of negative 7 over 9.

And then finally, D.

If we work out the gradient for negative 8, 12, we get negative 1.

5, and if we work out the gradient for negative 2, 4, we get negative 2.

The gradients are not the same, so therefore they are not parallel.

Great work on that.

Now let's move on then.

We're going to look at other ways that we can identify parallel vectors.

Here we've got Izzy and Jun.

Izzy says, "Hi, Jun.

"I've been thinking about parallel vectors." Hmm.

And Jun says, "Well, what have you been thinking about?" Izzy's response is, "Well, we have been finding the gradient, "and we know that linear graphs have an infinite nature." And Jun says, "Yes, that's right." Izzy says, "So surely any multiple of a vector is parallel." And Jun says, "Of course.

"We just use a longer or shorter section of the line "to find the gradient." Well, let's take a look.

We will consider the vectors 2, negative 3 and 10, negative 15.

Looking at the first vector and working out the gradient, we can see that the gradient of this line is negative 1.

Now let's consider the second column vector.

This time, I'm dividing by 10.

Negative 15 divided by 10 is negative 1.

Both vectors have a gradient of negative 1.

5, and so therefore they are, that's right, they're parallel, aren't they? They're parallel lines.

2, negative 3 and 10, negative 15.

Do you notice anything about those two vectors? To get from one to the other, I've actually multiplied by 5.

2 multiplied by 5 is 10, and negative 3 multiplied by 5 is negative 15.

Let's take a look and see what they're chatting about now.

Izzy says, "Could the multiplier of the vector "be a non-integer?" So could we replace the 5 with a non-integer? And Jun's response to that is, "Yes, it could be.

"We just chose the vertices of the squares on the grid, "as we know the exact lengths of displacement." When we're finding the gradient, remember, we do need to know the exact lengths of our changes in x and changes in y, which is why we choose the vertices of the squares on the grids.

Izzy says, "Yes, you're right, Jun.

"With that last example, "the gradient was negative 1.

5, "and that would be represented "by the vector 1, negative 1.

5." 2, negative 3 and 1, negative 1.

Here, we can see that we've multiplied by a half, or you may have decided to divide by 2.

That's the same thing.

We can see here that the multiplier is a non-integer.

True or false, negative 4, 3 and 0.

8 negative 0.

6 are parallel? Is that true or false? And as always, I want that justification please.

Justification is one of these two.

They are multiples of each other because the horizontal displacement and vertical displacement share the same multiplicative relationship, or they are not multiples of each other because the horizontal displacement and vertical displacement do not share the same multiplicative relationship.

Pause the video, make your decision, and then come back when you're done.

Super.

What did you decide, was it true or false? It was true.

And the reason is they are multiples of each other.

Vertical and horizontal displacement share the same multiplicative relationship.

If we take a look, negative 4 multiplied by negative 0.

2 gives us 0.

8, and 3 multiplied by negative 0.

2 gives us negative 0.

We can see that the horizontal and the vertical displacements share the same multiplicative relationship.

Shape ABCD is a quadrilateral.

Without drawing, decide whether it is a trapezium.

If this quadrilateral is a trapezium, then it must have one pair of parallel sides.

Are the lines AB and CD parallel? Let's take a look.

We've got AB is 3, 1 and CD is negative 1, 3.

We're looking for that multiplicative relationship.

3 multiplied by negative one third is negative 1, and 1 multiplied by 3 is 3.

We can clearly see here that the multiplicative relationship for horizontal and vertical displacement is not the same, so therefore they are not parallel.

One is not a multiple of the other.

Are the lines BC and DA parallel? Again, let's take a look.

BC is the vector 1, 2, and DA is the vector negative 3, negative 6.

My multiplicative relationship here is multiplied by negative 3, and my multiplicative relationship for the vertical displacement is also multiplied by negative 3.

We can see clearly now this time the relationship between the horizontal and vertical displacements is the same, and so therefore one is a multiple of the other, and this means that they are parallel.

And therefore it is a trapezium.

Your turn.

ABCD is a quadrilateral.

What type of quadrilateral is it? So I've given you the vectors of the lines AB, BC, CD, and DA.

I'd like you to decide please.

Remember, no guessing.

Is this a parallelogram, a rectangle, or a square? Pause the video and come back when you are done.

And what shape did you come up with? It is a rectangle.

We can see that the opposite sides are the same length, and we can see that AB and CD are vertical and BC and DA are horizontal, each with the same length.

Parallel vectors can also be identified in geometric situations, like the first example we looked at during this lesson.

Here is that example we looked at.

Can you think of any other shapes which have parallel sides? Pause the video, write down as many as you can.

I'm gonna challenge you to come up with at least three please.

Shapes with parallel sides.

Pause the video and then come back when you are done.

Okay, what did you come up with? Well, I came up with these, square, rectangle, parallelogram, trapezium, a regular hexagon, and a regular octagon.

And there are others as well.

This is a parallelogram.

We need to write down the column vectors of sides a and b.

Side a is both parallel to and has the same magnitude as the vector 2, 4.

If we look at side a, we can see it's parallel to the vector 2, 4.

It's also the same length.

It has to be because this is a parallelogram, and we know that opposite sides of a parallelogram are equal in length.

Therefore, the vector representing side a is the vector 2, 4.

Side b is both parallel to and has the same magnitude as the opposite side.

We can clearly see that the opposite side has a horizontal movement of six units.

What is the column vector that represents a horizontal movement only of six units? That's right, it's 6, 0.

This means then that b must also have the same vector because it's in exactly the same direction, and it is exactly the same length.

So b is 6, 0.

ABCD is a trapezium.

BC and AD are the parallel sides.

The magnitude of the vector joining A to D is double the magnitude of the vector joining B and C.

Draw the trapezium and write down the column vector joining D to C.

It tells us that the vector joining A to D is double the magnitude of the vector joining B to C.

Well, B to C, as we can see, has a displacement of one right, two up.

We need to double that.

So one right, two up, and one right, two up.

That now gives us point D, and then we can complete our trapezium.

The question also asked us to write down the column vector joining D to C.

We can see to go from D to C that I've moved two places to the left and one place up, represented by the vector negative 2, 1.

Sometimes we may get asked to draw something.

Draw a trapezium with one of the parallel sides represented by the vector 2, 4.

I can start by drawing this vector.

Here is my vector 2, 4.

I've moved two to the right and four up.

That's one of the parallel sides.

So to draw the other parallel side, I need to make sure it has the same gradient.

Doesn't have to be the same length, but it has to be the same gradient.

And I can choose to put that anywhere I like on my grid.

And then I would just join together the ends of each lines to create my trapezium.

So I could put it here, put it here.

There are lots and lots of different places it could go.

I could make the line longer as well.

Right then, Task B.

Which of the following pairs of vectors are parallel? And I'd like you this time, instead of using a gradient, I'd like you to use multiples to support your answer.

Pause that video, and then come back when you're ready.

Well done.

Question number 2.

This is a regular hexagon.

Write down the column vector of side a, side b, and side c.

Again, pause the video and come back when you are done.

Well done.

And question number 3.

I'd like you please to draw a trapezium with one of the parallel sides represented by the vector negative 3, 2.

Pause the video, draw me a trapezium, and come back when you are done.

Great work.

Now let's check those answers.

Question number one, A, scalar multiple of 3 from the first to the second.

Negative 3 multiplied by 3 is negative 9, and 4 multiplied by 3 is 12.

For B, there was a scalar multiple of one quarter from the first column vector to the second one.

4 multiplied by a quarter is one, and 14 multiplied by a quarter is 3.

So, yes, they were parallel.

C, they were not.

There was not a constant multiplicative relationship between the horizontal and vertical displacements.

And D had a scalar multiple of negative 6 from the first to the second.

5 multiplied by negative 6 is negative 30, and negative 9 multiplied by negative 6 is 54.

Question 2, A was 4, 3, B was 0, 5, and C was 4, negative 3.

And here are a couple of examples.

The green arrow shows the original vector, and then you had to draw a trapezium where one of the parallel sides was that green line.

Here are just a couple of examples.

You just need to make sure that your line is parallel to the green line.

Let's summarise our learning, shall we? Gradient can be used to identify parallel vectors, and we looked at that during the lesson.

Vectors with the same gradient are parallel to each other.

So here is an example of one that we looked at in the lesson.

We can see that both of them have a gradient of negative 0.

Therefore, they are parallel.

Parallel vectors have a multiplicative relationship.

They do not have to be in the same direction or have the same magnitude.

For example here, 1, 2, and negative 3, negative 6.

The multiplicative relationship between the first and the second vector is multiplying the horizontal and vertical displacement by negative 3.

You've super impressed me today with some fantastic learning.

Brilliant work, well done.

Hopefully, I will see you again really soon.

But for now, goodbye, and take care of yourself.

I've finished the video