(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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 in to 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

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# Homework Help: Point on plane closest to point

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