Point on line closest to origin given the line's parametric equations

Question

So I'm given the line described by:

$$\begin{cases} x = \frac{2}{3} + \lambda \\ y = \frac{1}{3} + \lambda \\ z = \lambda \end{cases}$$

And I'm asked to determine the point that is closest to the origin. I have the following formula for the distance between a point and a line (using parametric equations) in $\Bbb R^3$:

$$ d(\vec{p}, L) = \frac{\mid (\vec{p} - \vec{q}) \times \vec{r} \mid}{\mid \vec{r} \mid} $$ where $\vec{p}$ is the point and $\vec{x} = \vec{q} + \lambda\vec{r}$ are the parametric equations for the line.

Somehow I think I'm supposed to fill these in and find some sort of function for it, but I'm not really sure. Am I now supposed to calculate this for each $x, y$ and $z$ seperatly? I'm a little confused as I might go on about doing this.

Answer 1

If $(x,y,z)$ is a point on the line, then the distance to the origin is given by

$d(\lambda)= \sqrt{(\frac{2}{3}+\lambda)^2+(\frac{1}{3}+\lambda)^2+ \lambda^2}.$

Since $d$ is minimal $ \iff d^2$ is minimal, you have to determine $t $ such that $f(t)= \min f(\mathbb R)$, where

$$f(t)=(\frac{2}{3}+t)^2+(\frac{1}{3}+t)^2+ t^2.$$

Answer 2

Your formula is defective, it should read $$ d(\vec{p}, L) = \frac{\mid (\vec{p} - \vec{q}) \times \vec{r} \mid}{\mid \vec{r} \mid}. $$ The numerator is the area of the parallelogram spanned by $\vec p-\vec q$ and $\vec r$ and the denominator $|\vec r|$ is the length of its base, giving the height of the parallelogram, that is the distance from $\vec p$ to the line.