MHB Distance between two non parallel lines

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The discussion focuses on calculating the distance between two non-parallel lines using vector methods. The formula provided, d = |(v1 - v2) · (a1 × a2)| / ||a1 × a2||, is derived by considering the plane containing one line parallel to the other and projecting the vector between points on the lines onto the normal vector of the plane. Participants note that if the lines lie in the same plane and are not parallel, the minimum distance is zero, suggesting the context may involve skew lines in three-dimensional space. There's a call for clarification on the formulation of the question, indicating potential confusion about the geometric context. Overall, the thread emphasizes the importance of understanding the relationship between the lines and their spatial arrangement.
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Hey! :o

Using vector methods show that the distance between two non parallel lines $l_1$ and $l_2$ is given by $$d=\frac{|(\overrightarrow{v}_1 - \overrightarrow{v}_2) \cdot (\overrightarrow{ a}_1 \times \overrightarrow{a}_2)|}{||\overrightarrow{a}_1 \times \overrightarrow{a}_2||}$$ where $\vec{v}_1$ and $\vec{v}_2$ are random points of $l_1$ and $l_2$ respectively, and $\vec{a}_1$ and $\vec{a}_2$ are the directions of $l_1$ and $l_2$.

HINT:
We consider the plane that contains $l_1$ and is parallel to $l_2$. Show that $\frac{\vec{a}_1 \times \vec{a}_2}{\|\vec{a}_1 \times \vec{a}_2\|}$ is unit perpendicular to that plane. Then take the projection of $\vec{v}_2-\vec{v}_1$ to that perpendicular direction.
I have done the following: We consider the plane that contains $l_1$ and is parallel to $l_2$. That means that the plane passes through the point $\overrightarrow{v}_1$ and has as parallel vector the vector $\overrightarrow{a}$.

To find the distance between the two lines, we have to find the distance between the points $\overrightarrow{v}_1$ and $\overrightarrow{v}_2$.

The vectors $\overrightarrow{a}_1$ and $\overrightarrow{a}_2$ produce the plane, so the vector $\overrightarrow{a}_1 \times \overrightarrow{a}_2$ is perpendicular to the plane.

So, the unit perpendicular vector to the plane is $\frac{\overrightarrow{a}_1 \times \overrightarrow{a}_2}{||\overrightarrow{a}_1 \times \overrightarrow{a}_2||}$.

A vector from the plane to the point $\overrightarrow{v}_2$ is $\overrightarrow{v}_2-\overrightarrow{v}_1$.

The distance that we are looking for the length of the projection of this vector onto the normal vector to the plane.

So, $$d=\frac{|(\overrightarrow{v}_1 - \overrightarrow{v}_2) \cdot (\overrightarrow{ a}_1 \times \overrightarrow{a}_2)|}{||\overrightarrow{a}_1 \times \overrightarrow{a}_2||}$$
Is this correct?? (Wondering) Could I improve something at the formulation?? (Wondering)
 
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The formulation of the question is incomplete at best. If the two lines lie in a plane and they are not parallel, the minimum distance is 0 (by definition). And in spherical geometry all parallel lines meet at the poles.
 
I'm guessing that the intent here is to find the distance between two skew lines in ##\mathbb R^3##.
 
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