Planes are perpendicular is their normals are perpendicular. Find a vector lying on the plane and use this for the normal of the second plane (BTW - there is an infinite number of perpendicular planes since there are an infinite number of vectors lying on a given plane.)
I got confused by your answer as in the book there's a question with this answer:
Find the equation of the plane perpendicular to -1 + 3j + 2k and passing through the point (1,0,4):
-(x-1) + 3(y-0) + 2(z-4) = 0
which is the same way you approach the parallel problem... can you explain this to me?
I don't want to pick at you but I think it's high time that you read this here: http://www.mathhelpforum.com/math-he...l-posters.html
Two planes and are parallel if the normal vectors and are collinear. That means:
see attachment.
If you are looking for a plane perpendicular to a given plane, then there are an infinite number of those planes. Example: Open a door. Normally the door is perpendicular to the floor, but there isn't only one position the door can have.
see attachment.
If you have a vector ( )and a point Q with its stationary vector you can derive the equation of a plane which is perpendicular to the vector and contains the given point: Let R denote an arbitrary point in the plane with its stationary vector . Then the vector must be perpendicular to the vector . That means: For all points in the plane the equationand the example from the book that I present ask for perpendicular, and you solve both the same way.. why is that?
is true and therefore this equation describes the plane completely (Point-normal-form of the equation of a plane)
see attachment