# Algebraic Vectors

• Aug 27th 2009, 01:13 PM
skeske1234
Algebraic Vectors
ABCD is a parallelogram with vertices A=(-1,2,-1), B=(2,-1,3), and D=(-3,1-3), find the coordinates of C.

So this is my work below, BUT i cannot seem to get the correct coordinates of C, as my diagram does not seem to look exactly like a parallelogram..

I sketched A, B, D..
A is on the y-axis , D is below it to the right and in 4th quadrant, B is in second quadrant.. so I assumed that C would be in 2nd qadrant.

The vector AC= The vector DB
since they are parallel

(x+1,y-2,z+1)=(5,-2,6)
x+1=5
x=4
y-2=-2
y=0
z+1=6
z=5

C(4,0,5)
• Aug 27th 2009, 01:28 PM
Plato
First of all, be careful of the order of things.
We want $\overrightarrow {AD} = \overrightarrow {BC} \;\& \,\overrightarrow {AB} = \overrightarrow {DC}$. Note the order.
$\overrightarrow {AD} = \left\langle { - 2, - 1, - 2} \right\rangle = \overrightarrow {BC} = \left\langle {x - 2,y + 1,z - 3} \right\rangle$
• Aug 27th 2009, 01:32 PM
eXist
Plato is correct. The vector AC is not parallel with the vector DB. Look at this picture (this is as if it were flattened into $R^2$):
http://img175.imageshack.us/img175/8420/74953580.png
What I would do is find the equation of the line BC and the line DC, and then find their intersection. That intersection should be C.
• Aug 27th 2009, 02:06 PM
Plato
Quote:

Originally Posted by eXist
What I would do is find the equation of the line BC and the line DC, and then find their intersection. That intersection should be C.

Actually all one has to do is to solve: $\left\langle { - 2, - 1, - 2} \right\rangle = \left\langle {x - 2,y + 1,z - 3} \right\rangle$.
• Aug 27th 2009, 02:54 PM
eXist
That's probably a better idea :D thanks.
• Aug 27th 2009, 03:05 PM
skeske1234
I'm having trouble drawing the parallelogram.. actually plotting the point A, B, and D into 3-space :(

my parallelogram does not appear to be the same as yours, eXist, and AD and BC is not parallel in my drawing
• Aug 27th 2009, 03:12 PM
Plato
Quote:

Originally Posted by skeske1234
I'm having trouble drawing the parallelogram.. actually plotting the point A, B, and D into 3-space :(
my parallelogram does not appear to be the same as yours, eXist, and AD and BC is not parallel in my drawing

There is really no good way to make this drawing.
But if $\overrightarrow {AD} \;\& \,\overrightarrow {BC}$ are not parallel in your drawing, then it is wrong anyway.
• Aug 27th 2009, 03:25 PM
skeske1234
well actually I am having trouble plotting the points like A(-1,2,-1) B(2,-1,3) and D(-3,1,-3) in 3 space, my teacher said to plot the vertex of the point first, and then you should have 3 lines forming a corner of a cube.. and then draw your cube, so that the arrow goes from orgin to opposite side of cube. is there a better method of doing this? because I seem to mess up the points where I am plotting A, B, and D, so then my parallelogram vertices are in the wrong quadrant .. thus wrong order
• Aug 27th 2009, 03:49 PM
Plato
Quote:

Originally Posted by skeske1234
well actually I am having trouble plotting the points like A(-1,2,-1) B(2,-1,3) and D(-3,1,-3) in 3 space, my teacher said to plot the vertex of the point first, and then you should have 3 lines forming a corner of a cube.. and then draw your cube, so that the arrow goes from orgin to opposite side of cube. is there a better method of doing this? because I seem to mess up the points where I am plotting A, B, and D, so then my parallelogram vertices are in the wrong quadrant .. thus wrong order

Well that makes no sense to me.
There must be some good graphic program that would do that.
But even the CAS I use does it poorly.

But know this. Any three non-collinear points determine a triangle.
Any triangle is ‘one-half’ of a parallelogram.
• Aug 27th 2009, 05:37 PM
eXist
Also look at the way ABCD is defined. The ordering of the letters should determine how the parallelogram is formed. With A touching B and D, B touching A and C, C touching B and D, and D touch C and A.