Proving two lines don't intersect

Question

The line $l_1$ passes through the position vector $i-j-2k$ and is parallel to the vector $3i-4j-2k$.

The line $l_2$ passes through the position vector $(1+5\cos t)i-(1+5\sin t )j-14k$ where $0≤ t≤2\pi$ and is parallel to the vector $15i+8j-3k$.

Show that both lines don't intersect

I have solved it by showing that minimum distance between the two lines is greater than 0 by finding the minimum point of distance function of the perpendicular between them.

However I am interested in a method that involves comparing the vector components of the two lines and then somehow showing that two lines don't intersect. (Maybe by contradiction)

Is there a way to solve it through that method?

Answer 1

Suppose we have a line parallel to the vector $\mathbf{A}$ that passes through point $\mathbf{P}$ and a line parallel to vector $\mathbf{B}$ that passes through point $\mathbf{Q}$. For the lines to intersect, it is necessary that they lie in the same plane. This in turn implies the vector $\mathbf{P} - \mathbf{Q}$ also lies in that plane. Three vectors lie in the same plane if and only if the cross product of two of them is orthogonal to the third. Thus, if $(\mathbf{P}-\mathbf{Q})\cdot(\mathbf{A}\times\mathbf{B}) \ne 0$, the lines do not intersect. This condition is probably good enough for your problem.

However, note that $(\mathbf{P}-\mathbf{Q})\cdot(\mathbf{A}\times\mathbf{B}) = 0$ does not imply the lines intersect, as they might be parallel (that is, $\mathbf{A}\times\mathbf{B} = 0$). If this is the case, they will only intersect if $\mathbf{A}$ and $\mathbf{B}$ are also parallel to $\mathbf{P} - \mathbf{Q}$ (in which case the lines coincide).

Answer 2

Zwikker (see below) derives an equation the intersection of two lines in the complex plane. The lines have end points $(z_1,z_2)$ and $(z_3,z_4)$, respectively, and can be thought of as vectors $(z_1-z_2)$ and $(z_3-z_4)$. The point of intersection is given by

$$z=\frac{(z_1z_2^*-z_1^*z_2)(z_3-z_4)-(z_3z_4^*-z_3^*z_4)(z_1-z_2)} {(z_1-z_2)(z_3^*-z_4^*)-(z_1^*-z_2^*)(z_3-z_4)}$$

If the vectors $(z_1-z_2)$ and $(z_3-z_4)$ are parallel, then the denominator equals zero by the criterion of parallellity and then $z=\infty$. Unfortunatley, this does not address the problem you have posed in $\mathbb{R}^3$. However, if the lines are parallel, they must lie on a plane. Perhaps that will help you.

Ref: Zwikker, C. (1968). The Advanced Geometry of Plane Curves and Their Applications, Dover Press.