Plane through a point, parallel to line and perpendicular to plane

Question

I have to find the plane that is perpendicular to the plane

$\pi:x+y+z=0$

Parallel to the line

$r:x+z-1=y+2z=0$

and passes through

$P(2,1,3)$

This is a very easy problem but I can't find the solution of the system that I'm getting. I know that the normal to the given plane is $(1,1,1)$ and the vector of the line is $(1,2,-1)$

The system that you get is

\begin{cases} 2a+b+3c+d=0 \\ a+b+c=0 \\ a+2b-c=0 \end{cases}

but there should be a fourth equation for the $d$. What should I add to the system to get all the four parameters?

Answer 1

You are looking for $(a,b,c)$, a normal vector for your plane.

But this vector is only defined up to a (non-zero) multiple. This is why you're "missing" a fourth equation. There are actually infinitely many solutions to your system of equations (each yields the same plane in the end). You can solve by picking some parameter essentially at random.

First, add the bottom 2 equations: $2a+3b=0$. Now just pick something like $b=2$ so that $2a=-6$ and so $a=-3$. Now the 2nd equation says that $-3+2+c=0$ and so $c=1$. We have $(a,b,c)=(-3,2,1)$. Finally, get that $2(-3)+2+3(1)+d=0$ so that $d=1$.

Of course, if we had chosen something other than $b=2$, we would have found a different normal vector (for example: $b=1$ would lead to $(a,b,c)=(-3/2,1,1/2)$). Different but equivalent.

If you know about cross products, there is an easier approach. You can get a normal vector for your plane by computing the cross product of $(1,1,1)$ and $(1,2,-1)$ (the normal for the other plane and direction vector for the line). This way there are no equations to solve.

Answer 2

we have the conditions with the ansatz $$ax+by+cz+d=0$$ $$[a,b,c]\cdot [1,1,1]=0$$ (1) $$2a+b+3c+d=0$$ (2) $$[a,b,c]\cdot [-1,-2,1]=0$$ (3) can you proceed?