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 enjoy today's lesson, so let's start.

Hi everyone, and welcome to this lesson on fluency in arithmetic procedures with algebraic vector notation under the unit, "Vectors." And by the end of the lesson, you'll be able to carry out arithmetic procedures on vectors written algebraically.

Now our keywords consist of the word resultant vector, and a resultant vector is the single vector that produces the same effect as a combination of other vectors.

Today's lesson will be broken into two parts.

We'll be looking at operations with vectors first and then operations with vectors from diagrams. So let's make a start looking at operations with vectors.

Here, we have two vectors, vector AB and vector DC.

What do you notice about these two vectors? Well, the directions are the same.

They are both parallel because the gradients are the same.

The scalar multiplier is two because vector AB is two three in column form.

And if we multiply this by two, this means we get the vector DC, which is the vector four six in vector column form.

So that means two multiply by the vector AB is equal to vector DC.

We know they are parallel as one is a scalar multiplier of the other, and we know the direction of AB is also the same as DC.

Now remember, vectors can be placed anywhere.

So comparing vectors EF and the vector EG, what do you notice? Well, same again, we have the direction to the same.

They are both parallel because the gradients are the same.

And you may have spotted, two multiply by the vector EF is equal to vector EG.

Given this diagram, we can also say F is the midpoint of our vector EG.

What I want you to do, I want you to complete the sentences using this diagram.

C is the midpoint of what? 2BC is equal to what? And 2BD is equal to what? So you can give it a go.

Press pause if you need more time.

Well done.

Let's see how you got on.

Well, C is the midpoint of our vector BD and two lots of vector BC is equal to vector BD, and two lots of vector BD is equal to our vector AE.

Very well done if you got this.

Now using the same diagram, which of the following are true? So you can give it a go.

Press pause if you need more time.

Great work.

Let's see how you got on.

Well, it's this statement and this statement are true.

1/2 of vector BD is equal to vector CD and 1/2 of vector AE is equal to vector BD.

Great work.

Removing the grid and column vectors, we can identify parallel vectors and midpoints using scalar multipliers.

For example, the vector AB is equal to 2a and 12b, and two multiply by vector CD is equal to AB.

And we're asked to work out vector CD in terms of A and B.

Well, we know two lots of vector CD is equal to AB.

So we can rearrange to make CD the subject.

This means vector CD is 1/2 of vector AB.

Now we can substitute our vector AB in terms of a and b.

This is 1/2 of 2a and 12b.

So that means we know the vector CD is a plus 6b.

Just to let you know, we can have CD written in its factorised form or expanded form.

Now let's move on to a check.

I'm going to do the first question and I'd like you to try the second.

We know vector OA is equal to 3a subtract 4b, and X is the midpoint of OA.

The question wants us to work out the vec OX in terms of a and b.

Well, we know OX is equal to 1/2 of OA.

This is because X is the midpoint of OA.

So that means we have 1/2 multiply by 3a subtract 4b, working this out we have 1.

5a subtract 2b.

In its factorised form or expanded form is absolutely fine.

Now what I want to do is really embed this a little further by showing you where the diagram, and you can draw a diagram if it does help you with the visualisation.

Now we know the vector O to A is 3a take away 4b, and we know the vector X is the midpoint of OA.

So that means we simply halve the vector OA to give me the vector OX, which is 1.

5a subtract our 2b.

If it helps you, draw a diagram, because that visualisation will usually help students calculate the correct answer.

Now it's your turn.

Y is the midpoint of our vector AC.

Now if AY is equal to 6a subtract 12b, I want you to work out vector AC in terms of a and b.

If you want to draw a diagram to help with the visualisation, please do.

So you can give it a go.

Press pause if you need more time.

Well done.

Let's see how you got on.

Well, we know 1/2 of AC is equal to AY.

Given the fact that we need to work out AC, I'm going to rearrange this.

So the vector AC is the subject.

In other words, vector AC is two multiply by our vector AY.

Substituting what we know AY to b, we now have two multiply by our 6a subtract 12b, which is 12a subtract 24b.

Remember, factorised form or expanded form is absolutely fine.

I'm gonna draw a diagram now to help visualise this.

We know Y is the midpoint of AC.

So you can see I've drawn a vector A to C, I've identified the midpoint of Y, and I know AY is 6a subtract 12b.

Given the question wants us to find AC, that means I know the entire vector from A to C must be 12a subtract 24b.

Now let's move on.

The midpoint of AC is X.

Where do you think the midpoint of the vector AC would be in this diagram? Well firstly, we need to know what the vector AC looks like.

The vector AC looks like this.

Then, we can roughly locate the midpoint of our vector AC.

I'm gonna put it here.

Now let's move on to a question.

It states that the midpoint of vector AC is X.

And if vector AB is equal to 4p add q, and vector BC is equal to 3q, we can find our vector AXE in terms of p and q.

So let's identify how to find our vector AXE.

Well, we know AXE is 1/2 of our vector AC.

So we need to work out our vector AC in order to work out our vector AXE.

And vector AC can be found using this vector path, vector AB and vector BC.

So that works out to be 4p add q add 3q, which gives me vector AC to be 4p add 4q in its simplified terms. Now we can work out vector AXE.

Vector AXE is 1/2 of our vector AC.

We know vector AC is 4p add 4q, so therefore we simply work out our vector AXE to be 2p add 2q.

Great work, everybody.

Now it's time for your task.

For question one, I want you to fill in the blanks using the diagram.

For question two, read the question carefully, draw a diagram if it helps.

And for question three, same again, read the question carefully and draw a diagram if it helps.

So you can give it a go.

Press pause for more time.

Well done.

Let's move on to question four and five.

Question four wants you to work out the vector PX in terms of a and b.

And question five wants you to work out the vector OY in terms of e and f.

So you can give it a go.

Press pause if you need more time.

Great work.

Let's move on to these answers.

Question one, you should've had these vectors placed here.

Really, really well done if you've got this.

Question two, you should've had vector OY is equals to 6a and 7b.

And for vector three, vector AXE is 16a take away 4b.

Well done.

For question four, always draw on the diagram helps.

So that means we know, approximately, point X is here.

It's the midpoint of PQ.

So that means if you work out the vector PQ, it's simply 3a subtract 7b.

So therefore, the vector PX is 1/2 of PQ, which is 1.

5a subtract 3.

5b.

Now Y is the midpoint of OZ? Notice how I've drawn this on the diagram, as it helps me visualise.

We're asked to work out the vector OY.

So we know OZ is equal to the vector OX add XZ.

And summing these two vectors together gives us OZ to be negative 3.

5f.

Given the fact that we know OY is 1/2 of OZ, that means we know our vector OY is negative 1.

75f.

Really, really well done if you got this.

Great work, everybody.

Now it's time for the second part of our lesson where we're looking at operations with vectors from diagrams. Now we can use our knowledge of midpoints to express more complex vectors.

So for example, the vector OQ is equal to 2a subtract b, and the vector ON is equal to 2a add b.

We know N is the midpoint of vector OP, M is the midpoint of vector PQ, and Q is the midpoint of our vector OT.

Now we're asked to write the vector PQ in terms of a, b, and we're asked to work out the vector NM in terms of a and b.

So first of all, let's add all this wonderful information to the diagram.

You'll see here, the hash marks can be added to show which line segments have the same equal length.

Notice how we've used different hash marks to represent those different equal lengths.

Now I'm going to substitute in my vectors, as you can see here.

Now my vector diagram is complete, we can work out our vector PQ.

Now to work out the vector PQ, we write a vector pathway using known vectors.

So this is one example.

Vector P to O and vector OQ gives us our vector PQ.

Using the priority of operations, we can then sum these vectors together and then simplify, thus giving me vector PQ to be negative 2a subtract 3b.

You can draw that on the diagram if you want.

Now let's have a look at vector NM.

Same again, we'll go in to write our vector pathway using our known vectors and choosing N to P add 1/2 of vector PQ.

Substituting what we know once again using those priority operations and simplifying, we have the vector NM to be a subtract 0.

5b.

So identifying the known vectors on a diagram and adding that little bit of extra information really does help.

Now Laura says, "I wouldn't have worked out vector NM using that pathway." She would've used another pathway.

She says she would use vector NO and then sum it to vector OQ, and then sum it to 1/2 of vector QP.

Would Laura's approach get the same answer as what we got before? You can see my answers from before.

I want you to give this one a go.

Using Laura's vector pathway, will it give us the same answer for vector NM? So you can give it a go.

Press pause if you need more time.

Well done.

Let's have a look.

Well, the vector NM using Laura's pathway is given here, substituting what we know in terms of a and b, and then using that priority of operations identifies exactly the same answer, and that's what's great about resultant vectors.

If you have more known vectors, that means we have different pathways which will always give us the same resultant vector.

So therefore, yes, it's the same.

Now it's time for another check.

X is the midpoint of AB and Y is the midpoint of CD.

We have the vector AB is 4p add 2q and the vector BD is 8p subtract 4q.

We also know the vector AC is 4.

5p subtract 2q.

We're asked to write the vector CD in terms of p and q and write the vector XY in terms of p and q.

So you can give it a go.

Press pause if you need more time.

Well done.

Let's see how you got on.

A really good starting point is to label what we know in our diagram.

Notice how we have those midpoints who have indicated with those hash marks.

From here, I've identified unknown vectors which are labelled on the diagram with the correct direction.

Now let's work out vector CD.

Well, looking at vector CD, it can be found using this vector pathway.

C to A and A to B and B to D.

Substituting what we know from our diagram, that means I can work out the vector CD to be 7.

5p.

Well done if you got this.

Now let's have a look at vector XY.

Well, I'm going to use this pathway.

X to A and A to C and 1/2 of our C to D.

Substituting in what we know and then using that priority operations to work out vector XY to be 6.

25p subtract 3q.

Great work if you've got this.

Press pause if you need more time to have a look at the working out.

Well done.

I just want to point out that you may have used an alternative pathway.

For example, XY is equal to the vector XB add the vector BD, add 1/2 of the vector CD.

Substituting what we know, simplifying our terms gives us the exact same answer.

Great work, everybody.

So now it's time for your task.

Here, we have a triangle, O, X, and Z.

And we know X is the midpoint of OY.

We know OZ is given as 6b.

We know OX is given as the vector 2a.

Question 1a wants you to find vector ZX, and then B wants you to find vector MY.

So you can give it a go.

Press pause if you need more time.

Well done.

For question two, we have OACB is a parallelogram and M is the midpoint of OA, and N is the midpoint of OB.

We know vector OM is given as 4a and we know vector ON is given as 4b.

Want us to work out the vector AB in terms of a and b and then work out the vector MN in terms of a and b, and then find the missing multiplier when vector MN is equal to a multiplier of vector AB.

So you can give it a go.

Press pause for more time.

Well done.

Let's move on to question three.

M is the midpoint of our vector OY.

Q is the midpoint of our vector XM.

We're asked to find vector XM in terms of p and q and we're asked to find vector QO in terms of r.

This is a great question.

Writing different pathways will help you out here.

Give this one a go.

Press pause as you'll need more time.

Great work, everybody.

Let's go through these answers.

Well, for part A, finding vector ZX, I've used this vector pathway, and then it gives me the answer of 2a subtract 6b.

That will always be the vector ZX regardless of the vector pathway you've chosen.

For B, we need to find the vector MY.

Well, I've chosen this vector pathway, and I've got a final answer of 3a subtract 3b.

Once again, vector MY will always be 3a subtract 3b regardless of the vector pathway you've chosen.

For question two, let's see if we can work these out.

Well, substituting what we know into our diagram, I can work out the vector AB to be 8b subtract 8a.

For B, to find vector MN, I've used this vector pathway, so therefore the vector MN is 4b subtract 4a.

Lastly, for part C, hopefully you can spot the vector MN is 1/2 of vector AB.

Well done if you got this.

For question three, this was a tough one.

Really well done if you got any of this.

Well, we're labelling on our diagram, identify those equal length using those hash marks.

From here, we're asked to work out XM.

So I've chosen this vector pathway, and I've worked out vector XM to be 6p add 2q.

Next, vector QO in terms of r.

Well, first of all, we have to write P and Q in terms of r first.

To do this, one strategy would be to identify what the vector NQ is.

Well, we know the vector NQ can be formed using the vector pathway N to Y, Y to M, and 1/2 of MX.

And we also know the vector NQ is 2r.

So that means we can say 2r is equal to the algebraic terms of our vector NY add the algebraic terms of our YM, add the algebraic terms of our vector MX.

So from here, I simply expand and simplify to give me 2r is equal p add q.

So now I know r is equal to 0.

5p add 0.

5q.

So we can use this to find vector QO.

I'm using the vector pathway 1/2 of vector XM add 1/2 of vector MO.

Substituting in what we know, we know the vector QO is 3p add 3q.

Given we know vector r is 0.

5p add 0.

5q, that means I know the vector QO is 6r.

Fantastic work if you got that one, it was really tough.

Great work, everybody.

So remember, when using vectors, it's important to understand the notation.

For example, this diagram shows two times vector EF is equal to vector EG.

And this means the directions are the same, they're both parallel because the gradients are the same, and F is the midpoint of vector EG.

So EF is equal to 1/2 of vector EG.

We can express more complex vectors by forming vector pathways and using a fraction of a vector.

Really, really well done today.

It was great learning with you.