Find the equation of the plane through the points (1, -1, 2) and (2, -2, 2) and is perpendicular to the plane
6x - 2y + 2z = 9.
I need a figure for this
visualize a room where the floor is given by 6x - 2y + 2z = 9
and the perpendicular plane would be wall, with the two given points
on the wall.
The normal to your floor, the 6x - 2y + 2z = 9, would be <6,-2,2>
or in more simplified form <3, -1, 1>
This would be a direction number on the plane you want, we need
a second direction vector, which we can get from the 2 points
It would be <1, -1, 0>
Find the cross-product of these two directions vectors:
I get <1, 1, -2>
So the new plane is x + y - 2z = c
place (1, -1, 2) on it ....
1 - 1 - 4 = c
c = -4
your equation of the plane is x + y - 2z = -4
check: are both points on it? yes, easy to do in your head
are they perpendicular?
yes, if the dot product of their normals is zero ..
<1,1,-2> dot <6,-2,2> = 6 - 2 - 4 = 0
looks good!