How Do Lagrange Multipliers Help Calculate Distance from a Point to a Plane?

[EEWannabe]

URGENT - Lagrange Multipliers

Homework Statement

Using the method of lagrange multipliers prove the formula for the distance from a point (a,b,c) to a plane Ax + By + Cz = D

The Attempt at a Solution

Using the equation of the form;

H(x,y,z,L) = (x-a)^2 + (y -b)^2 +(z-c)^2 + L(Ax + By + Cz - D)

Therefore by differentiating, a relationship between x, y and z can be found with lamba, nameley;

x = (-LA/2) + a
y = (-LB/2) + b
z = (-LC/2) + c

By inserting these back into the equation for a plane I get to this stage for L;

[tex] L = \frac{2(Aa + Bb + Cc - D)}{A^2 + B^2 + C^2} [/tex]

Thus, putting this back into the Lagrange equation the value for D^2 I get is (D^2 =(x-a)^2 + (y -b)^2 +(z-c)^2);

[tex]D^2 to plane = \frac {2(Aa + Bb + Cc - D)(Ax + By + Cz - D)}{A^2 + B^2 + C^2} [/tex]

However this is at ends with;

[tex] D to plane = \frac{(Aa + Bb + Cc - D)}{(A^2 + B^2 + C^2)^_\frac{1/2}} [/tex]

I can't see what's gone wrong , I know i can't expect anyone to do it straight away or anything but I have a big exam tomorrow and there'll definatley be a question on something like this, any hints would be fantastic!

Thanks again

 

[Ray Vickson]

Do not use the same symbol, D, for both the parameter in the linear equation of the plane and for the distance to the plane. This _might_ be the source of the trouble.

PS. I get your expressions for x, y and z, and I get your first value of L. After that, we differ.

RGV

 

[EEWannabe]

Do not use the same symbol, D, for both the parameter in the linear equation of the plane and for the distance to the plane. This _might_ be the source of the trouble.

PS. I get your expressions for x, y and z, and I get your first value of L. After that, we differ.

RGV

But I only added that D as the distance to the plane right at the end :(, and by that D i simply mean

(D^2 =(x-a)^2 + (y -b)^2 +(z-c)^2) , have edited it now though..

Could you give me any hints as to why yours is different? :( To get so the D from the value of L, i simply put L back into the lagrange equation, the first one..

 

[HallsofIvy]

It's better to eliminate lambda (L) from the equations you get by differentiating.
x- a+ LA= 0, y- b+ LB= 0, z- c+ LC= 0 can be rewritten as x- a= -LA, y- b= -LB, z- c= -LC. Dividing the first equation by the second, (x- a)/(y- b)= A/B so B(x- a)= A(y- b). Dividing the first equation by the third, (x- a)/(z- c)= A/C so C(x- a)= A(z- c).
y= b+ (B/A)(x- a), z= c+ (C/A)(x- a).

Put those into Ax+ By+ Cz= D and solve for x.

 

[EEWannabe]

It's better to eliminate lambda (L) from the equations you get by differentiating.
x- a+ LA= 0, y- b+ LB= 0, z- c+ LC= 0 can be rewritten as x- a= -LA, y- b= -LB, z- c= -LC. Dividing the first equation by the second, (x- a)/(y- b)= A/B so B(x- a)= A(y- b). Dividing the first equation by the third, (x- a)/(z- c)= A/C so C(x- a)= A(z- c).
y= b+ (B/A)(x- a), z= c+ (C/A)(x- a).

Put those into Ax+ By+ Cz= D and solve for x.

Thank you so much sir

 

[Ray Vickson]

I mis-read what you wrote because use used the same symbol D for two different things and L for two different things---a bad practice! I _do_ get your final result for the distance, although you should remember that you got it by taking a square root of something of the form N^2/M^2 (where M > 0), so the distance, d, is d = +- N/M = |N|/M. (Your N = a*A + b*B + c*C - D could be < 0.)

RGV