Show that the planes 4x-3y+z-7=0 and 4x-3y+z+6=0 are parallel planes.?

Question

i get the second part which is to find the distance between the 2 planes which turns out to be distance= |16+9+1+6| / square root of 26=32/square root of 26. but i don't get how to show that they are parallel

Answer 1

The normal vector of both planes is <4, -3, 1>. If two planes have the same normal vector, or the normal vector of one is a non-zero multiple of the other, then the planes are parallel.

If the above is true and the constant terms of the equations of the two planes are related to each other by the same non-zero multiple as are the normal vectors of the planes, then the planes are coincident (i.e. they are the same plane). If this is not the case, then they are distinct parallel planes.

In this case you have distinct parallel planes.

Answer 2

You have two equations which can be written as 4x-3y+z=7 and 4x-3y+z=-6. The two equations are equivalent on the left hand side, but not on the right... thus you end up with the contradiction "7=-6". More generally, because there is no way to take one of the equations and turn it into the other (through multiplication, for example), the planes do not interesect, and thus they are parallel.

Answer 3

there is an answer in the back of the book if it is an even question.

turn to this page.
walk up to the teacher/professor.
point at the answer.
and you have shown that the two are parallel planes.

duh.