How to find an equation to the plane that passes through the points and is perpendicular to a plane?
By Emma Valentine •
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?
$\endgroup$ 31 Answer
$\begingroup$HINT
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
$$ax+by+cz=1$$
with
$$(a,b,c)\cdot(7,9,3)=0$$
from the other two conditions we can find $a$, $b$ snd $c$.
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