Determine if the plane is parallel to $x-2y+z=2.$

Question

Determine if the plane

$$\pi_1=\left\{ \begin{array}{rcr} x & = & 1+4s+7t \\ y & = & 2+5s+8t \\ z & = & 3+6s+9t \end{array} \right.$$

Is parallel to the plane $\pi_2=\{x-2y+z=2\}$.

My attempt: For the planes to be parallel, they can not intersect anywhere. This implies that the planes can not have any common point in space. Setting in the values of $x,y,z$ given in terms of $s$ and $t$ in $\pi_1$, we have that

$$\pi_2=(1+4s+7t)-2(2+5s+8t)+3+6s+9t=2.$$

If the above equation holds for any $s$ and $t$, then the planes are not parallel. If it doesn't hold for any $s$ and $t$, then the planes are parallel. So simplifying gives

$$\underbrace{4s-10s+6s}_{=0}+\underbrace{7t-16t+9t}_{=0}=2\Leftrightarrow0=2.$$

This means that $\nexists s, t$ that solves the equation, thus the planes are parallel.

Is this approach correct?

Answer 1

Another simple way would be to determine the normal vector of each plane and if they are multiples of each another, you have two parallel planes. This is easy to do for the second plane, and you should be able to do it for the first plane too with some rearrangements.

Answer 2

For planes to be parallel each vector from one plane should exist on another

So if $(4,5,6)\in \pi_1$ then must be $(4,5,6)\in \pi_2$ and similar for $(7,8,9)$. You can prove it if you add on any point from $\pi_2$ 4 to x, 5 to y, and 6 to z the equation stays true. Also for 7,8,9