TITLE: Find the point on a plane $3x + 4y + z = 1$ that is closest to $(1,0,1)$
QUESTION [6 upvotes]: Is anyone able to help me with regards to this question?
Find the point on a plane $3x + 4y + z = 1$ that is closest to $(1,0,1)$
http://i.imgur.com/ywdsJi7.png

REPLY [13 votes]: The normal vector to the plane is $\langle 3,4,1\rangle$. The point you seek would have to be some multiple of this vector added to $(1,0,1)$. $$P=(1,0,1)+c\langle 3,4,1\rangle=(1+3c,4c,1+c)$$ 
But this point has to satisfy the plane's equation:
$$\begin{align}
3(1+3c)+4(4c)+(1+c)&=1\\
26c+4&=1\\
c&=-\frac{3}{26}
\end{align}$$
So $$P=(1,0,1)-\frac{3}{26}\langle 3,4,1\rangle=\left(\frac{17}{26},-\frac{12}{26},\frac{23}{26}\right)$$

REPLY [3 votes]: With Lagrange Multipliers, we have the distance function $f(x,y,z)=(x-1)^2+y^2+(z-1)^2$ (squaring to simplify the calculations) and the constraint $3x+4y+z=1$. We get $$(x-1,y,z-1)=\lambda(3,4,1)$$ Therefore $$(x-1)/3=(y/4)=(z-1/1)=\lambda$$ and as the user above stated this gives the closest point.

REPLY [2 votes]: Yes. This is a multivariable minimization problem in which you want to minimize some function f(x,y,z) subject to the constraint g(x,y,z) - c = 0. 
The first thing to understand is that the function you are minimizing is the distance from the point (1,0,1). 
$$
f(x,y,z) = \sqrt{(x-1)^2+(y^2)+(z-1)^2}
$$  
and your constraint is
$$
3x+4y+z-1 = 0
$$
From here you can use the standard Lagrange Multipliers method. The wikipedia intro section here http://en.wikipedia.org/wiki/Lagrange_multiplier covers this nicely.
One last tip: minimizing the square of the distance function will give the same result, and is much easier :)