Question

Find the equation of the plane through the point (1,-2,3) and (2,0,4) which is perpendicular to the plane with 2x-2y+3z= 3

I found the vectors PQ and QP, I just need to know if I have to find the normal from the vectors or the two points I was originally given, as well as what to do with the plane being perpendicular to another plane.

Answer 1

hello, the plane is perpendicular to that plane, then its parallel to the normal line of that plane

Answer 2

The normal line of the plane is of the form
k*<2,-2,3> a plane that is parallel to this vector

Answer 3

Now we need to see a plane that is parallel to that vector as well as containing those points given

Answer 4

in other words, we know 2 vectors in the plane already: the one between the given points, and the normal to the stated plane.

Answer 5

Oh and there are a couple ways to see that is the normal line, one being The gradient of that equation 2x-2y+3z= 3
f(x,y,z)=2x-2y-3z
Grad F =<df/dx,df/dy,df/dz> <--- partial
another intuitive way is to get it from dot products

Answer 6

okay lets continue with finding the plane now...
(1,-2,3) and (2,0,4) are points on the plane so
(1,-2,3)-((2,0,4) is a vector on the plane
(1,-2,3) +k*<2,-2,3> <--- is another vector parallel to the normal line..
The cross of these will give us a normal line to the plane we are looking for

Answer 7

that normal line vector if its <a,b,c>
then our equation of the plane is
ax+By+cz=d
we can plug in any of those given points and see what d has to be, thats it. The end.

Answer 8

Hold on I feel like there's a much faster solution I might be missing, give me a second.