How does using cross product to find shortest distance work?

[ognik]

A method for finding the shortest distance between 2 skew, non intersecting lines is to 1st find the common normal, using $ \vec{n} = \frac{\vec{v_1} \times \vec{v_2}}{|\vec{v_1} \times \vec{v_2}|} $ I'm looking for a proof or intuition as to why this is true please?

Then apparently we get the shortest distance (d) by projecting any distance vector onto this normal? A distance vector here is any point on one vector minus any point on the other vector. I can obviously choose points to get distance vectors of any magnitude. So my 2nd question is, I assume the angles between any position vector and the skew lines effectively minimise the projection?

But I did a sketch - View attachment 4918 - which doesn't convince me at all about the above ..., here $|\vec{P_2P_1}| \gt |d|$, multiplying that by $Cos \theta $ produces the line from $P_2 \perp \vec{v_1}$ - still > d?

 

[ognik]

What I'm not quite sure of is - I know the cross product of 2 coplaner vectors is perpendicular to the plane those vectors are in, but why is that the same for 2 non-coplaner vectors?

 

[topsquark]

What I'm not quite sure of is - I know the cross product of 2 coplaner vectors is perpendicular to the plane those vectors are in, but why is that the same for 2 non-coplaner vectors?

The two vectors (unless they are colinear) form the basis vectors for a plane. Two distinct vectors are always coplanar.

-Dan

 

[ognik]

Thanks - I just can't visualise it in this case - where the vectors are skew and don't intersect ... the only surface I can visualise in this case is not flat?

 

[topsquark]

Thanks - I just can't visualise it in this case - where the vectors are skew and don't intersect ... the only surface I can visualise in this case is not flat?

We can always "move" the vectors so that they meet end to end.

-Dan

 

[ognik]

We can always "move" the vectors so that they meet end to end.

-Dan

Hi - after much thought I decided we must be talking about the directions of the vectors? IE there is a vector parallel to one of the vectors, which does intersect the other vector. Then the cross product makes sense as being perpendicular to both directions?

 

[topsquark]

Hi - after much thought I decided we must be talking about the directions of the vectors? IE there is a vector parallel to one of the vectors, which does intersect the other vector. Then the cross product makes sense as being perpendicular to both directions?

In a word: Yup! (Sun)

-Dan