# Vectorial geometry, linear algebra (planes, normals, distances)

## Homework Statement

I am given the following vectors :

Code:
p =   3       q =  2        r =  5
2            4             3
-4           -3            -1
1. a normal to the plane containing p, q and r.
2. the distance from the origin to the plane containing p, q and r
3. The distance from r to the line containing p and q
4. The matrix and translation of the affine transformation T:ℝ3 → ℝ3 which projects points orthogonally onto the plane containing p, q and r.

## Homework Equations

equation for a normal to a plane:
f(x,y,z) = ax + by + cz + d = 0

projection equation:
projvn = (|v . n|/|v|2) . v

## The Attempt at a Solution

I will have to edit this in since i need to leave quickly but I would still like to throw this out there, I know the equations, I know the normal is a vector which is orthogonal to the plane and thus who's dot product is equal to zero, but I do not know how to go about it.

As for #2 I do not know how to solve for an equation in which all vectors equal zero.

#3 I think the line containing p and q will be p - q = s, and i must project r onto s to find the distance

for #4 I do not understand what they are asking.

I like Serena
Homework Helper
Welcome to PF, Flowergirl! To find a normal vector, first you need to find 2 vectors that lie within the plane.
And to find a vector in the plane you need to subtract 2 vectors that have their end point in the plane.
Can you find 2 vectors that lie within the plane?

Thank you for the welcome :)

lets say p - q = n and q - r = m

n = (1, -2, -7)

m = (-3, 1, -4)
You said, originally,
I am given the following vectors :

p = 3 q = 2 r = 5
2 4 3
-4 -3 -1
1. a normal to the plane containing p, q and r.
So any two of p, q, and r are in the plane. Of course, any linear combination is then in the plane but it is not necessary to find new vectors in the plane.

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I like Serena
Homework Helper
Aha! It looks like you already know more or less what you need to do.

Now let's see... when I calculate p-q and q-r I get different vectors.
They are almost the same...
Is it possible you made a couple of mistakes with + and -?

Then you deduce a couple of equations from your matrix.
How did you arrive at those equations?
There appears to be a mistake in that step, since your final vector is not normal to your vectors n and m.
Can you check the dot product of your final vector with your vector m respectively your vector n?

oh oops, well i redid it and got (-1,-1,1) normal vector but .. how do I express it in equation form? like ax1 + bx2 + cx3 + d = n

EDIT: sorry, i'm retarded, it'd be -x1 - x2 + x3 + 9 = n

for b I have to find a vector from the plane to the origin? or do i just project to point zero... i'm really not quite sure how to go about this

I searched and found that I have to project a vector from the plane onto n, and that will give the distance between the origin and the plane.

I ended up finding d = ||projqn|| = |nTq| / ||n|| = |-9|/(3)1/2

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I like Serena
Homework Helper
Looking good. :)

Btw, that "n" in your equation should be zero.

The dot product is effectively a projection.
If you take the dot product of a vector in the plane with the normal vector you found, and divide by the length of the normal vector, you get the distance to the origin.
So yes, you got the right distance.

For #3, to find the distance from r to the line through p and q, you will need a different formula than the ones you've shown so far.
Do you have more formulas?
Or are you supposed to construct one yourself?

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