If the angle b/n2 hyperplanes is defined as the angle b/n their normals, r the 2 hyperplanes below orthogonal?

Question

Hyperplanes

3x + 2y + 4z - 2w = 5 and 2x - 4y + z + w = 6

Are these two planes orthogonal? How do you find the normal lines in these planes? Do I have to somehow manipulate these equations to set them equal to zero? Please help with sound explanations.

Answer 1

For a plane with the equation

ax+by+cz+dw=e

its normal is simply

ai+bj+ck+dl.

[Why? Skip this bit if you like.

Suppose you have a vector ai+bj+ck+dl and want to find the plane of all points to which it is normal. Let general point, P, in the plane be (x,y,z,w).

Suppose you know one point in the plane P0=(e1,e2,e3,e4).

Then a general vector lying in the plane is (x-e1,y-e2,z-e3,w-e4) and this vector must be orthogonal to ai+bj+ck+dl for all x,y,z,w.

=>a(x-e1)+b(y-e2)+c(z-e3)+d(w-e4)=0

=>ax+by+cz+dw=e

for some constant e (=ae1+be2+ce3+de4).]

So the normal to 3x+2y+4z-2w=5 is 3i+2j+4k-2l

and the normal to 2x-4y+z+w=6 is 2i-4j+k+l

Now (3i+2j+4k-2l).(2i-4j+k+l) = 6-8+4-2=0

so the normal to the planes are orthogonal, hence the planes are orthogonal.