# Find a plane that is parallel with $z$-axis and the line $r = (\lambda+ 1, \lambda - 1, 1 -\lambda)$, passes through $A(-2, 3, 0)$.

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...

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 $$x-y=-5$$
• Thank you for responding. I do have one question though: How did you get from the parametric equation (where $x = -2 + t$, $y = 3 + t$ and $z = s$) to $x - y = -5$? I tried expressing $t$ and $s$, but the problem is I don't use both $t$ and $s$ in one of my coordinates. – NumberSymphony Mar 26 '17 at 7:22
• You don't need to express t and s, only to obtain an equation of the form $Ax+By+CZ=D$ without t and s. In this case $1x+(-1)y+0z=-5$. Note that $x+2=t$ and $y-3=t$, so $x+2=y-3$, that is $x-y=-5$ – boaz Mar 26 '17 at 12:49