Find the equation of a plane with a normal vector of N(3,1,2,-1) and going through the point P(4,6,3,2)?

Answer 1

Did you have in mind a three dimensional hyperplane (one dimension less than the 4-space in which it resides)? If so the equation is:

3(x-4) + 1(y-6) + 2(z-3) - 1(w-2) = 0
3x - 12 + y - 6 + 2z - 6 - w + 2 = 0
3x + y + 2z - w - 22 = 0

Answer 2

♣ normalize vector N: n=N/|N|, where
|N| =sqrt((N·N)) =sqrt(3^2 +1^2 +2^2 +(-1)^2) = sqrt(15);
n=(3,1,2,-1)/sqrt15;
♦ now distance from origin D=(P·n) =(4*3+6*1+3*2-2*1)/sqrt15 =22/sqrt15;
♥ the demanded equation is: (n·r) =D, where r=(x,y,z,t);
or just 3*x+y+2z-t=22 if it suits your teacher.

Answer 3

3(x1-4)+1(x2-6)+2(x3-3)-1(x4-2)