Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Can Any Genius Answer

  1. Sep 27, 2006 #1

    Plz plz tell me the solutions of these question :

    I have tried the questions many time but not getting the exact logic plz help me

    1): Find equation of straight line perpendicular to x-1=(y-1)/2=(z+2)/3 and (x+2)/2=5-y=(z+3)/2 and passing through their point of intersection .

    2):Find equation of plane passing through the point (1,0.-1) and (3,2,2) and parallel to the line x-1=(1-y)/2=(z-2)/3

    3):Find equation of straight line passing through midpoint of AB ,perpendicular to AB and lies in the plane x=z where A(1,-1,1) and B(2,1,2)
  2. jcsd
  3. Sep 27, 2006 #2


    User Avatar
    Science Advisor

    Geniuses? Will you settle for me?

    The line x-1= (y-1)/2= (z+2)/3 has parametric equation x= 1+ t, y= 1+ 2t, z= -2+ 3t (setting t to be the common value above).
    The line (x+2)/2= 5- y= (z+3)/2 has parametric equation x= -2+ 2s, y=5- s, z= -3+ 2s.

    Set x= x, y= y, z= z in the two above to determine where they intersect. (That's three equations in two unknown variables, s and t. In general two lines do not intersect in 3 dimensions but you are told that these do.) Vectors in the direction of the two lines are i+ 2j+ 3k and 2i- j+ 2k. The cross product of those will give you a vector perpendicular to both.

    To find the equation of a plane, you need a vector perpendicular to that plane. A vector in the the plane is (3-1)i+ (2-0)j+ (2-(-1))k= 2i+ 2j+ 3k. Another vector in the plane is i+ 2j+ 3z (from the given line). Their cross product is perpendicular to the plane. Use either of the given points to determine the equation of the plane.

    The midpoint of AB is ((1+2)/2, (-1+ 1)/2, (1+ 3)/2)= (3/2, 1, 3/2) and a vector in the direction of AB is (2-1)i+ (1-(-1))j+ (2-1)k= i+ 2j+ k. The plane perependicular to that vector, containing that point (x- 3/2)+ 2(y- 1)+ (z- 3/2)= 0. The line you are looking for is the intersection of that plane and z= x. You can solve those two equations for 2 of x, y, z in terms of the third and then use that third as a parameter.
    Last edited by a moderator: Sep 27, 2006
  4. Sep 27, 2006 #3
    thz HallsofIvy for helping me thz a lot
  5. Sep 27, 2006 #4
    Hey hallsoivy.. check out my post.. wana help me out?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook