Find an equation of a plane that passes through the point $(0, 1, 0)$ and is parallel to the plane $4x – 3y + 5z = 0$
I first plugged the "missing" variable
$4x – 3y + 5z – d = 0$
then calculated $d$
$d = 4(0) – 3(1) + 5(0)=-3$
and wrote my final answer as
$4x – 3y + 5z + 3 = 0$
Are these to the correct steps to solving these type of problems? My textbook is a bit sparse in this area.
Best Answer
This method yields the correct solution.
The plane with equation $$ax+by+cz=0$$ goes through $(0,0,0)$. Now consider the plane with equation $$ax+by+cz=d.$$
If $a \neq 0$, then its also the plane with equation $$a(x-d/a)+by+cz=0$$ is formed by shifting the plane $d/a$ units along the $x$-axis in the positive direction.
If $b \neq 0$, then its also the plane with equation $$ax+b(y-d/b)+cz=0$$ is formed by shifting the plane $d/b$ units along the $y$-axis in the positive direction.
If $c \neq 0$, then its also the plane with equation $$ax+by+c(z-d/c)=0$$ is formed by shifting the plane $d/c$ units along the $z$-axis in the positive direction.
(If all three of $a,b,c$ are zero, then we don't have a plane to begin with.)
In this particular case, we have the second item above $$ax+b(y-d/b)+cz=0$$ with $a=4$, $b=-3$, $c=5$ and we want to shift $d/b=1$ unit along the $y$-axis in the positive direction.