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 

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 $f(x,y)$.
