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 problem solving with vectors under the unit vectors, and by the end of the lesson, you'll be able to use your knowledge of vectors to solve problems. Today's lesson will include these key words.

A vector can be used to describe a translation.

So the vector 2, -5 shows the translation of two units to the right and five units down.

Displacement is the distance from the starting point when measured in a straight line, and a resultant vector is a 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 vector problems involving polygons first and then solving equations involving vectors.

So let's make a start looking at vector problems involving polygons.

Now when a polygon is identified in the question, we can use its properties to identify other vectors, for example, write what vectors you immediately know, given this parallelogram and this regular hexagon.

See if you can give it a go.

Press pause if you need more time.

Well done.

Well, hopefully you've spotted, as it's a parallelogram, these vectors will be exactly the same.

And for our regular hexagon, we have these vectors which will be exactly the same.

Really well done if you spotted this.

So we can solve more complex problems when vectors are not given, as we're able to use the properties of any known polygons.

For example, here are two regular hexagons.

From the diagram, let's work out in terms of PQ, the following vectors, vector OA, vector OB, vector OC, vector OD, and vector OE.

Well, firstly, it's always good practise to label other vectors if it helps.

So notice how I've labelled all these wonderful vectors, given what we know about regular hexagons.

Now from here, we can easily find that vector OA.

Vector OA is 2P.

Vector OB has to be 2Q.

Vector OC is 2Q - 2P.

Vector OD is 2P - 2Q.

And vector OE is -Q.

Well done if you got this.

Now here is a star with a centre O and it's made up of six congruent parallelograms. Sam said these must be -4B, as they're the other way round from 4B.

Explain if Sam is correct.

No, Sam is incorrect because the -4B means it's parallel to the direction of vector 4B, but it's in the opposite direction.

This is an example of -4B.

So let's have a look at this same shape again.

Here's the star still with the same centre O and it's still made up of our six congruent parallelograms. I want you to write the simplified vectors of the following in terms of A and B.

As a little hint, label other vectors that you may know from our diagram.

So you can give it a go.

Press pause if you need more time.

Well done, let's see how you got on.

Well, labelling all our vectors that we know, we have this and now we can easily find our vector OX, which is -3A.

Our vector OY is -3A - 4B.

And our vector OZ is 3A add our 4B.

Really well done if you got this.

Great work, everybody.

So now it's time for your task.

The Oak pupils programme and electronic spider to traverse a vector diagram which is made of three regular hexagons.

Each programme starts in the centre.

Where does the spider finish for each programme? Now Izzy programmes the vector 3A.

I want you to mark X on the diagram where that spider finishes for Izzy's programme.

For Aisha, Aisha programmes 2A - 2B.

Mark with a Y, where does the spider finish for Aisha's programme? Laura programmes -2A add 2B.

I want you to mark with a Z, where does the spider finish with Laura's programme? Sam programmes -3A.

I want you to mark that with a P, where does the spider finish with Sam's programme? And Sophia programmes -3A add 3B and I want you to mark that with a Q, where does the spider finish using Sophia's programme? See if you can give it a go.

Press pause if you need more time.

Well done, let's move on to question two.

Now I want you to write the programme in terms of A and B.

Each programme starts in the centre and finish at these different points.

So you can give it a go.

Press pause for more time.

Well done.

Now let's have a look at these answers.

For Izzy's programme, the spider should end up there.

Remember it starts at the centre O and it finishes there.

Well done if you got this.

For Aisha's programme, remember it starts at zero and it's going to finish here, which we've marked with a Y.

Well done if you got this.

For the next programme, which is Laura's, remember we're starting at O and we finish here.

Well done if you got this.

And for Sam's programme, it should be here.

Well done.

And for Sophia's programme, we should be here.

Fantastic work, everybody, if you got this.

Next, for point A, it should be 3B.

For point B, it should be A - B.

For point C, it should be 2A - 2B.

And for point D, it should be 3A - 3B.

Great work.

Fantastic work, everybody.

So let's move on to solving equations involving vectors.

From these common vectors, I want you to work out the resultant vector.

We're given the vector A in column form is 4, 5.

We are given the vector B in column form is 3, -2.

I want you to work out the resultant vector of 2A and 2B, and I want you to work out the resultant vector of 3A - 2B.

Press pause as you'll need more time.

Well done.

Well, let's see what you've got.

Working out the resultant vector, we should have had 14, 6.

Working out the resultant vector of 3A - 2B, we should have had 6, 19.

Well done if you got this.

So if we're given the resultant vector, it's possible we may be able to work out an unknown component, or scaler multiplier.

For example, how do you think we can form an equation if we have this, two lots of the vector XY, add three lots of the vector 1, 3 gives us the final answer of 11, 13.

How do you think we can form an equation? Well, we can identify the equation from each component.

Let's look at that horizontal component first.

Looking at the horizontal components, you can see we multiply two by the horizontal component of X, add three multiply by the horizontal component of one, will equal the horizontal components of 11, so therefore 2X + 3 = 11.

Another equation can be formed looking at the vertical components.

Two multiply by Y, add three multiply by three, will give us the resultant vertical component of 13, thus we have the equation 2Y + 9 = 13.

From here we can solve for X and Y.

We can work out X to be four, and we can work out Y to be two.

Knowing this, let's see if you can work out the values of X and Y for each question.

See if you can give it a go.

Form those equations and solve.

Press pause as you'll need more time.

Well done.

Let's see how you got on.

Well for A, looking at those horizontal components, we should have these equations.

3X + 10 = 22, and 3Y + 2 = 8.

Working out X, we know X is four, and working out Y, we know Y is equal to two.

For B, let's identify those equations.

Well, looking at the horizontal components, we have 5X - 6 = 4.

And looking at the vertical components, we have 15 + 3Y = 24.

From here, we can solve.

Solving for X, we have X = 2.

And solving for Y, we have Y = 3.

Lastly, here we have the horizontal components of 6X - 9 = 9, and the vertical components of 4X + 3 = Y.

Therefore, we know solving for X, X is three, and solving for Y, we have Y = 15.

Great work if you've got these.

So what would happen if we formed equations with more than one variable in each? For example, we're asked to form the equations for the horizontal and vertical components.

Here we have R is multiplied by the vector 2, 3 and T is multiplied by the vector 3, 8 and that gives us the resultant vector of 4, 13.

We know the horizontal equation would be 2R + 3T = 4, and the vertical equation would be 3R + 8T = 13.

So what have we just formed? Well, we've just formed simultaneous equations so that means we can use our knowledge on solving simultaneous equations.

This is why I love simultaneous equations 'cause there's so many different ways in which you can solve them.

For me, I'm going to look at this R term.

You might notice in equation one we have 2R, and in equation two we have 3R, so what we need to do is identify the lowest common multiple of R two and three.

Well, the lowest common multiple of two and three is six, so that means if I multiply equation one by three, I have 6R as my term in my equation.

And if I multiply equation two by two, I also have 6R in my equation.

This is what I want because later on I'm going to subtract these terms because I want to eliminate the R term, ensuring that we multiply the whole equation by the same value.

And remember the second equation is being multiplied by a different value to the first equation.

So what we now have is 6R + 9T = 12.

And then 6R + 16T = 26.

Now I'm going to subtract.

Subtracting these gives me 7T = 14, so I've sold for 4T.

T is now two.

Next, let's substitute this value of T into our equation one.

You really can substitute into any equation here.

I'm just choosing equation one.

Substituting R value of T is two, I now have 2R + 6 = 4, so that means I've worked out R to be -1.

This is a great question as we're using our knowledge on vectors and simultaneous equations.

So what we're gonna do is I'm going to do the question on the left and I'd like you to do the question on the right.

Here, we're given column vectors and we're asked to solve for A and B.

Looking at our equation form from our horizontal components, we have 4B - 5A = 7.

Looking at the equation form from our vertical components, we have -5B + 5A = -5.

From here, using our knowledge on simultaneous equations, I can simply sum equation one and two to give me B to be -2.

Next, I'm gonna substitute it in into any equation.

I'm gonna choose the first one and then solving for A to give me A = -3.

Now it's your turn.

I want you to solve for A and B.

Take your time and press pause if you need.

Well done.

Let's see how you got on.

While forming those equations from the horizontal component, the vertical component, we should have 2B - 6A = 16, which is formed using the horizontal components.

And we should have equation two being B + 6A = -1, which is formed using the vertical components.

And then solving our simultaneous equation should identify B to be five.

Substituting into any equation you want, really, I chose the first one, gives me A to be -1, so notice how we've solved for A and B, well done.

Great work, everybody, so now it's time for your task.

I want you to solve the following unknowns.

For question 1A, two is multiplied by the column vector X5, add six is multiplied by the column vector 2Y, giving the resultant vector of 18, 22.

For 1B, three is multiplied by the column vector 1Y, add five is multiplied by the column vector X5, giving the resultant vector of 23, 16.

And lastly for C, 2/3 is multiplied by the column vector 9Y, add two multiplied by the column vector X-4.

This gives us the resultant vector of 4, 0.

Press pause if you need more time.

Well done.

Let's have a look at question two.

Solve for X and Y.

Here we're given X is multiplied by the column vector 4, 6, add Y is multiplied by the column vector 3, -5, giving us the resultant vector of 27, -7.

Press pause as you'll need more time.

Well done.

And for question three, solve for A and B where A is multiplied by the column vector 3, 4, and we're adding it to B being multiplied by the column vector 2, 3, giving us the resultant vector of 330, 445.

Press pause as you'll need more time.

Great work.

Let's have a look at these answers.

Well, hopefully you've identified the horizontal and vertical components found in equation where the equation formed from the horizontal component is 2X + 12 = 18, and the equation formed from the vertical component is 10 + 6Y = 22, and solve for X and Y where X is three and Y is two.

Question B, forming those equations using the horizontal and vertical components from the horizontal components and from the vertical components, so the equation you should have formed using the horizontal components is 3 + 5X = 23, and the equation formed from the vertical components is 3Y + 25 = 16, and solve them simultaneously to give you X = 4, and Y = -3.

And for C, once again, forming those equations using those horizontal and vertical components, we can form those equations using the horizontal components, we simply have 6 + 2X = 4, and forming equation from the vertical components, we have 2/3Y - 8 = 0.

We can solve using our knowledge of simultaneous equations to find X = -1 and Y = 12.

For question two, I've structured it a little bit more for you to see, so you can see forming our equations, we have this which can be solved simultaneously, giving me an answer for Y to be five, and giving me an answer for X to be three.

Well done.

Next, let's form our equations using the horizontal and vertical components.

You should have 3A + 2B = 330, and our second equation is 4A + 3B = 445.

And then from here, one method to solve this pair of simultaneous equations is to look at the B term and make the coefficient of the B term the same.

Here you can see I have a B term of 2B, and a B term of 3B.

So identifying the lowest common multiple of 2 and 3B, it simply says so that means I need to multiply equation one by three to get me 9A + 6B = 990, and I need to multiply equation two by two, giving me 8A + 6B = 890.

Notice how both equations now have 6B as a term in their equation.

We can then subtract, giving me A to be 100.

This is why I like simultaneous equations 'cause there's lots of different ways in which you could have solved for B.

For me, I'm choosing to substitute that value of A to be 100 into equation one.

You could have substituted it into equation two if you wanted as well.

From here, I've simply solved to give me B = 15.

Really well done if you got this.

Fantastic work, everybody.

And remember, the properties of polygons allow us to solve problems when only some vectors are indicated, as we are able to use properties of polygons.

Simultaneous equations can be formed to help us work out the unknown multiplier or multipliers and/or horizontal and vertical components.

Great work, everybody.

It was wonderful learning with you.