# How to find an equation to the plane that passes through the points and is perpendicular to a plane?

Find an equation of the plane with the given characteristics. The plane passes through the points (4, 3, 1) and (4, 1, -6) and is perpendicular to the plane 7x + 9y + 3z = 13.

I had trouble solving this problem because the plane's equation has an equal 13. It is usually 0 for most equations so I can't come up with an answer. Help?

• change the 13 to 0, exact same answer – Will Jagy Sep 4 '18 at 23:48
• let's see, $7x+9y+3z = 13$ and $7x+9y+3z = 0$ are parallel planes. A line that is perpendicular to one is also perpendicular to the other; a plane that is perpendicular to one is also perpendicular to the other. – Will Jagy Sep 4 '18 at 23:54
• @Mary Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details HERE – user Oct 23 '18 at 21:01

The plane $7x + 9y + 3z = 13$, exactly as the parallel plane through the origin $7x + 9y + 3z = 0$, has normal vector $n=(7,9,3)$ therefore the plane we are looking for has equation
$$ax+by+cz=1$$
$$(a,b,c)\cdot(7,9,3)=0$$
from the other two conditions we can find $a$, $b$ snd $c$.