Lesson video

Hi everyone, my name is Ms. Ku, and I'm really happy that you're joining me today.

Today, we'll be looking at vectors, a great topic.

And chances are you've already used vectors today already, whether that be playing a game or using satellite navigation.

I hope you enjoyed today's lesson.

So let's start.

Hi everyone and welcome to this lesson on parallel vectors in algebraic vector notation.

And it's under the unit, Vectors.

And by the end of the lesson, you'll be able to identify vectors written algebraically which are parallel.

Now let's have a look at some key words.

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

Today's lesson will be broken into two parts.

We'll be looking at parallel vectors first and then problem solving with parallel vectors.

So let's make a start looking at parallel vectors.

Here, we have a grid.

And I know vector A is in column vector form 1, 2, and vector B in column vector form is 3, 2.

And I want you to use a square grid and draw the following resultant vectors, A add B.

Then I want you to draw 2A add 2B, and then I want you to draw -3A subtract 3B.

So you can give it a go.

Press pause as you'll definitely need more time.

Great work.

Let's see how you got on.

Well, the vector AB is given here.

You can see it in purple.

Question 2, we have the vector 2A add 2B, is given here, and we have the vector -3A subtract 3B, is given here.

Well done if you've got this.

Now what I want you to do is write the column vectors for A and B.

Write the column vectors for 2A and 2B, and write the column vectors for -3A and subtract 3B.

So you can give it a go.

Press pause if you need more time.

Well done.

Let's see how you got on.

Well, A plus B gives us the vector 4, 4 in column vector form.

2A and 2B gives us 8, 8.

And for question 3, we have -12, -12.

So, what do all three of the drawn vectors have in common? Have a little look.

Well, they're all parallel.

They're all parallel because they have the same gradient.

You can see it here.

I've calculated the gradient to be 4 over 4, which simplifies to 1.

Here, I've got the gradient of 8 over 8, which simplifies to 1.

Here, I have the gradient of 12 over 12, which simplifies to 1.

So they are all parallel because they have the same gradient.

Now let's have a look at these column vectors.

These column vectors, what do they all have in common? Have a little think.

Well, they're all scalar multiples of one another.

If you were to multiply our column vector 4, 4 by 2, you have the column vector 8, 8.

If you were to multiply our column vector 8, 8 by -1.

5, you have the column vector minus 12, minus 12, or you could even multiply our vector 4, 4 by -3 to give you -12, -12.

So you can see, they're all scalar multiples of one another.

Now let's have a look at these algebraic vectors.

What do you notice? Well, they all have A and B as a factor.

But we know A and B is a factor of A and B.

2A add 2B is 2 multiplied by the factor A + B.

<v ->3A subtract 3B</v> is -3 multiplied by our factor A add B.

Therefore, two vectors are parallel if one vector can be written as a scalar multiple of the other.

And the scalar multiples can be seen in algebraic form.

For example, if you have the vector AB, is given as A + 3B, it's going to be parallel to vector CD because vector CD can be written as 5 multiplied by A + 3B.

In other words, five lots of vector AB is equal to our vector CD.

Another way that you can see parallel vectors is when they're given in column form.

You can see the scalar multiples.

For example, if vector AB is 2, -6, we know it's going to be parallel to vector CD, which is 10, -30 because five lots of our vector AB is equal to our vector CD.

Lastly, parallel vectors can also be identified by their gradients.

Remember parallel vectors have equivalent gradients.

What I want you to do is have a look at this check.

Which of the following are parallel to 3A subtract 4B? And I want you to explain how you know.

So you can give it a go.

Press pause one for more time.

Well done.

Let's see how you got on.

Well, these are all parallel to 3A subtract 4B.

Let's find out why.

Well, 12A subtract 16B is equivalent to 4 multiplied by our 3A subtract 4B.

Next, -9A add 12B is equivalent to -3, multiply by 3A subtract 4B.

Lastly, we have 2A subtract 8/3 of B.

Well, this is equivalent to 2/3 multiplied by our 3A subtract 4B.

In other words, all of the following which we've ticked have 3A subtract 4B as a factor.

Really well done.

Next, I want you to explain which of the following are parallel to PQ, which is -9, 12.

And then I want you to explain how you know.

See if you can give it a go.

Press pause for more time.

Well done.

Let's see how you got on.

Well, all of these are parallel to vector PQ.

Let's have a look at why.

Well, we know vector MT is negative 1/3 of vector PQ.

Or we could say vector PQ is equal to -3 multiplied by our vector MT.

In either case, you can see one is a scalar multiple of the other.

For C, well, vector XY is equal to 5/3 of vector PQ, or you could even say 3/5 of vector XY is equal to vector PQ.

Once again, you can clearly see one vector in a scalar multiple of the other.

For JK, well, JK is equal to -PQ, or PQ is equal to -JK.

Really well done if you've got this, in particular, the explanation.

Now Laura says, "They can't be parallel because the directions are different." I want you to explain if Laura is correct.

See if you can give it a go.

Press pause if you need more time.

Well done.

Well, hopefully you've spotted.

We're focusing on the gradient.

Well, the gradient of our first vector is 8 over 6.

The gradient of our second vector is 4 over 3.

Therefore, they are parallel given that 8/6 is equivalent to our 4/3.

So that means she's incorrect, as they are parallel.

Now remember, the direction of the vector does not affect the gradient.

Great work, everybody.

Now it's time for your task.

I want you to pair the vectors that our parallel.

So you can give it a go.

Press pause if you need more time.

Well done.

Let's move on to question 2.

For question 2, we're given the following column vectors.

A is 2, 3, and B is 4, -3.

And I want you to identify which of the following are parallel to 3A subtract 2B.

So you can give it a go.

Press pause if you need more time.

Well done.

Let's move on to question 3.

Name the quadrilaterals and explain how you know, and I want you to explain using vectors.

Great question.

See if you can give it a go.

Well done.

Let's move on to the next part of question 3.

Same again, I want you to name the quadrilateral and explain how you know it's this quadrilateral, but you've got to use vectors.

See if you can give it a go.

Press pause if you need more time.

Well done.

Let's go through these answers.

Well, for question 1, you should have had this pairing.

Really good work if you've got this.

Press pause if you need.

For question 2, we should have worked out what is the vector 3A subtract 2B in column vector form first.

Well, it's -2, 15.

So we need to identify which of the following column vectors here have -2, 15 as a factor.

Well, it's definitely going to be A 'cause we simply multiply the vector by -3.

It's going to be vector D because we multiply our vector by two.

It's going to E because we multiply the vector by -1.

5, and it's going to be F because we multiply our vector by five.

And it's going to be G because we multiply our vector by 2.

5.

Great question.

Well done if you got this one right.

For question 3, it's a parallelogram.

And it's a parallelogram because we know vector DA is parallel to vector BC.

AB is parallel to DC.

And also, DA is parallel to BC, and AB is parallel to DC.

Great work.

Have you got this? Question 3, it's an irregular quadrilateral.

There are no vectors that are parallel or equal in length.

For question 3, it's a trapezium.

Now the reason why it's a trapezium is because we only have one pair of parallel lengths.

IH is parallel to JK, and this is because IH is equal to -2 multiplied by our vector JK.

Great work if you've got this.

Great work, everybody.

So now it's time to move on to the second part of our lesson, problem solving with parallel vectors.

Sometimes we're asked to show two vectors of parallel, even though a diagram may make it appear that they're not.

And since we know vectors are parallel, if one is a scalar multiple of the other, we can show vectors are parallel by calculating them either algebraically or in column form.

And then we can identify if there is a scalar multiplier.

For example, here we have quadrilateral.

AOBC is given, and Y is the midpoint of AC.

We have all these beautiful vectors identified on our diagram, and we need to show that the vector OY is parallel to 4P add Q.

So I'm simply gonna identify the vector OY here and then form a vector pathway to find OY.

I've chosen the vector OA add a half of vector AC.

Substituting that, we know OA is 12P, and then we have to work out what the vector AC is.

Now I've worked out the vector AC to be the resultant vector of vector AO and vector OB add vector BC, which I've substituted here.

Then I'm going to work it all out and simplify to find the vector OY.

The vector OY is 10P add 2.

5Q.

So OY is parallel to 4P add Q because if we multiply 2.

5 by 4P add Q, we get the vector OY.

In other words, one is a scalar multiplier of the other, so they're parallel.

I just want to say there are often multiple ways to approach this type of problem.

Can you find another way? Well done if you did, and you would've got the same answer.

You would've shown that the vector OY is parallel to 4P add Q.

Well done.

Now let's move on to a check.

OACB is a parallelogram.

And M is the midpoint of OA, and N is the midpoint of OB.

We know the vector OM is 4A, and the vector ON is 4B.

And you may need to show that the vector AB is parallel to MN.

See if you can give it a go.

Press pause as you'll need more time.

Great work.

Let's see how you've got on.

Well, we know the vector AB is 8B subtract 8A.

I've chosen that vector pathway and worked it out.

Then I've worked out vector MN.

I've worked out vector MN using this pathway and worked it out to be 4B subtract 4A.

I can clearly see that the vector MN is one half of our vector AB.

So that means they are parallel.

Great work if you've got this.

Fantastic work, everybody.

Now it's time for your task.

Here, we have OPZ.

It's a triangle.

And T is the midpoint of our vector OZ.

M is the midpoint of our vector YZ.

Vector OX is equal to vector TY, and that is equal to 2A.

The vector OT is equal to 2B, and we're asked to show that the vector TM is parallel to vector OY.

So you can give it a go.

Press pause if you need more time.

A little hint, just label on that diagram as it really does help, and don't forget your direction.

Well done.

Let's move on to question 2.

Question 2 shows a quadrilateral, A, B, C, and D.

X is the midpoint of AB.

Y is the midpoint of BC, and Z is the midpoint of CD, and W is the midpoint of AD.

We know the vector AW is 5A.

We know the vector of AXE is 5B, and we know the vector DZ is 5C.

and we're asked to show that XWZY is a parallelogram.

Great question.

See if you can give it a go.

Same again.

Label the diagram and don't forget your direction.

Great work, everybody.

So let's go through these answers.

Firstly, I'm gonna label my diagram with all those beautiful vectors and their direction.

Notice how I've also used those hash marks to identify those equivalent banks and the midpoint.

From here, we can work out the vector TM using this vector pathway.

I've chosen vector TY add a half of vector YZ.

Substituting what we know, I know the vector TM is A + B.

Now let's find the vector O to Y.

I've chosen this vector pathway.

O to T, add T to Y.

I've worked this out to be 2A add 2B in its factorised form.

This would be 2 multiply by A + B.

I do like things in their factorised form as they show the factors and the scalar multipliers a little bit easily.

So that means you can see vector OY is equal to two lots of vector TM.

So yes, they are parallel.

Well done if you got this.

Question 2 is a great question.

Let's start by labelling directions and vectors.

Well done if you've got this.

And let's work out our vector XW first.

Well, for me, I've chosen this vector pathway, and I've got -5B add 5A.

For vector XY, I've chosen this vector pathway.

And I've worked out vector XY to be 5A add 5C.

For vector WZ, I've worked it out to be vector 5A add 5C.

And for vector YZ, I've chosen this vector pathway and worked it out to be -5B add 5A.

Now, let's identify if it's a parallelogram.

It is because vector XY is equal to vector WZ, and vector XW is equal to vector YZ.

So therefore, we know it is a parallelogram.

Fantastic work, everybody.

So in summary, two vectors are parallel if one is a scalar multiple of the other.

And remember the direction of the vector does not affect whether two vectors are parallel.

Great work, and it was wonderful learning with you.