# Point on plane closest to point

1. Jul 8, 2012

### JeeebeZ

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

2. Jul 8, 2012

### Simon Bridge

Hmmm... I'd have found the equation of the line through A in direction of the normal ... where this intersects the plane is the closest point.

But there are several ways to do it. You could, for instance, find the equation of the distance from an arbitrary point on the plane to point A, then minimize it.

The trick to being confident that your answer is the correct one is to see if you can explain the geometry of your method, like the two examples above.

You can also do some basic tests to check - eg. is B a point in the plane?

3. Jul 8, 2012

### ehild

You miscalculated something. ((3*1) + (5*2) + (6*3))/14)=31/14

ehild

4. Jul 8, 2012

### JeeebeZ

That would have been the smartest thing to do...

3 + 2*6 + 3*9 != 1... /sigh

5. Jul 8, 2012

### Simon Bridge

I think the meta-skill here, of checking your answers and choosing methods that allow you to be confident of your results, is actually more important than solving the actual puzzle in front of you.

6. Jul 8, 2012

### JeeebeZ

So I should be looking for A - ProgCA(n)...

( 25/14, 8/14, -9/14 )

(25 + 2*8 - 9*3)/14 = 1
(25 + 16 - 27) / 14 = 1
14 / 14 = 1
1 = 1

So the point B on the plane that is closest to A would be (25/14, 8/14, 9/14)

http://desmond.imageshack.us/Himg513/scaled.php?server=513&filename=junkh.png&res=landing [Broken]

Last edited by a moderator: May 6, 2017
7. Jul 8, 2012

### JeeebeZ

Hmm.. See I still don't think that's right.. b/c when i calculate the distance it comes out to be 7.825.. and I know it is 8.819 based off of calculators for distance from a plane. And i've used more then one just to make sure that the calculator wasn't wrong. :p

8. Jul 8, 2012

### JeeebeZ

Code (Text):

x    4     1
y = 5 + t2
z    6     3

x = 4    t
y = 5 + 2t
z = 6    3t

4 + t + 2( 5 + 2t ) + 3( 6 + 3t ) = 1
4 + 10 + 18 + t + 4t + 9t = 1
32 + 14t = 1
t = -31/14

Which would give me ( 25/14, 8/14, -9/14 )

Same answer as the last one.. these are the only 2 methods i know and they are both wrong :(

Last edited: Jul 8, 2012
9. Jul 8, 2012

### ehild

You meant -9/14 for z... It is correct. You might have some rounding errors when calculating the distance. The projection of CA onto n is CAdot n /||n||=31/√14=8.28

Have you learnt about the Lagrange multiplier method to find conditional extrema?

http://www.encyclopediaofmath.org/index.php/Lagrange_multipliers.

You have to find the minimum distance between a point in the plane P(x,y,z) and point A, that is the minimum of d2=(x-4)2+(y-5)2+(z-6)2 with the condition that x + 2y +3z -1 =0. Multiply the condition with the Lagrange multiplier λ and add to the function to be minimised. Find the values of x,y,z which make the partial derivatives of (x-4)2+(y-5)2+(z-6)2+λ(x + 2y +3z -1)
all zero and fulfil the condition.

ehild

Last edited: Jul 8, 2012
10. Jul 8, 2012

### JeeebeZ

wow...

I've been entering the calculation as a point P(4,5,6) and equation 1x + 2y + 3z + 1 = 0... not -1 but 1.
So I was entering the equation x +2y +3z = -1... that's why i kept getting 8.8...

I get 8.2851 when i calculate ||A - B||... the online one gives 8.285098499285...

I swear if I could go one day without a sign error somewhere in my problems... I would be happy

11. Jul 8, 2012

### ehild

Never mind...It happens with everybody.

ehild