- #1

Oxygenate

- 7

- 0

## Homework Statement

Find the equation of the plane which contains the line r(t) = (2-t, 1+t, t) and the point (1,0,1).

## The Attempt at a Solution

Since r(t) = (2-t, 1+t, t), we know that vector a, which is parallel to the line, is (-1, 1, 1).

I'm assuming that since the point is (1,0,1), that means if we look to that point from the origin (0,0,0) we will have vector b = (1,0,1).

Okay so with two vectors a and b, I can find a vector n such that n is orthogonal to vectors a and b. So a x b = (1-0,1+1,0-1) = (1,2-1) = vector n.

The scalar equation of a plane is a(x-x

_{0}) + b(y-y

_{0}+ c(z-z

_{0}). So x + 2y -z = 0...

But somehow I have a feeling that my result is wrong. Is there supposed to be a "y" at all in the equation for this plane? Please advise. Thanks in advance!