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Hi, everyone.
My name is Miss 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 dividing vectors into ratios under the unit Vectors.
By the end of the lesson, you'll be able to solve geometric problems in 2D where vectors are divided into a given ratio.
Let's have a look at some keywords.
A vector can be used to describe a translation.
And 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.
Today's lesson will be broken into two parts.
Dividing vectors into a ratio first and then moving on to further forming vector pathways.
So let's make a start dividing vectors into a ratio.
Here we're asked to divide a line segment AB into the ratio 3:1.
What do you think that would look like? Well, it means we would divide the line segment into four equal parts or lengths.
And this now represents our three parts, and this now represents our one part.
But what fraction of the line segment does this represent? Well, it represents 3/4 of our line segment AB.
What fraction of the line segment do you think this represents? Well, it represents 1/4 of the line segment AB.
There is no difference when using vectors.
For example, this represents 3/4 of our vector AB.
This represents 1/4 of our vector AB.
We use our knowledge of fractional line segments and apply this to vectors.
Therefore, we can divide a vector into a ratio whether it be in common form or algebra form.
For example, here we have vector OB, and A divides OB in the ratio of 2:3.
Given the fact that we know OB is equal to 15, 20 in column vector form, we're asked to write the column vector for OA and AB.
Well, to do this, let's indicate what fraction of our vector OB is represented.
We know OA is 2/5 of the entire vector OB.
We know AB is 3/5 of our entire vector OB.
So I've just labelled them here.
From here, let's substitute what we know.
We know OA is 2/5 of OB, so that means we're working out 2/5 of our vector 15, 20.
This gives me vector OA to be 6/8.
Next we know vector AB is 3/5 of vector OB, so that means we work out 3/5 of our 15, 20.
This gives me vector AB to be 9, 12.
The same applies with algebraic form.
For example, here we have the vector OB, and A divides OB in the same ratio 2:3.
But now we're given the vector OB in algebraic form.
So we know OB is 12a + 15b.
We're asked to write vector OA and vector AB in terms of a and b.
Well first of all, let's label it on our diagram.
Always helps.
You can see our identified vector OA to be 2/5 of vector OB and vector AB is 3/5 of our vector OB.
Now knowing this we can simply substitute.
Vector OA is 2/5 of our OB, so 2/5 of our vector 12a + 15b is simply 24/5a + 6b.
We know vector AB is 3/5 of OB, so that means we work out 3/5 of 12a + 15b.
This works out to be 36/5a and our 9b.
It's important to note you can provide the answers in factorised form or expanded form.
The use of fractions may help with more complex problems. So let's have a look at two check questions.
Here we have vector XY, and it's given as 21, -9 in column vector form.
The point W lies on the line XY and divides the line in the ratio 2:1.
We have to write the vector of XW as a column vector.
See if you can give these a go.
Press pause if you need more time.
Here we have vector AB, and vector AB is given as 4a + 6b.
Point C lies on vector AB and divides the line in the ratio of 3:1.
And you're asked to work out the vector of AC in terms of a and b.
See if you can give these a go.
Press pause if you need more time.
Well done.
Let's see how you got on.
Well, we know XW is two thirds of XY, so substituting this in, 2/3 of our 21, -9 gives us 14, -6.
Well done if you got this.
For the question, we know AC is 3/4 of AB, so we work out 3/4 of our 4a + 6b, which is simply 3a + 9/2b.
Really well done if you've got these two.
Now let's have a look at another check.
OL is divided into 12 equal parts.
And you're asked to fill in the missing information giving the ratio in its simplest form.
See if you can give this a go.
Take your time.
Press pause if you need.
Great work.
So let's see how you got on.
Well hopefully you can spot we're looking at the line segment OA:AL, and it wants us to write it in the ratio of one to something.
It should be 1:11, which fits the fraction of OL.
O to A is 1/12 of OL.
And then we can identify well that means the vector AL is 11/12 of our vector OL.
Next we have OH:HL.
Writing in a simplified ratio gives us 2:1.
This makes sense.
If you look at the vector O to H, it's 2/3 of our vector OL.
You can also see the vector H to L is 1/3 of our vector OL.
OF:FL is in the ratio of 1:1.
Well done if you got this.
That means the vector O to F is 1/2 of vector OL, and the vector F to L is 1/2 of vector OL.
Really, really well done if you got this.
Great work, everybody.
So now it's time for your task.
Here you're given vector XY.
XY is 24p + 8q.
And we're given the following ratios.
XP:PY is in the ratio 3:1, and we're asked to write vector XP in terms of p and q.
Part b gives you another ratio.
XR:RY is in the ratio of 5:3, and you're asked to write the vector XR in terms of p and q.
Lastly, given the ratio XT:TY is in the ratio of 1:1, you are asked to write XT in terms of p and q.
And given this ratio, what do you notice about the position of T on XY? See if you can give these a go.
Press pause if we need more time.
Well done.
Let's move on to question two.
Two Oak pupils have made errors with these questions.
Find their errors, and I want you to work out the correct answer.
Take your time and press pause if you need.
Well done.
Let's move on to question three and four.
We have two vectors here.
Vector O to A is 12p + 15q, and we know X lies on OA such that OX is 8p + 10q, but you're asked to write the ratio of OX:XA.
See you can give these a go.
Press pause if you need more time.
Question four.
You have the vector X to Y, which is 10p - 15q.
Now N lies on that vector XY, such that vector X to N is 6p - 9q, and you're asked to write the ratio of XN:NY.
See if you can give these a go.
Press pause as you'll need more time.
Well done.
Let's move on to these answers.
For question one, you should have had these answers.
And hopefully you spotted T is the mid point.
Well done if you got this.
For question two, well, let's have a look at what Alex did.
Vector XW is 1/4 XY.
Substituting in what we know XY to be means the vector XW is 3a + 6b.
Next let's have a look at what Jun did.
The NB is 3/5 of AB and AN is 2/5 of AB.
So that means we can write our vector A to N is equal to 2/5 AB, and then we can substitute what we know to identify vector AN is 6a + 8b.
For question three, one approach would be to add the vector OX and the vector XA, which gives us 12p + 15q.
Now given that OX is equal to 8p + 10q, this means X to A must be 4p + 5q.
So that means we know vector OX is 2/3 of our 12p + 15q.
And we know vector XA is 1/3 of our 12p + 15q.
So that means the ratio is 2:1.
For question four, one approach again would be to sum vector XN and vector NY, giving me 10p - 15q.
Given that vector XN is equal to 6p - 9q, this means that NY is 4p - 6q.
This means we know vector XN is 3/5 of 10p - 15q, and we know vector NY is 2/5 of our 10p - 15q.
So the ratio is 3:2.
Great work, everybody, so let's move on to further forming vector pathways.
Now we can apply our knowledge of dividing vectors into ratios with writing vector pathways in order to form a resultant vector.
For example, here's a parallelogram OACB.
And what we know is the vector OA is 2a and the vector OB is 2b.
The question says P divides the line AB in the ratio 3:1, and we're asked to write the expression for OP in terms of a and b.
So let's have a look at this.
First, let's form a vector pathway.
I know to work out vector OP I can sum the vector OA and 3/4 of vector AB.
So substituting in what we know, I know vector OA is 2a and I'm working out 3/4 of vector AB.
Now vector AB is -2a + 2b.
So let's simplify and find out the vector OP is 1/2a + 3/2b.
Vectorizing is 1/2 bracket a + 3b.
Either of these forms is absolutely fine.
Now let's have a look at a check.
Laura says, "I would've used this pathway: vector OP is equal to vector OB and 1/4 of vector BA." I want you to work out the answer using Laura's vector pathway and show it gives the same answer of vector OP being 1/2 bracket a + 3b.
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, using Laura's pathway, let's substitute what we know.
We know vector OB is 2b + 1/4 of vector BA is -2b + 2a.
Expanding and simplifying, I get 1/2a + 3/2b, which is exactly the same.
1/2 bracket a + 3b.
And that's what's great about resultant vectors.
You can use multiple vector pathways and still get the exact same answer.
There may be more than one ratio to consider.
For example, OACB is a quadrilateral.
And we're given the vector OB to be b, BC to be 5a + 2b, and AC to be 3b.
Now D is a point on OA such that the vector OD:DA is in the ratio 2:3.
E is on a point AC such that the vector AE:EC is 1:2, and we're asked to work out the vector DE in terms of a and b.
Well, let's identify roughly where D and E is, which is around about here.
And I'm going to identify those vectors on our diagram.
They always help.
And from here, we can form a vector pathway.
I'm using the vector pathway of 3/5 of OA + 1/3AC.
This will give me the vector DE.
Substituting what we know.
Well, 3/5 of the vector OA means I need to work out vector OA.
OA is simply found by b + 5a + 2b - 3b, and then we're adding 1/3 of all vector AC.
Vector AC is given as 3b.
Simplifying, we know vector DE is 3a + b.
Now let's have a look at a check.
Only one vector path can ever be formed between two points.
True or false? And I want you to justify your answer with one of these solutions.
There is only one shortest path, or b, multiple vector paths can lead to the same resultant vector.
Well done.
Well hopefully you spotted it's false, and it's simply because multiple vector paths can lead to the same resultant vector.
And that's what's great about resultant vectors.
You can use multiple vector pathways and still get the exact same answer.
Fantastic.
Great work, everybody.
So now it's time for your task.
This is triangle OAB, and the vector OA is 9a and the vector OB is 6b.
The point P lies on AB such that AP:PB is 1:2.
Vector OP can be written as k multiply by 3a + 3b.
And the question wants you to work out the value of k.
Great question here.
I'm gonna give you a little hint.
Label what you can on that diagram including vectors, and direction, and positions.
See if you can give it a go.
Press pause for more time.
Well done.
Let's move on to question two.
Great question from question two.
We have OCD and OMB are triangles.
Vector OC is 9p and vector OB is 8q.
We know AMB is a straight line, and M is the midpoint of that line AB.
We have a ratio OA:AC is in the ratio 2:1, and N is a point on OB such that the ratio of ON:NB is in the ratio 1:3.
Question a wants you to write vector AN in terms of p and q, and vector b wants you to show vector MN is -3p - 2q.
Great question here.
I'm gonna give you a little hint.
Label what you can on that diagram including vectors, and direction, and positions.
Let's see if you can give it a go.
Press pause if you need more time.
Well done.
Let's look at these answers.
Well, for question one, you should have worked out vector OP means that k = 2.
Well done if you got this.
And for question two, great question.
I've just labelled all the information I can on my diagram and then we can work out the following.
I know vector AN is -6p + 2q, and I've justified vector MN using the vector path of MA + AN.
I've shown that vector MN is -3p - 2q.
Great work.
Fantastic work, everybody.
So in summary, we can divide a vector into a ratio, whether it be in column or algebraic form.
More complex questions can involve multiple vectors being divided into differing ratios.
Really well done, everybody.
It was great learning with you.
