Thread: Lines in a Plane

The distance between the skew lines r=OP + t d_1 and r= OQ + s d_2 is |Proj(PQ onto n)| or |PQ . n|/|n| wherer n=d_1 x d_2.

Find the distance between the lines x=6, y=-4-t, z=t, and x = -2s, y=5, z=3+s.

Bolded = vectors (arrow on top)

My answer is as follows:

x=6
y=-4-t
z=t

therefore in parametric eqtn form
(x-6)/0
(y+4)/-1
z/1
d_1 = (0,-1,1)

x/-2
(y-5)/0
(z-3)/1
d_2=(-2,0,1)

PQ=d=(-2,1,0)

(-2,1,0) . (-1,0,-2)
-------------------
sqrt5

final answer
1/sqrt5

is this correct??
please let me know.
thanks!

Does anyone know how to do this question=?

..?

DO NOT BUMP. It is rude.

Suppose that + tD\;\& \;\ell _2 :q + tE" alt="\ell _1 + tD\;\& \;\ell _2 :q + tE" /> are two skew lines.
The distance between them is given by:

ok, sorry about that Plato

From my work, i think it looks like I have vector PQ wrong only. Is that correct? Here let me try that again:

l1 P (6,-4,0)
l2 P(0, 5,3)
so, PQ would be
(-6,9,-3)

D=(-6,9,-3) . (-1,0,-2)
----------------------
sqrt5
=6+5
-----
sqrt5
=11/sqrt5

is this correct now? thanks for your help by the way

From the given: 6,-4,0) + t\left\langle {0, - 1,1} \right\rangle \;\& \;\ell _2 0,5,3) + t\left\langle { - 2,0,1} \right\rangle " alt="\ell _1 6,-4,0) + t\left\langle {0, - 1,1} \right\rangle \;\& \;\ell _2 0,5,3) + t\left\langle { - 2,0,1} \right\rangle " />.

So, .

What you have above is incorrect: .

Now all you need is

EDIT: Correct

how did you get
p(6,4,0)?

Given:
y=-4-t

(x,y,z)=(6,-4,0)+t(0,-1,1)

It was a typo.

Ok, then I get

D= (-6,9,3) . (-1,-2,-2)
----------------------
3

therefore

6 units..

(hmm :S turns out same as another q i had, i think)

Originally Posted by skeske1234

Ok, then I get

D= (-6,9,3) . (-1,-2,-2)
----------------------
3

therefore

6 units..

Yes that is correct.