0
$\begingroup$

Let's say there are two lines of the form: $$p_0=\frac{x-x_0}{l_0}=\frac{y-y_0}{m_0}=\frac{z-z_0}{n_0}$$ $$p_1=\frac{x-x_1}{l_1}=\frac{y-y_1}{m_1}=\frac{z-z_1}{n_1}$$ For $p_0$ and $p_1$ to intersect following conditions need to be set:
1. $p_1$ and $p_2$ can't be parallel
2. $\overrightarrow{T_0T_1}\cdot\left(\vec{s_0}\times\vec{s_1}\right)=0$, where $T_0$ and $T_1$ are arbitrary points on lines $p_0$ and $p_1$, respectively, and $\vec{s_0}$, $\vec{s_1}$ their vectors of direction

Does this mean that this triple product will also equal $0$ if the lines are parallel? Why is triple product equal to $0$ if lines intersect? How can we prove this?

Any method of proof is fine, also a graphical explanation would be very much appreciated.

$\endgroup$

1 Answer 1

Reset to default
1
$\begingroup$

If the lines are parallel, then $$\overrightarrow{s_0}=\lambda\overrightarrow{s_1}\implies\overrightarrow{s_0}\times\overrightarrow{s_1}=0$$ and so the triple product would be zero.

The expression $$\frac{\overrightarrow{T_0T_1}\cdot\left(\overrightarrow{s_0}\times\overrightarrow{s_1}\right)}{|\overrightarrow{s_0}\times\overrightarrow{s_1}|}$$ is the formula for shortest distance between the two non-parallel lines which is obviously zero if the lines intersect. So if the numerator is zero, the distance is zero.

This formula can be readily seen as the shortest distance as it is the projection of the vector $\overrightarrow{T_0T_1}$, joining two points on the respective lines, onto the vector perpendicular to both lines.

$\endgroup$

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.