As the title says, I have to find a plane parallel to the z-axis and the line $r = (\lambda + 1, \lambda – 1, 1 – \lambda)$. The plane also passes through the point $A(-2, 3, 0)$.
Now, I was thinking, since the plane is parallel to the $z$-axis, then the normal vector of the plane would be perpendicular to the $z$-axis. Am I wrong here?
Could someone solve this task step-by-step, explaining the reasoning as best as possible? I can't seem to figure it out…
Thanks in advance!
Best Answer
First write the line in vector form: $$ r(\lambda) = (\lambda+ 1, \lambda - 1, 1 -\lambda) =(1,-1,1)+\lambda(1,1,-1) $$ Hence, your plane is parallel to the direction vector of the line $r$, namely to the vector $(1,1,-1)$. In order to determine the plane completely, you need another direction vector. Since the plane parallel to $z$-axis, it follows that another direction vector is $(1,0,0)$. Since your plane pass trough the point $A(-2,3,0)$, it follows that , the parametric equation of the plane is $$ \pi(t,s)=(-2,3,0)+t(1,1,-1)+s(0,0,1) $$ To find the plane equation, set $$ (x,y,z)=(-2,3,0)+t(1,1,-1)+s(0,0,1). $$ Thus $$ \begin{cases} x=-2+t\\ y=3+t\\ z=s \end{cases} $$ so the plane equation is $$ x-y=-5 $$