Question

Let ℓ1 and ℓ2 be lines given in vector form by ℓ1: (-5, -1, 1)+t(2; -3; 1) and ℓ2 : (1; 3; 1) + t(0; 0; 5) respectively. There exists precisely one line in R3 which intersects and is perpendicular to both of these. Find this line in vector form.

Answer 1

the cross product is the vector perpendicular to both lines. the problem is to find the one of the two points intersected by the required line.

Answer 2

oh cross product :P yah that makes sense. hmm

Answer 3

@SithsAndGiggles , any suggestion on finding a particular point?

Answer 4

If the lines \(l_1\) and \(l_2\) have exactly one line which intersects and is perp to both of them, then should \(l_1\) and \(l_2\) be skew?

Answer 5

@Erin001001 , not necessarily. the two lines may be coplanar. but for the given lines, they are skew.

Answer 6

Ah yes, I meant if they didn't intersect (I'd already checked for that).

Answer 7

I think I have a highly tedious way to do this. Using cross product, we can find a vector perp to both \(l_1\) and \(l_2\). Now divide out its length, giving a unit vector \(\underline{n}\) in that direction. Now find the distance \(d\) between the two skew lines and multiply \(\underline{n}\) by \(d\). Let \(\underline{v}=d\underline{n}\).

Now there should be a unique point on \(l_1\), say (-5; -1; 1) + s(2; -3; 1),
and a unique point on \(l_2\), say (1; 3; 1) + t(0; 0; 5),
such that (-5; -1; 1) + s(2; -3; 1) + \(\underline{v}\) = (1; 3; 1) + t(0; 0; 5).
(Or perhaps \(-\underline{v}\)?) Solve for s, which gives one of the points of intersection.

Can someone check if my method works? Or preferably, come up with a neater one....

Answer 8

\[\vec{PR}=\vec{PQ}\times\vec{RS}=<2,-3,1>\times<0,0,5>\]
also, \[\vec{PR}=<6-2t,4+3t,5s-t>\]

Answer 9

now, \[\vec{PR}\cdot\vec{RS}=0\]
or \[(6-2t)(0)+(4+3t)(0+(5s-t)(5)=0\]
which implies that \[5s=t\]

Answer 10

@sirm3d I'm not sure about this line:
\(\vec{PR}=\vec{PQ} \times \vec{RS}=<2,−3,1> \times <0,0,5>\).
The length of \(\vec{PQ} \times \vec{RS}\) will give a vector in the direction of \(\vec{PR}\), but the length is necessarily right, surely?

Answer 11

The way I did it was basically to find the length of \(\vec{PR}\) by finding the (shortest) distance between the two skew lines. Then I multiplied it by the unit vector in the \(\vec{PQ} \times \vec{RS}\) direction, which gives \(\vec{PR}\).

As this point your method is an improvement on mine. You can just equate \(\vec{PR}\) with \(\vec{PR}=<6−2t,4+3t,5s−t>\), and find s and t. Since we have s, we have one of the intersection points which is what we set out to find.

Answer 12

@Erin001001 , i was doing it your way

Answer 13

Oh. Still it's an improvement: you don't need to deal with the +v or -v nonsense I was going on about...

Answer 14

in the end, we have to solve the equation between \(|\vec{PR}|\) (expressed in terms of t or s) and the actual distance between the skew lines.

Answer 15

http://answers.yahoo.com/question/index;_ylt=AmOu8_nGaxbug8O3gAA0P3QazKIX;_ylv=3?qid=20130818173407AAbQjPh