How Do You Calculate the Distance Between Two Skew Lines in 3D Space?

[menal]

Homework Statement

Determine the distance between L1=(-2,1,0)+s(1,-1,1) and L2=(0,1,0)+t(1,1,2)

how would this question be done?

 

[SammyS]

What have you tried?

Where are you stuck?

 

[menal]

i tried taking the scalar
product of the vector joining the two given points with the unit
vector of the common perpendicular.

|i j k|
|1 1 2|
1 -1 1| but i get stuck after that...will it equal i+j+k?

 

[SammyS]

What did you get for the common perpendicular?

 

[menal]

I got 1,1,1
so i+j+k

 

[menal]

and the vector joining the 2 given points would be" (2,0,0)

 

[menal]

i tried the question and got 5 as my answer, is tht correct?

 

[SammyS]

I get <3, 1, -2> = 3i + j -2k for the common perpendicular.

 

[menal]

may i ask how you got that? using cross product?

 

[SammyS]

I evaluated the following determinant.

\begin{vmatrix}<br /> \hat{i}&amp;\hat{j}&amp;\hat{k}\\<br /> 1 &amp; 1 &amp; 2\\<br /> 1 &amp; -1 &amp; 1<br /> \end{vmatrix}

 

[menal]

oh yup, i see what i did wrong there. i switched up the values while calculating.
alright, so i used your common perpendicular, and the vector joining the two points (2,0,0) and did this:

[(-2,1,0)-(0,1,0)]dotproduct(3,1,-2)/1
=[(2,0,0)*(3,1,2)/1]
=(2)(3)+(0)(1)+(-2)(0)/1=6/1
=6

would that be correct?

 

[SammyS]

The distance distance between L1 & L2 can't be greater than the distance between any two points on the respective lines, so it can't be greater than 2.

You need to divide your result by the magnitude of the common perpendicular.
You want the component of <2, 0, 0> in the direction of <3, 1, -2> .

 

[menal]

so your saying: 6/3/2=1 would be my answer?

 

[SammyS]

No, the magnitude of <3, 1, -2> is given by:

\left|\left\langle3,\,1,\,-2\right\rangle\right|=\sqrt{3^2+1^2+(-2)^2}=\sqrt{14}

.