Finding distance between a point and parametric equations This problem is from the parametric and trigonometric coordinate systems of the Art of Problem Solving Precalculus book:
Find the smallest distance between the point $ (1,2,3) $ and a point on the graph of the parametric equations $ x = 2-t, y=4+t, z=3+2t.$
I tried eliminating $ t$ by adding the first two equations together, yielding $ x+y=6$, and then multiplying the first equation by two and adding that to the third, yielding $ 2x+z=7$, and then I added these two equations together, and I got $ 3x+y+z=13$.
My textbook only covered simple 3d graphs using spherical and cylindrical coordinates, graphing cones, cylinders, spheres, and planes, so I didn't know what the graph of $ 3x+y+z=13$ was, however using a 3d graphing calculator I saw that it was a plane. I don't know how to proceed any further, any help would be appreciated.
 A: Hint: $x = 2-t, y=4+t, z=3+2t$ is a map from $R\to R^3$ so you know the image is a 3d curve.
Since $x,y,z$ are a linear funtion of $t$, so what is the shape of its image?
Hint2: the shortest path from a point to a line shall be perpendicular to the line. Could you use the perpendicular condition to solve the point $(x(t),y(t),z(t))$ that is closest to $(1,2,3)$
A: You have the given point $P_0 = (1, 2, 3) $
And the curve (which is a line) is given by
$Q(t) = (2,4,3) + t ( -1, 1 , 2) $
The vector extending from $P_0$ to a point on the line $Q(t) $ is
$V(t) = Q(t) - P_0 = (2, 4, 3) + t (-1, 1, 2) - (1, 2, 3) = (1, 2, 0) + t (-1, 1, 2) $
And you want this vector to be perpendicular to the line, i.e. perpendicular to its direction vector $(-1, 1, 2) $.  The condition for perpendicularity is that the dot product between the two vectors is $0$.  Hence, we have
$ V(t) \cdot (-1, 1, 2) = 0 $
i.e.
$ \big( (1,2,0) + t (-1, 1, 2) \big) \cdot (-1, 1, 2) = 0 $
And this simplifies to
$ 1 + 6 t = 0 $
So $ t = \dfrac{-1}{6} $
Plug this into $V(t)$,
$V(\dfrac{-1}{6}) = (1,2,0) - \dfrac{1}{6} (-1, 1, 2) = \dfrac{1}{6} (7, 11, -2) $
The magnitude of this vector is the shortest distance
$\| V \| = \dfrac{1}{6} \sqrt{ 7^2 + 11^2 + (-2)^2 } = \dfrac{\sqrt{174}}{6} $
One could verify this result by finding the magnitude of the vector $W$ given by
$ W = \dfrac{(P_0 - (2,4,3) ) \times (-1, 1, 2)}{\| (-1 , 1, 2 ) \|} $
This is equal to
$ W = \dfrac{ (-1, -2, 0) \times (-1, 1, 2) }{\sqrt{6} } $
$ = \dfrac{(-4 , 2, -3 )}{\sqrt{6}} $
The magnitude of $W$ is
$\| W \| = \dfrac{\sqrt{ 29 }}{\sqrt{6}} = \dfrac{\sqrt{29 (6) }}{6} = \dfrac{\sqrt{ 174 }}{6} $
So the results match.
