Linear algebar distance between two lines,help for part(b)

  1. Linear algebar!!!distance between two lines,help for part(b)

    1. The problem statement, all variables and given/known data

    (a)use cross product to find the distance d from the line K determined by the two points(4,4,4) and (-1,-2,-5) to the line L determined by the parametric equations x=3,y=-1,z=5-5t

    (b)find two points P and Q on the lines K and L respectively in part(a) such that the idstance from P to Q is d.

    2. Relevant equations

    projection formula, corss product

    3. The attempt at a solution

    i slove part(a)

    i set A as (4,4,4) and B as (1,-2,-5), then we can get -->AB=(-3,-6,-9)

    from x=3,y=-1,z=5-5t , we get C as (3,-1,-5),then -->AC=(-1,-5,1)

    and we know l(distance between L and K)=area of parallelogram/length of AB

    which is

    l= ||-->AB*-->AC|| / ||-->AB||

    now i use cross product -->AB*-->AC=(-3,-6,-9)*(-1,-5,1)=(-39,-12,21)

    ...................

    ...................

    FINALLY i get l= sqrt(117/7)

    i think i did right way for part(a)

    -------------------------------------------------------------------------------------

    but for part(b)

    i find the projection of -->AC first

    proj(AC)={AC*AB/AB*AB}AB

    =(3+30-9)/126 (-3,-6,-9)

    =4/21(-3,-6,-9)

    ok, till here,

    can i use (4,4,4)as point P

    then use C(3,-1,-5) minus (4/21)(-3,-6,-9) to get point Q?

    if i did wrong way ,plz give me a hand

    thanks!!!
     
  2. jcsd
  3. Re: Linear algebar!!!distance between two lines,help for part(b)

    i get some ideas from another topic

    https://www.physicsforums.com/showthread.php?t=274638

    i use the eqution wut Dick offered X_p(s)-X_q(t)=u*W

    then i get

    (4,4,4)+t[-3,-6,-9] - (3,-1,5)+s[0,0,-5] = u[30,-15,0]<--cross product by deriction vectors

    i solve it and get t=11/15 s=38/25 u=-1/25

    wuts next?

    PLZ HELP
     
  4. Re: Linear algebar!!!distance between two lines,help for part(b)

    any1 help me !! plz!!
     
  5. Re: Linear algebar!!!distance between two lines,help for part(b)

    Your original posted solution to (a) gives the dist from C to K, but how do you know C is the closest pt to K? Your original (a) is wrong because K and L don't lie in the same plane.

    Look at your second post.

    The skew lines K and L lie on two parallel planes P1 and P2. The distance between K and L is the distance between the two planes P1 and P2.

    Your vector [30,-15,0] is perpendicular to both planes P1 and P2.

    (4,4,4)+t[-3,-6,-9] when t=11/15 is pt M on line K

    (3,-1,5)+s[0,0,-5] when s=38/25 is pt N on line L

    Vector MN is parallel to [30,-15,0].

    You are ready to write down the answer to both parts now, do you see it?
     
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