Proving that ABE is a Straight Line: Vector Method

[Natasha1]

Homework Statement

The diagram shows parallelogram ABCD. (you don't really need the diagram)

vector AB= (2 above, 7 below) and vector AC= (10 above, 11 below)

The point B has coordinates (5, 8)

(a) Work out the coordinates of the point C.

The point E has coordinates (63, 211)
(b) Use a vector method to prove that ABE is a straight line.

The Attempt at a Solution

(a) Work out the coordinates of the point C.

I did this:
vector AB= (2 above, 7 below)
So
5-2 = 3 and
8-7 = 1

So point A (3, 1)

vector AC= (10 above, 11 below)
So
10+3 = 13
11+ 1 = 12

The coordinates of the point C are (13, 12)

I did this:
The point E has coordinates (63, 211)
(b) Use a vector method to prove that ABE is a straight line.

I know I need to prove that ABE a collinear by proving that vector AE is a multiple of vector AB

vector AB= (2 above, 7 below)

2 x (63/2) = 63
7 x (63/2) = 220.5

I'm sure there's a better way? But how? Please help...

 

[berkeman]

(you don't really need the diagram)

But I'm a very visual person! Please use the UPLOAD button in the lower right of the Edit window to attach the diagram (in PDF or JPEG format) to your OP or as a reply. Thanks.

 

[DaveE]

I know I need to prove that ABE a collinear by proving that vector AE is a multiple of vector AB

Yes! So what is the vector AE? What is the vector AB? Can you express AE = k * AB, where k is some scalar constant?

 

[Natasha1]

Yes! So what is the vector AE? What is the vector AB? Can you express AE = k * AB, where k is some scalar constant?

Well AE = 31.5 x AB

Is that it?

 

[DaveE]

How do you find the vector from point A to point E?

 

[Natasha1]

I was hoping you might explain this...

 

[DaveE]

I think you are confusing points with the vectors between points.
"The point E has coordinates (63, 211)". This could also be thought of as the vector from the origin (0,0) [let's call that point O] to E. So the vector OE = (63,211).
Point A is (3,1), so the vector OA = (3,1) [or (3 above, 1 below)]. Point B is (5, 8), so the vector OB = (5,8)
"vector AB= (2 above, 7 below)" This is AB = OB - OA, you could also think of this as point B - point A.
So, what is the vector AE then?

 

[Natasha1]

I think you are confusing points with the vectors between points.
"The point E has coordinates (63, 211)". This could also be thought of as the vector from the origin (0,0) [let's call that point O] to E. So the vector OE = (63,211).
Point A is (3,1), so the vector OA = (3,1) [or (3 above, 1 below)]. Point B is (5, 8), so the vector OB = (5,8)
"vector AB= (2 above, 7 below)" This is AB = OB - OA, you could also think of this as point B - point A.
So, what is the vector AE then?

Is AB not BO + OA?

 

[Natasha1]

I get
AE = OE - OA
= (63, 211) - (3, 1)
= (60, 210)

How does this prove that ABE is a straight line?

 

[DaveE]

Is AB not BO + OA?

No. The vector notation AB means a vector that starts at point A and ends at point B. The vector sum BO + OA will give the vector BA, not AB. You may also think of this as going from B to O then from O to A which is that same as going from B to A.
Also note that the vector AB = -1*BA, so OA +BO = OA - OB = BA.

 

[DaveE]

I get
AE = OE - OA
= (63, 211) - (3, 1)
= (60, 210)

How does this prove that ABE is a straight line?

Yes, AE =(60,210) is correct. You also know that AB = (2,7).
As you said before "I need to prove that ABE a collinear by proving that vector AE is a multiple of vector AB".
So can you find a number k that satisfies AE = k*AB? This is the same as saying that AE is a multiple of AB.

 

[Natasha1]

Yes, AE =(60,210) is correct. You also know that AB = (2,7).
As you said before "I need to prove that ABE a collinear by proving that vector AE is a multiple of vector AB".
So can you find a number k that satisfies AE = k*AB? This is the same as saying that AE is a multiple of AB.

Ah yes yes yes!

AE = 30 x AB

Thanks so much DaveE. Super helpful... Much appreciated!