# Homework Help: Equations of a Plane: Determining if a line lies within a plane

1. Apr 2, 2012

### Buzzlastyear

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

Does the line with equation (x, y, z) = (5, -4, 6) + u(1,4,-1) lie in the plane with equation (x, y, z) = (3, 0, 2) + s(1,1,-1) + t(2, -1, 1)? Show algebraically.

2. Relevant equations

3. The attempt at a solution

I really have no idea where to start so I put the equations into parametric form to try to jog my brain but still got me nowhere!

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

(x,y,z)=(3,0,2) + s(1,1,-1) + t(2,-1,1)
x=3+s+2t
y=s-t
z=2-s+t

EDIT: I was looking back in some of my notes and although there was nothing on how to determine if a line lies within a plane, I found how to determine if a point does. So I discovered that the point from the first vector equation (5,-4,6) does indeed lie within the plane (x,y,z)= (3, 0, 2) + s(1,1,-1) + t(2, -1, 1). After discovering this however, i still don't see how i can use this information to see if the line (x,y,z)=(5,-4,6) + u(1,4,-1) lies within the plane

Not sure but maybe i have to sub in a value for s or t and solve the equations? i really have no leads so any help is very much appreciated thank you!

EDIT 2: I think i may have figured it out, here is what i've done:

substitute a value for u, ex. let u = 0

solve (x,y,z)=(5,-4,6)+0(1,4,-1)
therefore (5,-4,6) is a point when u=0

sub the values of x,y, and z into the parametric equations of the plane:

x=3+s+2t, y=s-t, z=2-s+t
6=3+s+2t, 0=s-t, 5=2-s+t
................s=t

sub s=t in 6=3+s+2t

6=3+t+2t
t=1

Sub t=1 in s=t

s=t
s=1

Now sub s=1 and t=1 into 5=2-s+t

LS=5 RS=2-s+t
RS=2-1+1
RS=2

Therefore the line (x,y,z)=(5,-4,6) + u(1,4,-1) does not lie in the plane???????

Last edited: Apr 2, 2012
2. Apr 2, 2012

### Staff: Mentor

Find two points on your line and determine whether they satisfy the equation of your plane.
If both points on the line satisfy the plane equation, the line is in the plane.
If only one point satisfies the plane equation, the line intersects the plane but doesn't lie in it.
If neither point on the line satisifies the plane equation, the line is parallel to the plane and doesn't lie in it.

3. Apr 2, 2012

Thank you
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