Finding the Cartesian Equation of a Plane

Question

Find the Cartesian equation of the plane containing points $A$ $(6,2,1)$ and $B$ $(3,-1,1)$ and perpendicular to the plane $x+2y-z-6=0$.

I get the vector $AB$ $=(-3,-3,0)$ and the vector of the plane $x+2y-z-6=0$ which is $(1,2,-1)$. I know that the equation of the plane can be defined by the equation $A(x − x_0)$ $+$ $B(y − y_0)$ $+$ $C(z − z_0)$ $=$ $0$. How can I use this concept to find the equation of the plane?

Answer 1

Since the two planes are perpendicular, the normal vector of the second plane (i.e. $(1,2,-1)$) is parallel to the first. To find the normal vector of the first plane, take the cross product: $(1,2,-1)×(-3,-3,0)=(-3,3,3)$. Finding the constant in the plane equation is now a matter of substitution: $k=-3\cdot6+3\cdot2+3\cdot1=-9$. So the plane's equation is $-3x+3y+3z=-9$ or $x-y-z=3$.

Answer 2

HINT

A good way to proceed is to find the normal vector to the plane $n(a,b,c)$ and then the equation by:

$a(x-x_0)+b(y-y_0)+c(z-z_0)=0$

To find the normal $n(a,b,c)$ let’s consider that it has to be orthogonal both to AB and to the normal vector to the given plane m(1,2,-1).

To find the normal vector to two given vectors, cross product is a very effective way.

Answer 3

You know that your new plane must be parallel to both $(-3,-3,0)$ and $(1,2,-1)$. Thus, you want a plane perpendicular to the vector perpendicular to these two vectors, let’s call it $\vec w$. You can find w by doing the cross product of the two vectors, then dotting it with $\vec x - A = (x-6,y-2,z-1)$ and setting equal to zero.