# Equation of a plane?

1. Aug 26, 2009

### fball558

equation of a plane??

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

Find an equation of the plane through the point and perpendicular to the given line.
(-2, 8, 10)
x = 4 + t, y = 3t, z = 4 - 4t

3. The attempt at a solution

i know the point (-2,8,10) is the starting positon (call it R0)
i know that N (normal) is perpendicular to R - R0 where R is ending position
R = (x,y,z) so i said
R= (4+t,3t,4-4t)
the N (normal) = (a,b,c) and i dont know how to get these numbers. if i can find out these i can plug it into the scalor equation of a plane formula and be set.

any help would be great!!
thanks

2. Aug 26, 2009

### rock.freak667

Re: equation of a plane??

You need to find the direction of x = 4 + t, y = 3t, z = 4 - 4t, this direction is parallel to the normal of your plane

eg. x=2t,y=3t,z=t

[x] [0+2t] [0] [2t] [0] [2]
[y] = [0+3t] = [0] + [3t] = [0] =t[3]
[z] [0+t] [0] [t] [0] [1]

so <2,3,1> is the direction of that line

Do the same for your line.

3. Aug 26, 2009

### HallsofIvy

Staff Emeritus
Re: equation of a plane??

In general, if vector <A, B, C> is perpendicular the plane and (p,q,r) is a point in the plane, then, for a general point (x,y,z) in the plane, <x- p, y- q, z- r> is a vector in the plane and so <A, B, C>.<x- p, y- q, z- r>= A(x- p)+ B(y- q)+ C(z- r)= 0.

In your case, a vector in the direction of the line x = 4 + t, y = 3t, z = 4 - 4t is <1, 3, -4>, the coefficients of t in each component.

4. Aug 26, 2009

### fball558

Re: equation of a plane??

thanks alot everyone. i got in a little study group with a few of my friends and we strugled for a while but finally found it out.
(5 hours for 30 problems... not too bad) haha