# Basic plane-geometry issue

1. Jan 29, 2010

### kenewbie

1. The problem statement, all variables and given/known data

This should be simple, but I cant figure out what I did wrong.

A(1,2,1)
B(1,1,3)
C(-1,1,-1)
D(1,-2,1)

a) Give a "parametric representation" of the plane a which goes through the points A, B and C.
b) Give a "parametric representation" of the plane b which goes through the points B, C and D.
c) Give a representation of the line which describe the intersection between the planes a and b.
d) Calculate the angle between planes a and b.

I'm not sure if the term "parametric representation" looses meaning in translation, but the idea is to give a formula using 2 variables so that all values give describe a point on the plane.

3. The attempt at a solution

a)

Vector AB = [ 0, -1, 2 ]
Vector AC = [ -2, -1, -2 ]

Using point A with vectors AB and AC I get the following representation of the plane:

x = 1 - 2t
y = 2 - s - t
z = 1 + 2s - 2t

But my book says the plane is supposed to be

x = 1 + 2t
y = 2 - s + t
z = 1 + 2s + 2t

This seems to be using vectors AB and CA along with the point A. Why would they use CA? Do I describe the same plane as the book?

b)

Using the point B and vectors BC and BD, i get:

x = 1 - 2s
y = 1 - 3t
z = 3 - 4s -2t

which again is different from the answer in my book, which is

x = 1 + s
y = 1 + 3t
z = 3 + 2s + 2t

c)

I would try to set the x-value of plane a equal to the x-value plane b and so forth, and see what values of t and s that solves for, but since my plane-representations are wrong I get bogus values.

d)

using AB x AC as the normal for plane a and BC x BD as the normal for plane b I get an angle of 43.5 between the planes. My book says 58.4

All help appreciated.

k
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jan 29, 2010

### HallsofIvy

Yes, of course. The fact that the three given points satisfy both sets of equations tells you they are the same plane. They are just using a different parameter. Their "t" is the negative of yours. And there is no reason not to use CA rather than AC. Both are vectors in the plane and that is all that is required.

Again, just a different paremeterization. Here, their "s" is -2 times yours and their "t" is the negative of yours.

No, your equations are NOT wrong. One thing you have to be careful to do is to recognize the the "s" and "t" in each of your sets of equations may represent different things- don't use the same name when solving them together.

You have
x = 1 - 2t
y = 2 - s - t
z = 1 + 2s - 2t
and
x = 1 - 2a
y = 1 - 3b
z = 3 - 4a -2b

where, for the second equation, I have replaced "s" and "t" by "a'" and "b'" to distinguish them.

No reason why that shouldn't work. Show your work and we will try to check it.

3. Jan 29, 2010

### kenewbie

Ok, good. I started thinking there might be some "right-handed system" issue here that I didn't understand.

Aaah, see, I used "t" and "s" for both of them, that is why I got nonsense then. What you are saying makes sense. Since both parameterizations use different points on the plain and different vectors, they will point to different parts of the plane given the same values as paramters! Thanks.

As for the angle, this is my work (slightly abbreviated)

AB x AC = [0, -1, 2] x [-2, -1, -2] = [4, -4, -2]
BC x BD = [-2, 0, -4] x [0, -3, -2] = [12, -8, 6]

(AB x AC) * (BC x BD) = [4, -4, -2] * [12, -8, 6] = 68

|AB x AC| = 6
|BC x BD| = sqrt(244)

so, the angle = inv cos( 68 / ( 6 sqrt(244) ) ) = 43.5

My book says 58.4

k