# Thread: Lines in a Plane

1. ## 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)

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

1/sqrt5

is this correct??
thanks!

2. Does anyone know how to do this question=?

3. ..?

4. DO NOT BUMP. It is rude.

Suppose that $\displaystyle \ell _1 + tD\;\& \;\ell _2 :q + tE$ are two skew lines.
The distance between them is given by: $\displaystyle D\left( {\ell _1 ,\ell _2 } \right) = \frac{{\left| {\overrightarrow {pq} \cdot \left( {D \times E} \right)} \right|}}{{\left\| {\left( {D \times E} \right)} \right\|}}$

5. 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

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

So, $\displaystyle p = (6,-4,0)\;,\,D = \left\langle {0, - 1,1} \right\rangle \;,\,q = (0,5,3)\;\& \,E = \left\langle { - 2,0,1} \right\rangle$.

What you have above is incorrect: $\displaystyle \overrightarrow {pq} = \left\langle { - 6,9,3} \right\rangle$.

Now all you need is $\displaystyle \left( {D \times E} \right) = ?$

EDIT: Correct

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

Given:
y=-4-t

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

8. It was a typo.

9. 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)

10. Originally Posted by skeske1234
Ok, then I get

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

therefore

6 units..
Yes that is correct.