Equation of a plane2. Heres the steps and answer. Can someone check this to see if it is correct?

Question

Just want to see if my answers are right, took a second stab at it and this is what i got.

here was the question.

Find the equation of plane including the line r(t) = (2,1,2) + t(1,-1,1) and perpendicular to the line s(t) = (1,-1,2) + t(-1,0,1)

My answer:

find a point on r(t):
r(0)=(2,1,2)

Find a point on s(t):
s(0)=(1,-1,2)

then:
1(x-2) -1(y-1) +2(z-2) = 0 Simplify

answer:

x-y+2z = 5

*do i have to use the points r(0) and s(0)??

Thanks

Answer 1

I'll start from the end: t just stretches the vectors but doesn't change the relationship of the line and plane.

My intuition says that your answer is not correct. That's because you can change the vector that t multiplys, thus changing the direction of the lines, and the answer doesn't change.

Also, to answer the * question: those are the only points that you've used!

Answer 2

the correct answer is:

the directional vector of the line s(t) = (1,-1,2) + t(-1,0,1) is
the normal vector to the plane you are looking for,
in this case is
(-1,0,1)
so the correct equation is:
-1(x-2)+0(y+1)+1(z-2)=0
-x+2 +z-2=0
-x+z=0

so your answer is NOT correct