As the title states, going over a review here and I can't figure this one out.
It asks to give the equation of a plane perpendicular to the plane x+y-2z=1
Any insight as to how I should start this.
As the title states, going over a review here and I can't figure this one out.
It asks to give the equation of a plane perpendicular to the plane x+y-2z=1
Any insight as to how I should start this.
I'm only a n00b, so I may be wrong, but this is how I'd attempt it:
Firstly, what you have there is the Cartesian form of a plane. When you have a plane in Cartesian form, the co-efficients amazingly form the normal of the plane. In this case, the vector x = 1, y = 1, z = -2 or (1, 1, -2).
Of course, planes are linear combinations of two vectors. We want a plane perpendicular to the given plane. We have the normal, which by definition, is perpendicular. So, we can use that to construct a perpendicular plane. If you convert the given plane into parametric form, you'll easily see two vectors that make up its span. Basically, take one of those vectors, and create a linear combination of that and the normal.
For instance, (-1, 1, 0) is a vector which lies on the given plane. And the normal is perpendicular to this. Thus a linear combination of the two produces a perpendicular plane.
The easiest way to do this is to find a direction vector of the plane (which is perpendicular to the normal) and make this you normal - bada boom bada bing you're done.
From Inspection it can be seen that the points (1,0,0) and (0,1,0) satisfy the equation of the plane.
Therefore a direction vector of the plane is (1,0,0)-(0,1,0) = (1,-1,0)
Therefore the perpendicular plane has a normal of (1,-1,0), so a set of perpendicular planes are in the form of x - y = d.
Taking the dot product of the normals of these planes is (1,-1,0) dot (1,1,-2) = 1 -1+0 = 0. So they are perpendicular.
There are heaps though. Another way to do this even quicker is see that the plane has a normal of (1,1,-2) so a perpendicular vector would be (1,-1,0) or (0,2,1) or (2,0,1)
Cheers,