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Say I have these 2 lines represented as parametric equations like so:

$L1 : x= 1 + t, y = 0, z = 2 - t$

$L2 : x = t , y = t, z = 1 + t$

My question is:

Is there a way to find the distance (minimal) between those 2 lines only by using the formula listed below?

$$\frac{|| AB \times AP || }{ ||AB||}$$ (Distance of a point from a line passing through two points.)

$A$ is a point on $L1$

$B$ is a point on $L1$

$P$ is a point on $L2$

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    How do you define the distance between two lines **that aren't parallel** ? – 2017-02-14
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    probably OP means to say "minimal distance" between two lines – 2017-02-14
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    You can see: https://en.wikipedia.org/wiki/Skew_lines#Distance – 2017-02-14
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    @DonAntonio the minimal distance – 2017-02-15

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No, there is no way. $$ \frac{\lVert AB \times AP \rVert}{\lVert AB \rVert} = \frac{\lVert AB \rVert \lVert AP \rVert \sin \alpha}{\lVert AB \rVert} = \lVert AP \rVert \sin \alpha $$ This formula only depends on the distance between $A$ and $P$ and that angle between the first line and $AP$ which is the shortest connection between $A$ and $P$, so it leaves out the direction of the second line.

We could rotate the second line around $P$ without changing the value of that formula, while that rotation might change the minimal distance between both lines.

For an example minimizing the (squared) distance over two parameters see here.

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    That would depend how literally you take the “only by using the formula …” part of the question. You could plug in a general point $P$ on ghe second line given by its parametrisation, square it, and get a quadratic polynomial in the parameter $t$. This is, of course, easy to minimise. – 2017-02-14
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    @HaraldHanche-Olsen can you please explain more – 2017-02-15
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    @HashWizard See the link in the last sentence of my answer. – 2017-02-15
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    Simply this: Pick two points on $L_1$, and call them $A$ and $B$. For example, the one with $t=0$ and the one with $t=1$. Write $P=(t,t,1+t)$. Calculate the cross product $AB\times AP$. It will be some vector in which each component is a first degree polynomial in $t$. Calculate the squared norm of the cross product. It will be a quadratic polynomial in $t$. Find the $t$ value that minimizes it. Take the square root, since we used the squared norm before. Divide by $\|AB\|$. Done. – 2017-02-15