# Find a plane that passes through a point and is parallel to a given plane

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.

• Do you mean normal to that plane, rather than parallel? Commented Sep 8, 2013 at 20:12
• Your procedure is right. The plane you produced is parallel to the given plane, and passes through the target point. For completeness you should perhaps have said that the required plane has an equation of shape $4x-3y+5z=d$. Plug in, you get $d=-3$. Commented Sep 8, 2013 at 20:16

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.

The general equation of a plane is $$\vec r\cdot \hat n=0$$ where $\vec n$ is a unit vector perpendicular to the plane and $\vec r$ is any point on the plane. Since The required plane is parallel to the given plane. The normal of the required plane is parallel to the normal of the given plane. Hence the direction ratios of the normal is $(4,-3,5)$.Let $(x,y,z)$ be any point on the required plane. Since the plane passes through $(0,1,0)$ we must have $$((x-0)\hat i+(y-1)\hat j+(z-0)\hat k)\cdot(4\hat i-3\hat j+5\hat k)=0$$ $$\Rightarrow 4x-3(y-1)+5z=0$$ $$\Rightarrow 4x-3y+5z+4=0$$ is the required equation of the plane. Hope this helped.

$\langle(4,-3,5) | p - (O,1,O)\rangle = O$ is parallel to $\langle(4,-3,5) | p \rangle= O$; and so is $\langle(4, -3, 5) | p - c\rangle = O$. Parallel planes have the same normal.

Equation of any plane parallel to the given plane is 4x-3y+5z+k=0 (0,1,0) will satisfy eqn of this plane. so, -3+k=0 k=3 therefore required eqn is 4x-3y+5z+3=0