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Easy Vector algebra

  1. Oct 19, 2009 #1
    p = (2,6,3)
    q = (1,0,1)
    r = (4,1,-1)

    part a)
    Find a parametric equation of the line that passes through p with direction q
    x=p + t(q)
    x=(2,6,3) + t(1,0,1)

    part b)
    find the point X on the line in part a so that RX is perpendicular to the line

    I'm having trouble doing this one
    please help
     
  2. jcsd
  3. Oct 20, 2009 #2

    Office_Shredder

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    What way can you test to see if two vectors are perpendicular?
     
  4. Oct 20, 2009 #3
    if a.b = 0
    or a=0
    or b=0

    -

    It's asking me to find the point

    p = (2,6,3)
    q = (1,0,1)
    r = (4,1,-1)

    pq=p + t(q)
    pq=(2,6,3) + t(1,0,1)

    part b)
    find the point X on the line in part a so that RX is perpendicular to the line

    can you please give some more direction?

    I already know the answer i just need to know how to get to it!!
    (1,6,2) is the point
     
  5. Oct 20, 2009 #4
    cmon PF don't fail me now!
     
  6. Oct 20, 2009 #5

    HallsofIvy

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    Let (x, y, z) be the point X. Then a vector from r to X is (x- 4, y- 1, z+ 1). That is the vector that must be perpendicular to (1, 0, 1) and so must have dot product with it equal to 0.

    That will give you one equation for x, y, and z. Since X is "on the line in part a" it must also satisfy the equation of that line.
     
  7. Oct 20, 2009 #6
    (x-4,y-1,z+1).(1,0,1)=0
    (2+t , 6 , 3+t)

    (x-4).1 + (y-1).0 + (z+1).1 = 0
    (2+t-4).1 + 6-1.0 + ((3+t) + 1).1 = 0
    (2+t-4) + (3+t)+1 = 0
    t= -1

    into original equation,
    2+(-1) , 6 , 3+(-1) = (1,6,2)
    OH SHI-

    XD =] :)
     
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