Quick distance between skew lines question

[kankerfist]

If one of the skew lines in the question is:

\(\displaystyle \frac{{x - 3}}{2} = \frac{{y + 1}}{4} = \frac{{1 - z}}{3}\)

Should I rewrite it to be:

\(\displaystyle \frac{{x - 3}}{2} = \frac{{y + 1}}{4} = \frac{{z - 1}}{-3}\)

And consider that a point on the line would be (3,-1,1) and N = <2,4,3>? I am not sure exactly how to handle the \(\displaystyle \frac{{1 - z}}{3}\) part when considering a point on the line and its N vector. Any help is appreciated!

 

[tkhunny]

Should I rewrite it to be:

Who cares? Have you been told to write things with the variable first? Have you been instructed to keep positive values in the denominator? In the absence of direction, write it any way you and the rest of the world can understand it.

 

[kankerfist]

Well the way it is written on the test gives an N value of <2,4,3> and my way it gives an N value of <2,4,-3>. And I need to know if my way is accurate even though the value of N changes, otherwise I cant determine the distance between the 2 skew lines.

 

[pka]

You may have confused some basic formulae.
To find the distance between two skew lines:
1. Find a point on one of the kines.
2. Find the equation of a plane that contains the other line and is parallel to the first line.
3. Find the distance between the first point and the plane.

So P is a point on the first line and D<SUB>1</SUB> is its direct vector; and Q is a point on the second line and D<SUB>2</SUB> is its direct vector; then the distance between the two lines is \(\displaystyle \L
D(l_1 ,l_2 ) = \frac{{\left| {\vec{PQ} \cdot \left( {D_1 \times D_2 } \right)} \right|}}{{\left\| {\left( {D_1 \times D_2 } \right)} \right\|}}\).