# Finding the last corner of a square

H4NS

## Homework Statement

Show that there is a C point to these points A(-2|2|3), B(2|10|4) and D(5|-2|7), so that the quadrangle ABCD a square is. Determine the coordinates of C.

## The Attempt at a Solution

All I have done is measure AC, CD and BC, which are 9, 9 and 12.78. But how do I get the C point from this data?

Homework Helper
Gold Member

## Homework Statement

Show that there is a C point to these points A(-2|2|3), B(2|10|4) and D(5|-2|7), so that the quadrangle ABCD a square is. Determine the coordinates of C.

## The Attempt at a Solution

All I have done is measure AC, CD and BC, which are 9, 9 and 12.78. But how do I get the C point from this data?

The length of AB = 9 by calculation and the length of AD = 9 by calculation, so AD and AB are adjacent sides, with BC a diagonal of length 12.78. That means AC must be a diagonal and its length can't be 9. Something is wrong with your figures.

H4NS
The length of AB = 9 by calculation and the length of AD = 9 by calculation, so AD and AB are adjacent sides, with BC a diagonal of length 12.78. That means AC must be a diagonal and its length can't be 9. Something is wrong with your figures.

Oh Sorry, I meant AC = 12.78, since it's a square, BC and CD must be 9 too.

Homework Helper
Gold Member
Usually when you give a quadrilateral ABCD that would give the vertices in order as you go around the square. In this case with AB and AD adjacent sides, the order around the square can't be ABCD. Is that what you meant to have or do you still have typos? It might be good to state carefully exactly what you are given.

Have you studied vectors and how to manipulate them?

H4NS
Usually when you give a quadrilateral ABCD that would give the vertices in order as you go around the square. In this case with AB and AD adjacent sides, the order around the square can't be ABCD. Is that what you meant to have or do you still have typos? It might be good to state carefully exactly what you are given.

Have you studied vectors and how to manipulate them?

This is what I mean

I have also missed some classes, and that's why I require help from you

Homework Helper
Gold Member
Have you studied vectors and how to manipulate them?

This is what I mean

I have also missed some classes, and that's why I require help from you

You didn't answer my question whether you have studied vectors. Think about what you would get if you add the vectors AB and AD. What direction would it point?

H4NS
You didn't answer my question whether you have studied vectors. Think about what you would get if you add the vectors AB and AD. What direction would it point?
AB+AD would point in the direction of AC, but I don't know those vektors, I only know the points ABD

I know how to perform certain operations with vectors, but as I said, I missed some classes, and probably the key to resolve this problem

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AB+AD would point in the direction of AC, but I don't know those vektors, I only know the points ABD

I know how to perform certain operations with vectors, but as I said, I missed some classes, and probably the key to resolve this problem

To get a vector between points P and Q you subtract the coordinates of P from the coordinates of Q to get the components of the vector PQ.

H4NS
To get a vector between points P and Q you subtract the coordinates of P from the coordinates of Q to get the components of the vector PQ.
Oh man, thanks a lot, then I suppose I have to do the following:

Then AC: (4+7|8-4|1+4)

Thus the C point is (11|4|5), right?

But this is somehow wrong because when doing calculating CD or BC I get don't get 9!

What am I doing wrong?

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Oh man, thanks a lot, then I suppose I have to do the following:

Then AC: (4+7|8-4|1+4)

Thus the C point is (11|4|5), right?

But this is somehow wrong because when doing calculating CD or BC I get don't get 9!

What am I doing wrong?

No, that is not the point C. It is a vector pointing towards C. You have a vector in the right direction, but its length is whatever it came out. How long is it? How long does it need to be to fit the diagonal? How do you change its length?

H4NS
No, that is not the point C. It is a vector pointing towards C. You have a vector in the right direction, but its length is whatever it came out. How long is it? How long does it need to be to fit the diagonal? How do you change its length?
It's lenght is as far as I know is |C|, 12.72, which is the measure of the diagonal

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It's lenght is as far as I know is |C|, 12.72, which is the measure of the diagonal

OK, so it is the right length. Now, if you are at the point A, how do you use that vector with the coordinates of A to figure out the coordinates of C? And, by the way, it would be best to use exact values instead of decimal approximations if you want the exact coordinates of C.

H4NS
Thank you very much, I got it

I did it like this: AC(11|4|5) = (cx-(-2)|cy-2|cz-3), then separated it in 3 simple equations: cx+2=11; cy-2=4; cy-3=5. And got C(9|6|8).

Also, you kind of implied with your last post, that it could be measured with help of the vector's lenght, how?

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Thank you very much, I got it

I did it like this: AC(11|4|5) = (cx-(-2)|cy-2|cz-3), then separated it in 3 simple equations: cx+2=11; cy-2=4; cy-3=5. And got C(9|6|8).

Also, you kind of implied with your last post, that it could be measured with help of the vector's lenght, how?

When you added the two vectors from the sides to get a vector in the direction of the diagonal, you needed to check its length to make sure it is the right length for the diagonal. In this case it checked OK. Had it not been the right length, when you added its components to the coordinates of A, you wouldn't have gotten the coordinates of C.

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Thank you very much, I got it

I did it like this: AC(11|4|5) = (cx-(-2)|cy-2|cz-3), then separated it in 3 simple equations: cx+2=11; cy-2=4; cy-3=5. And got C(9|6|8).

Also, you kind of implied with your last post, that it could be measured with help of the vector's lenght, how?

You are making things much more complicated than they need to be. Just look at the displacement from A to B; that is, starting from A, how do you reach B? You start from x = -2 at A and get to x = +2 at B, so you go 4 units in the x-direction. You start from y = 2 at A and get to y = 10 at B, so go 8 units in the y-direction. You start at z = 3 and end at z = 4, so go 1 unit in the z-direction. So, AB = (4,8,1). Thus, in order that ABCD be a parallelogram you need DC = AB --- draw a picture to see this --- so DC (4,8,1), hence C = D + DC = (5,-2,7) + (4,8,1) = (9,6,8).

RGV