Plane through (4,9,-3) perpendicular to 3x - 5z + 3 = 0

[vybear]

Find the scalar equation of a plane that contains the point P(4,9,-3) and is perpendicular to the plane 3x - 5z + 3 = 0

I know that the normal vector of the given plane is 3,0,-5. I also know that in order for two planes to be perpendicular, their normal vectors must also be perpendicular. I think i should use the normal vector (3,0,-5) and the point to make a line but I don't know where to go from there.

 

[Roni1985]

I think you are right, you need to form a line on the new plane and find another vector,
and if you do the cross product with these two vectors, you will find the vector normal to the new plane.

you can check your answer by doing the dot product between the two normal vectors and see if you get zero.

I think this is how its done.

 

[Mark44]

Find the scalar equation of a plane that contains the point P(4,9,-3) and is perpendicular to the plane 3x - 5z + 3 = 0

I know that the normal vector of the given plane is 3,0,-5. I also know that in order for two planes to be perpendicular, their normal vectors must also be perpendicular. I think i should use the normal vector (3,0,-5) and the point to make a line but I don't know where to go from there.

Notice that the problem asks for "a plane" and not "the plane." If you think about this awhile, you should realize that there are an infinite number of planes that are perpendicular to the plane 3x - 5z + 3 = 0. There are even an infinite number of planes that pass through (4, 9, -3) that are perpendicular to the given plane.