I'm working on a question that asks me to find the vector equation of the plane in $\mathbb R^3$ that contains the point P(-1,6,0) and is parallel to the plane x+2y-z=5. I'm trying to make sure my answer is right versus the answer that I was provided with, which I think is wrong.
I pulled out the normal vector n = (1,2,-1). I proceeded to find two vectors v and u that are orthogonal to n but are not parallel to each other. I followed the theorem that states that when the dot product of 2 vectors equals zero, those vectors are orthogonal. I came up with v = (1,1,3) and u = (4,1,6). So I ended with the vector equation (x,y,z) = (-1,6,0) + t(1,1,3) + s(4,1,6).
It's my understanding that there are a number of different answers for this problem as there are a number of different vectors, besides the ones I found, that are orthogonal to the normal vector. The answer I was provided for this question stated that the vector equation is (x,y,z) = (-1,6,0) + t(3,-4,1) + s(1,-4,-1). I'm fairly certain that this answer is incorrect as the vectors in the vector equation are not orthogonal to the normal equation. Is my thinking correct?
Best Answer
The plane parallel to $x+2y-z=5$ containing $(-1,6,0)$ is $x+2y-z=-1+2(6)-0=11$.
Your parametrization satisfies that: $(-1+t+4s)+2(6+t+s)-(3t+6s)=11$.
The other answer does not: $(-1+3t+s)+2(6-4t-4s)-(t-s)\not=11$ generally.