find the point on the plane closest to the point Find the point on the plane $ \frac x2 + \frac y4 + \frac z4=1$ that is closest to the point (0, 0, 1)
To minimize the distance, I have the function $f(x, y, z)=(\frac x2)^2+(\frac y4)^2+(\frac z4 -1)^2$ to start with.
However my professor put an X next to this, indicating that it is wrong. I can't think of what else to do. 
 A: So you should have a distance formula like
$$
f(x,y,z)=x^2+y^2+(z-1)^2
$$
However, now you must enforce that $(x,y,z)$ is a pair of numbers that satisfies $x/2+y/4+z/4=1$, one way to do this is solve for $z$, $z=4-2x-y$ and substitute into the distance formula.
$$
d(x,y)=x^2+y^2+(4-2x-y-1)^2
$$
The short answer is that you did not enforce the constraint $x/2+y/4+z/4=1$. 
A: Hint:
The euclidean equation of this plane, gives you a normal vector of the plane $\vec n=(2,1,1)$ (I removed any denominators). The closest point is the orthogonal projection of the point onto the plane. So parameterise the line passing throught the point $(0,0,1)$ which is perpendicular to the plane:
$$M=\begin{pmatrix}0\\0\\1\end{pmatrix}+t\vec n$$
and write the coordinates of $M$ satisfy the equation of the plane to determine the value of $t$.
A: As Bernard said, the shortest distance from a point to a plane is along the perpendicular to the plane.  And here, of course, a perpendicular to the plane is <1/2, 1/4, 1/4> or, multiplying by 4 because I don't like fractions <2, 1, 1>.  So the shortest distance from the point (0, 0, 1) to this plane is along the line x= 2t, y= t, z= 1+ t.  Find the point where that line intersects the plane (i.e. where 2t/2+ t/4+ (t+1)/4= 1) and then calculate the distance from that point to (0, 0, 1).
A: The vector that is normal to the plane 
$$\frac x2+\frac y4 + \frac z4 =1\tag 1$$
is
$$(z_x’, z_y’, -1) = (-2, -1,-1)$$
and the line that passes through the point (0,0,1) and has the shortest distance from the plane is $(0,0,1)+t(2,1,1)$, or
$$x=2t, \>\>\>y=t,\>\>\>z=1+t \tag 2$$
Then, substitute the line (2) into the equation of the plane (1) to get $t=\frac12$ for the point of their intersection. Plug $t=\frac12$ back into (2) to obtaIn the point ($1,\frac12, \frac32$) on the plane that is closest to the given point.
