Does anyone know how to do this question=?
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!
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