Lesson video

Hi, everyone.

My name is Ms. Coe, 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.

Hello everyone and welcome to this lesson on geometric proofs with vectors, under the unit vectors.

And by the end of the lesson you'll be able to produce geometric proofs to prove that points are collinear, or that vectors are parallel.

So let's have a look at some keywords.

A vector can be used to describe a translation.

The vector 2 -5 shows the translation 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 the single vector that produces the same effect as a combination of other vectors.

We'll also be looking at the word collinear.

When three or more points lie on a single straight line, these points are said to be collinear.

Today's lesson will be broken into two parts.

We'll look at geometric proofs with vectors, including polygons first, and then proving points are collinear, using vectors.

So let's make a start looking at geometric proof with vectors involving polygons.

I want you to have a look at this diagram and tell me what shapes are these, and how can you be certain.

See if you can give it a go.

Press pause if you need more time.

Well done.

Well hopefully you've spotted that the vector C to A is -2 2, written as a vector in column form.

Vector BD is 5 -5, written as a vector in column form.

Now we can actually see there's a relationship between these two vectors.

If you multiply vector CA by -2.

5, you get vector BD.

So what does it mean? It means that these two vectors are parallel, and we also know there's no scaler multiplier connecting vector AB to vector CD.

So I'm gonna simplify it as vector AB is not a scaler multiple of CD.

So that means we know quadrilateral ABCD is a trapezium.

Next I know vector OZ is equal to vector XY as they both are the same column vector, 3 -1.

So that means they are parallel.

Then I'm gonna look at vector ZY.

Now vector ZY is exactly the same as vector OX, which has the column vector form 5 3.

So they're parallel.

So that means we have OXYZ is a parallelogram, because vector OZ is parallel to vector XY, and we know vector OX is parallel to vector ZY.

Well done.

Vectors can be used to demonstrate that a shape has particular properties, and these properties can help us identify the shape.

For example, here we have OAB and ABC are two triangles.

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

And you can see I've indicated it here on the diagram.

We're going to identify some vectors.

Now we know the vector OM is 4a and the vector ON is 4b.

So notice how I've labelled this on my diagram, because we know M is the midpoint of OA.

That means I know OM is 4a, but also MA is 4a.

And because we know N is the midpoint of OB, that's why I've written ON is 4b and NB is 4b.

So what type of quadrilateral is MABN? Well to do this, let's work out vector AB first.

Vector AB, using the vector path OA at OB, is 8b - 8a.

Now let's look at the vector MN.

Well finding the vector MN can be found using this vector path, M to O, O to N.

So from here I can work out the vector MN to be 4b - 4a.

Now you might be able to spot a relationship.

Vector MN is equal to half of vector AB, so therefore they are parallel.

And you might also notice that the vector MA is not a scalar multiple of NB.

So that means we know MABN must be a trapezium.

Now it's time for a check.

We have a quadrilateral, AOBC, and Y is the midpoint of AC.

And what we need to do is show that OYCB is not a trapezium.

See if you can give it a go.

Press pause if we need more time.

Well done.

Let's see how you got on.

Well first of all, the vector OY can be found using this vector path, vector OA and a half of vector AC.

Substituting what we know, well, I know OA is 12p, but I need to work out vector AC.

So vector AC can be found by summing -12a and 8q and 8p -3q, and then half of this vector AC can be found.

Expanding and simplifying then gives me vector OY to be 10p + 2.

5q.

Now we know OY, and we know BC, and we know vector YC, and we know vector OB.

That means we can make an informed decision.

None of these vectors show they are parallel to another vector.

So therefore OYCB is not a trapezium, as we do not have a single pair of parallel sides.

Great work everybody.

So let's move on to your task.

Here we have a wonderful diagram plotted on a square grid.

Using the column vectors only, explain if one or the following are true or false.

Is OABD a trapezium? We're going to explain it using those column vectors.

Is ABEF a trapezium? Same again, I want you to explain using column vectors.

Is OACD a parallelogram? Same again, I want you to explain using column vectors.

So you can give it a go.

Press pause if you need more time.

Well done.

Let's move on to question two.

Question two says ABCD is a kite.

X is the midpoint of AB, Y is the midpoint of BC, and Z is the midpoint of CD.

W is also the midpoint of AD.

We're given the vectors AXE is 5b, AW is 5a, and DZ is 5c.

And the question wants us to work out what type of quadrilateral is XWZY.

See if you can give it a go.

My hint to you, add all these vectors to this diagram, as it really helps you work out that final answer.

Give it a go.

Press pause if you need one time.

Well done.

Let's see how you got on.

Well for question one here is my quadrilateral OABD.

I've pulled it out of the diagram.

From here I can see it's true.

And it's true because the vector OA is not a scalar multiple of the vector BD.

It's true because vector AB, in column form, is 3 -1, and vector OD, in column form, is 6 -2.

So that means two lots of vector AB is equal to vector CD.

So we only have one pair of parallel size.

So therefore it is a trapezium.

Well done if you've got this.

Fantastic work if you use the correct notation as well.

Now for the second part, ABEF is a trapezium.

Let's find out if it's true or not.

I've pulled it out the diagram so you can see it clearly here.

Hopefully you can spot it's false.

It's false because the vector AB is exactly the same as the vector FE, which in common form is 3 -1.

Then we know vector FA is exactly the same as vector EB, which in common form is 1 7.

So therefore we have two pairs of parallel sides.

So it's not a trapezium, it's a parallelogram.

Lastly, let's have a look at C.

All I've done is pull out this image to show what OACD looks like.

Is it a parallelogram? Yes it is.

So let's show it using those column vectors.

I know vector OA is exactly the same as vector DC, as in column form, it is 2 3.

Vector AC is the same as vector OD, because in column form it's 6 -2.

So therefore we have two pairs of parallel sides.

So it must be a parallelogram.

Great work if you got this.

Now for question two, great question.

First of all, always draw that wonderful information on our diagram.

So you can see all my vectors and my points plotted.

We can now identify what type of quadrilateral XWZY is.

Well first of all, let's work out WZ as a vector.

I've used this vector path, and I've worked out vector WZ to be 5a + 5c.

Next let's work out vector XY.

Using this vector path, I've worked it out to be 5a + 5c.

So that means vector WZ is exactly the same as vector XY.

From here let's work out vector XW.

I've used this vector pathway to work it out to be -5b + 5a.

Then let's work out vector YZ.

Using this vector path, I've worked it out to be 5a - 5b.

So that means I can see vector XW is exactly the same as vector YZ.

So I have a parallelogram.

It's a parallelogram, because vector WZ is exactly the same as vector XY, and vector XW is exactly the same as vector YZ.

Great work if you've got this.

Fantastic work everybody.

So let's move on to proving points are collinear using vectors.

When three or more points lie on a single straight line, these points are said to be collinear.

For example, to show that three points are collinear, we choose two vectors that share a common point, and show that one is a scalar multiple of the other.

Here you can see we have the vector OX is equal to two lots of vector XY.

Since the two vectors have a common direction, and share a point, we know O, X, and Y are collinear.

Now let's have a look at a check.

I want you to select the correct conclusion statement, O, B, and C are collinear because.

Which one do you think is correct? Is it because they're parallel, they're on the same line, vector OB is equal to two lots of vector BC, so the vectors are parallel and share a common point, or is it simply because vector OB is equal to two lots of vector BC, so there is a common direction? What do you think? Well, it is C.

This is a fantastic conclusion statement, identifying that points O, B, and C are collinear.

We recognise that they are parallel and share a common point.

Really well done if you've got this.

So moving on, here we have a diagram.

OBA and OBC are triangles.

A is the midpoint of OC, and Y is the midpoint of AB, and X is on the line OB.

We know vector OA is p, and we know vector OX is 2q.

We also know the ratio of OX to XB is two to one.

And what we need to do is show that C, X, and Y are collinear.

Well to do that, let's identify these vectors and points on our diagram.

From here, let's work out vector XY.

Well I'm gonna use this vector pathway, vector XY can be found by summing vector XB and half of vector BA.

We know vector XB is q, and we're adding it to half of vector BA.

So a half of vector BA is 1/2 multiplied by -3q + p.

From here we can expand and simplify.

I can work out vector XY to be a 1/2 p - 1/2 q.

Now let's work out vector YC.

Well, I know vector YC can be found using this vector pathway.

Same as before, let's substitute what we know.

Well half of vector BA is 1/2 multiplied by -3q + p.

And we're adding it to vector AC.

Well we know vector AC is p.

Expanding and simplifying will give us the final answer.

I know vector YC is 3/2 p - 3/2 q.

So now we can see a relationship between our vectors.

Vector YC is equal to three lots of vector XY.

Since the two vectors have a common direction and share a point, we know that C, X, and Y are collinear.

Well done everybody.

So let's move on to a check.

We have AOBC is a parallelogram, and ACD is a straight line.

Vector AC to CD is in the ratio of one to one.

Now N divides AB in the ratio of two to one.

Show that O, N, and D are collinear.

See if you can give it a go.

Press pause if you need more time.

Well done.

Let's see how you got on.

Well, first of all we know it's a parallelogram, so let's identify our vectors.

We know it's a parallelogram, so therefore we know the vector OB is exactly the same as vector AC, and we know vector OA is exactly the same as vector BC.

Given that we know C is the midpoint, I can also add in this vector.

Well done if you added this to your diagram.

Now from here I'm gonna add some more information, given the ratio.

I know the vector A to N is 2/3 of our vector AB, and I know the vector N to B is 1/3 of our vector AB.

So I've just illustrated it here.

Now let's work out that vector ON.

Working out the vector ON, it's vector OB + 1/3 of vector BA.

So let's substitute.

We know vector OB is 2b, and we're adding 1/3 of vector BA.

This is 1/3 multiplied by our 2a - 2b.

This works out to be 2/3 a + 4/3 b.

Next, let's work out that vector ND.

Well, to find vector ND it's simply 2/3 of vector BA add vector AD.

Well, 2/3 of vector BA is 2/3 multiplied by 2a - 2b.

And we're adding our vector AD, and vector AD is 4b.

From here we simply expand and simplify.

Vector ND is 4/3 a + 8/3 b.

Hopefully you can spot two lots of vector ON gives your vector ND.

So therefore we know, since these two vectors have a common direction and share a point, we know O, N, and D are collinear.

Great work if you've got this.

Fantastic work everybody.

So let's move on to your task.

For question one, it says OAB and OZB are triangles.

X is the midpoint of OA and W is the midpoint of OB.

Y divides AW in the ratio AY to YW is equal to two to one.

We know vector OX is 3a and vector OW is 3b, and we're asked to prove X, Y, and B are collinear.

See if you can give it a go.

Press pause if you need more time.

Great work.

Let's move on to question two.

Question two says OYZ is a triangle.

Vector OY is equal to 10p, vector OZ is equal to 6q.

C divides the line OY such that OC to CY is in the ratio four to one.

M is the midpoint of YZ.

Now, Z divides the line OT such that the ratio of OZ to OT is three to four, and we're asked to prove that point C, M, and T lie on the same straight line.

See if you can give it a go.

Press pause if you need more time.

Well done.

Let's see how you got on.

Well for question one, to work out vector XY, it's simply vector XA + 2/3 of vector AW.

Well we know vector XA is 3a, and working out 2/3 of vector AW, it's simply 2/3 X -6a + 3b.

From here we simply expand and simplify.

Gives us -a + 2b.

Working out vector YB is made by adding 1/3 of vector AW with vector WB.

Well, 1/3 of vector AW is 1/3 multiplied -6a + 3b, and with summing it to vector WB, which is 3b.

Gives us -2a + 4b.

You can spot two lots of vector XY gives you YB, so that means they're parallel and collinear, as they have a common point.

Well done if you got this.

Next, it's always important to label our diagram.

So look at all these wonderful vectors that I've labelled on here, as well as those key points.

From here we can work out the vector CM.

The vector CM is -3p + 3q.

We can work out the vector MT.

So this gives us -5p + 5q.

Then we can work out a relationship between these vectors.

5/3 of vector CM is equal to vector MT.

Therefore we know it's parallel and collinear, as they have a common point.

Great work everybody.

So in summary, the properties of various polygons can be applied to vector questions.

We can prove whether three or more points are collinear by showing that two vectors are parallel to each other and share a common point.

Great work, everybody.

It was wonderful learning with you.