- #1
JeeebeZ
- 40
- 1
Homework Statement
Find the point B on the plane x + 2y +3z = 1 that is closest to the point A (4,5,6).
2. The attempt at a solution
norm vector = n = [ 1 2 3 ]
A = (4,5,6)
a point C on the plane would (1,0,0)
CA vector = [ 3 5 6 ]
Projection of CA on to n = (CA.n / ||n||^2) n
= (((3*1) + (5*2) + (6*3))/14) n
= 3n
ProgCA(n) = [3 6 9]
So... That's where I am... I don't know if that would be the answer? B = (3, 6, 9)?
The reason I'm confused is if I enter the plane and point into different on-line calculators that are for distance between a point and a plane I get 8.8196 as the shortest distance.
But distance is the ||BA Vector|| = 3.316
I tried swapping the CA vector to AC Vector and I got B = (-3,-6,-9)
But then distance was 19.874.
So that didn't equate either. So.. I'm stuck and don't know what to look for next