Buzzlastyear
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Homework Statement
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.
Homework Equations
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??
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