Find the point on a plane $3x + 4y + z = 1$ that is closest to $(1,0,1)$ 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 
 A: 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.
A: 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 :)
A: 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 $$P=(1,0,1)-\frac{3}{26}\langle 3,4,1\rangle=\left(\frac{17}{26},-\frac{12}{26},\frac{23}{26}\right)$$ 
A: Let $(x, y, z)$ be the point in question. The distance is given by $\sqrt{(x - 1)^2 + y^2 + (z - 1)^2}$. By Cauchy Schwarz, $\left((x-1)^2 + y^2 + (z-1)^2\right)(3^2 + 4^2 + 1^2) \geq (3x + 4y + z - 4)^2$, so $\left((x-1)^2 + y^2 + (z-1)^2 \right) \geq \frac{9}{26}$
Equality is reached when $\frac{x-1}{3} = \frac{y}{4} = \frac{z-1}{1}$. Solving using $3x + 4y + z = 1$ gives $\left(\frac{17}{26}, -\frac{6}{13}, \frac{23}{26} \right)$
A: For the completeness of discussion, I would give an approach using multivariable calculus but without Lagrange Multiplier.
The square of the distance between any point $(x,y,z)$ and $(1,0,1)$ is given by:
$$
d^2=(x-1)^2+y^2+(z-1)^2 \tag{1}
$$
Sub. $z=1-3x-4y$ into eq. (1) and define a function $f(x,y)$ as follows:
$$
f(x,y):=(x-1)^2+y^2+(1-3x-4y-1)^2
$$
$$
f(x,y)=(x-1)^2+y^2+(3x+4y)^2
$$
$f(x,y)$ can be interpreted as the square of the distance between points on the plane and the point $(1,0,1)$.  Our objective is to find a local minimum point of $f(x,y)$ if exists. As $f(x,y)$ is minimized, $d$ will be minimized.  As such, we calculate the gradient and Hessian matrix of $f$ as follows:
$$
\nabla f(x,y)=
\begin{bmatrix}
2(x-1)+2\cdot3(3x+4y) \\
2y+2\cdot4(3x+4y)
\end{bmatrix}
$$
$$
\nabla f(x,y)=
\begin{bmatrix}
20x + 24y - 2 \\
24x+34y
\end{bmatrix}
$$
$$
H=
\begin{bmatrix}
20 & 24 \\
24 & 34
\end{bmatrix}
$$
The Hessian is always $104>0$.
$$
\nabla f(x,y)=\mathbf{0}
$$
$$
\begin{bmatrix}
20x + 24y - 2 \\
24x+34y
\end{bmatrix}=\begin{bmatrix}
0\\
0
\end{bmatrix}
$$
By solving the above system of linear equations, we obtain $x=17/26$ and $y=-6/13$ as well as $z=3/2$.   As $\det (H)=104>0$ and $f_{xx}=20>0$, thus $(17/26, -6/13)$ is a local minimum of $f(x,y)$.
