How Do You Find the Equation of a Plane Perpendicular to the YZ-Plane?

[plutolover]

find the general form of the equation of the plane with the given characteristics:

passes through (0,2,4) and (-1,-2,0) and is perpendicular to yz-plane


I know what the general form of an equation is, but I was wondering, do I set the direction vector to be <0,1,1> or <1,0,0>? How do I use this to determine the general form of the equation?

I started by determining the vector given by the two points and then used this vector with the cross product of the two vectors mentioned above, but I am not sure what to do after that?...

 

[Mark44]

find the general form of the equation of the plane with the given characteristics:

passes through (0,2,4) and (-1,-2,0) and is perpendicular to yz-plane


I know what the general form of an equation is, but I was wondering, do I set the direction vector to be <0,1,1> or <1,0,0>? How do I use this to determine the general form of the equation?

The displacement vector between your two given points can be crossed with <1, 0, 0> to produce a vector that is normal to your plane.

I started by determining the vector given by the two points and then used this vector with the cross product of the two vectors mentioned above, but I am not sure what to do after that?...

When you post a problem, don't delete the template elements. They are there for a reason.