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Vectors and ratios- difficult to grasp

Proofing parallelograms and rhombus in vectors comes quite easy to me as well as midpoints but as soon as ratio is introduced, my mind goes blank.

The Cambridge A-Level edition maths I’m studying doesn’t teach me much in regards to ratio in vectors, therefore I presume it was taught in GCSE, however I don’t wish to buy a GCSE book just for one section.

Question I’m confused with is:

Points A and B have coordinates, A(10,1) and B(2,7). Point C lies on the segment AB such that AC:BC = x: 1-x, where 0<x<1

a. find the coordinates of C.

I’ve drawn a line and found the displacement AB vector to be = [-8,6]

Honestly, I have I have no clue where to start.

If someone can explain, I’d be very appreciative.

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Reply 1
Original post by KingRich
Proofing parallelograms and rhombus in vectors comes quite easy to me as well as midpoints but as soon as ratio is introduced, my mind goes blank.

The Cambridge A-Level edition maths I’m studying doesn’t teach me much in regards to ratio in vectors, therefore I presume it was taught in GCSE, however I don’t wish to buy a GCSE book just for one section.

Question I’m confused with is:

Points A and B have coordinates, A(10,1) and B(2,7). Point C lies on the segment AB such that AC:BC = x: 1-x, where 0<x<1

a. find the coordinates of C.

I’ve drawn a line and found the displacement AB vector to be = [-8,6]

Honestly, I have I have no clue where to start.

If someone can explain, I’d be very appreciative.

If thats the full question, youll have to get the coordinates of C in terms of x.
As the ratio sums to 1, just think of it as a weighting variable. So (1-x)% of A and x% of B. Obviously, x and (1-x) are fractions rather than percentages, but you should get the idea?
(edited 2 years ago)
Original post by KingRich
Proofing parallelograms and rhombus in vectors comes quite easy to me as well as midpoints but as soon as ratio is introduced, my mind goes blank.

The Cambridge A-Level edition maths I’m studying doesn’t teach me much in regards to ratio in vectors, therefore I presume it was taught in GCSE, however I don’t wish to buy a GCSE book just for one section.

Question I’m confused with is:

Points A and B have coordinates, A(10,1) and B(2,7). Point C lies on the segment AB such that AC:BC = x: 1-x, where 0<x<1

a. find the coordinates of C.

I’ve drawn a line and found the displacement AB vector to be = [-8,6]

Honestly, I have I have no clue where to start.

If someone can explain, I’d be very appreciative.

Imagine AB as the hypotenuse of a right angled triangle. You could use Pythag to work that the length of it is 10. So, you need to split 10 up into an x:x-1 ratio. You could set up a formula, but an easier way to do it would be to divide 10 into to parts such that one part is 1 unit longer than the other.

Now you know the magnitude of vector BC and it's going in the same direction as vector BA, so there are a couple of different ways to work out its terminal point, C. I think the simplest way is to use similar triangles.

Imagine two similar, right angled triangles, one with BA as the hypotenuse and one with BC as the hypotenuse. Set up the ratios to work out the side lengths and then use that to help you work out C.

Note: Edited because I gave too complete of a solution!
(edited 2 years ago)
Reply 3
Original post by mqb2766
If thats the full question, youll have to get the coordinates of C in terms of x.
As the ratio sums to 1, just think of it as a weighting variable. So (1-x)% of A and x% of B. Obviously, x and (1-x) are fractions rather than percentages, but you should get the idea?


Ratio sum is 1? Is that from the fact that AC:BC = x:x-1 and therefore AC:CB = x:-x+1.. therefore x cancelling out leaving 1??

I had approach it from the point where the displacement vector being [-8,6] and dividing -8/1 and 6/1 to give 1 ratio multiplying that by x would give me the value AC? At least I think.

I’m not sure why I’m having such trouble with this. I’m sure it’s pretty simple.
Reply 4
Original post by solvemyproblems
Imagine AB as the hypotenuse of a right angled triangle. You could use Pythag to work that the length of it is 10. So, you need to split 10 up into an x:x-1 ratio. You could set up a formula, but an easier way to do it would be to divide 10 into to parts such that one part is 1 unit longer than the other.

Now you know the magnitude of vector BC and it's going in the same direction as vector BA, so there are a couple of different ways to work out its terminal point, C. I think the simplest way is to use similar triangles.

Imagine two similar, right angled triangles, one with BA as the hypotenuse and one with BC as the hypotenuse. Set up the ratios to work out the side lengths and then use that to help you work out C.

Note: Edited because I gave too complete of a solution!

Length 10? The length of AB= b-a, how did you come up with 10?
Reply 5
Original post by KingRich
Ratio sum is 1? Is that from the fact that AC:BC = x:x-1 and therefore AC:CB = x:-x+1.. therefore x cancelling out leaving 1??

I had approach it from the point where the displacement vector being [-8,6] and dividing -8/1 and 6/1 to give 1 ratio multiplying that by x would give me the value AC? At least I think.

I’m not sure why I’m having such trouble with this. I’m sure it’s pretty simple.

x + (1-x) = 1
So the ratio "combines" the points A and B (joins the dots) to produce C.

With a question like this, it can help to put a value in for x and think about what it means. So the ratio
0.5:0.5
would mean you want to take the average of A and B so 0.5*A + 0.5*B.

A ratio of 3/4:1/4 would mean AC was 3 times larger than BC so the point C would be closer to B. Similarly if its 1/4:3/4, the point C is closer to A. C equals A when its 0:1 and equals B when its 1:0.

So
C = A*(1-x) + B*x = A + x*(B-A) = A + x*AB = B + (1-x)*BA
The A+x*AB formula is probably the "easiest" way to view it, but any of the expressions are obviously equivalent. x is a "slider" which has a value of zero at A and a value of one at B.
(edited 2 years ago)
Original post by KingRich
Length 10? The length of AB= b-a, how did you come up with 10?

Draw an additional coordinate, R, at (2, 1). The triangle ABR is a right angled triangle with hypotenuse AB, which would have a length of 10. Thinking of AB as a vector, it you were to travel along that vector, you'd travel 10 units.

With ratio problems, I tend to ignore the vector aspect and focus in on the geometry, which is why I started by thinking about the length of AB.

Imagine the problem had the segments AC:CB divided in a 3:7 ratio. This would mean that C is 3/10 of the way along the line AB, starting at A. Since triangle ABR has side lengths of 6, 8, and 10, I want a similar triangle with side lengths that are 3/10 of those, so lengths 1.8, 2.4, and 3. From point A, I would go up 1.8 and left 2.4 to get to (7.6, 2.8)

Out of curiosity, when the problem states "AC:BC = x: 1-x", are AC and BC written as vectors or are they representing line segments? I originally read the ratio as x:x-1, which gives a coordinate for C, but would only work if AC and BC are vectors, because then the ratio of AC:CB would be what I was working with.
Original post by KingRich
Length 10? The length of AB= b-a, how did you come up with 10?

I suggest you focus on mqb's responses here.
Reply 8
Original post by DFranklin
I suggest you focus on mqb's responses here.

Yeah, MQ is like my superhero when it comes to asking maths questions on here.
i do not think you can find the coordinates of C as numbers, the answer will have to include x, unless there is further information given somewhere.

:hmmmm2:
Reply 10
Original post by the bear
i do not think you can find the coordinates of C as numbers, the answer will have to include x, unless there is further information given somewhere.

:hmmmm2:

Yes, I have to give the solution to C in terms of x. It’s why I’m thrown of by this. Ordinary if I had a value for x, finding c would be a lot simpler to find.
Original post by KingRich
Yes, I have to give the solution to C in terms of x. It’s why I’m thrown of by this. Ordinary if I had a value for x, finding c would be a lot simpler to find.

What are you still unsure about? In #6, there are some simple values ot gain intuition. Feel free to choose your own values.
(edited 2 years ago)
Reply 12
Original post by mqb2766
What are you still unsure about? In #6, there are some simple values ot gain intuition. Feel free to choose your own values.

I’m just replying back to other people. I will be revisiting the question shortly and I’ll apply what you’ve written above and hopefully my brain figures it out.
Original post by KingRich
I’m just replying back to other people. I will be revisiting the question shortly and I’ll apply what you’ve written above and hopefully my brain figures it out.

ok. Its easier than you're thinking/expecting. Slighly unusual, hence the plugging values in in the previous post.
Reply 14
Original post by mqb2766
x + (1-x) = 1
So the ratio "combines" the points A and B (joins the dots) to produce C.

With a question like this, it can help to put a value in for x and think about what it means. So the ratio
0.5:0.5
would mean you want to take the average of A and B so 0.5*A + 0.5*B.

A ratio of 3/4:1/4 would mean AC was 3 times larger than BC so the point C would be closer to B. Similarly if its 1/4:3/4, the point C is closer to A. C equals A when its 0:1 and equals B when its 1:0.

So
C = A*(1-x) + B*x = A + x*(B-A) = A + x*AB = B + (1-x)*BA
The A+x*AB formula is probably the "easiest" way to view it, but any of the expressions are obviously equivalent. x is a "slider" which has a value of zero at A and a value of one at B.

I believe I understand it a little better. My question though: the formula A+x*AB, does this refer to the ratio AC?

And if it does, why is it not x*AB-A to give the value of C?
Original post by KingRich
I believe I understand it a little better. My question though: the formula A+x*AB, does this refer to the ratio AC?

And if it does, why is it not x*AB-A to give the value of C?

x is the variable in the question so AC:BC = x: 1-x.
C(x) = A+xAB is the equation of points on the line segment in terms of x, A and B. I was hoping you'd put the numbers in so for A (10,1), B (2,7), AB = B-A = [-8,6]
C(x) = (10,1) + x*(-8,6) = (10-8x, 1+6x)

If C(x) = x*AB-A
C(0) = 0*AB-A = -A = (-10,-1)
C(1) = 1*AB-A = B-2A = (-18,5)
Neither point lies on the line segment AB.

For C(x) = A+x*AB
C(0) = A+0*AB = A = (10,1)
C(1) = A+1*AB = B = (2,7)
Both points lie at the ends of the line segment AB. Therefore all other points C(x) lie in AB for 0<x<1.

Try and put simple numbers in like this if you're unsure about what an expression represents. Its easy to misuse algebra if you dont have a clear understanding of what it represents. Note #2 has all the information you need to solve the question. Its a couple of simple lines, at most. So
C(x) = A*(1-x) + B*x = (10,1)*(1-x) + x*(2,7) = (10-8x, 1+6x)
(edited 2 years ago)
Reply 16
Original post by mqb2766
x is the variable in the question so AC:BC = x: 1-x.
C(x) = A+xAB is the equation of points on the line segment in terms of x, A and B. I was hoping you'd put the numbers in so for A (10,1), B (2,7), AB = B-A = [-8,6]
C(x) = (10,1) + x*(-8,6) = (10-8x, 1+6x)

If C(x) = x*AB-A
C(0) = 0*AB-A = -A = (-10,-1)
C(1) = 1*AB-A = B-2A = (-18,5)
Neither point lies on the line segment AB.

For C(x) = A+x*AB
C(0) = A+0*AB = A = (10,1)
C(1) = A+1*AB = B = (2,7)
Both points lie at the ends of the line segment AB. Therefore all other points C(x) lie in AB for 0<x<1.

Try and put simple numbers in like this if you're unsure about what an expression represents. Its easy to misuse algebra if you dont have a clear understanding of what it represents. Note #2 has all the information you need to solve the question. Its a couple of simple lines, at most. So
C(x) = A*(1-x) + B*x = (10,1)*(1-x) + x*(2,7) = (10-8x, 1+6x)

Hey, I’ve just come back from a week away at my parents house.

I don’t know why it took me so long to grasp this but it makes a lot of sense. I believe I was misunderstanding the position vector and displacement vector meaning.

346F90F2-1F2E-4E78-94F4-A34E65CA6A4A.jpeg

By introducing the origin, it’s easy to see path to follow to find the value of c 😅
Original post by KingRich
Hey, I’ve just come back from a week away at my parents house.

I don’t know why it took me so long to grasp this but it makes a lot of sense. I believe I was misunderstanding the position vector and displacement vector meaning.

346F90F2-1F2E-4E78-94F4-A34E65CA6A4A.jpeg

By introducing the origin, it’s easy to see path to follow to find the value of c 😅

Agreed, a sketch nearly always helps.
Original post by KingRich
Hey, I’ve just come back from a week away at my parents house.

I don’t know why it took me so long to grasp this but it makes a lot of sense. I believe I was misunderstanding the position vector and displacement vector meaning.

By introducing the origin, it’s easy to see path to follow to find the value of c 😅

I have some concerns with this working. Following the logic, if x = 0.5, then AC should be equal to BC. Using the formulas in the working, x = 0.5 leads to a value for AC of (-4, 3) and a value for OC of (6, 4). If OC is (6, 4) then BC is (4, -3). But with x = 0.5, BC is supposed to be equal to AC, not -AC......
Reply 19
Original post by old_engineer
I have some concerns with this working. Following the logic, if x = 0.5, then AC should be equal to BC. Using the formulas in the working, x = 0.5 leads to a value for AC of (-4, 3) and a value for OC of (6, 4). If OC is (6, 4) then BC is (4, -3). But with x = 0.5, BC is supposed to be equal to AC, not -AC......

I don’t know how you found the x to be 0.5. The value of x=6/25.

This can be found with the help of coordinate d(3,2) and CD= √26

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