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Find perpendicular vectors and angles

  1. Nov 28, 2008 #1
    1. The problem statement, all variables and given/known data
    I'm awful with vectors. I know the dot product rule and the uses to find perpendicular vectors and angles. I know about multiplying vectors (a1b1 + a2b2 + a3b3). I know vector arithmetic.

    But i cant do:

    r = 8i + 12j + 14k + t(i + j - k)

    where t is a parameter

    The point A has co-ordinates (4,8,a), where a is a constant. The point B has co-ordinates (b,13,13), where b is a constant. Points A and B lie on the line l1.

    a) Find the values a and b



    2. Relevant equations



    3. The attempt at a solution

    Well i presume i need to find the parameter t first... how? I dont have enough information do i?

    thanks :)
     
  2. jcsd
  3. Nov 28, 2008 #2

    Doc Al

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    Staff: Mentor

    Can't do what?

    What line? And what does this have to do with r above?
     
  4. Nov 28, 2008 #3
    sorry the question is

    The line l1 has vector equation:

    r = 8i + 12j + 14k + t(i + j - k)

    where t is a parameter

    The point A has co-ordinates (4,8,a), where a is a constant. The point B has co-ordinates (b,13,13), where b is a constant. Points A and B lie on the line l1.

    a) Find the values a and b

    Thanks :)
     
  5. Nov 28, 2008 #4

    Doc Al

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    Staff: Mentor

    My bad. (I realized what you meant shortly after I posted.)

    Think of it this way: All points on that line must satisfy the parametric equations:
    x = 8 + t
    y = 12 + t
    z = 14 - t
    (where t is just a parameter that takes on all real values.)

    Since points A and B are on that line, they must satisfy those three equations. See what you can deduce from that. Hint: Start by finding the value of the parameter t for each point. (You have all the information needed.)
     
  6. Nov 29, 2008 #5
    4 = 8 + t
    8 = 12 + t
    a = 14 + t

    b = 8 + t
    13 = 12 + t
    13 = 14 + t


    point a t = -4
    point b t = 9

    a = 18, b = 9

    Okay cheeers :)

    Next question:

    Given that point O is the origin, and that the point P lies on l1 such that OP is perpendicular to l1

    b) Find the coordinates of P

    i've sketched a graph of it. using the dot product rule, a.b =0 (0 as it's cos(90))

    so does that mean vector a x vector b = 0?

    Where exactly now?

    Thanks :)
     
  7. Nov 30, 2008 #6

    Doc Al

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    Staff: Mentor

    Sorry for the delay--I forgot that you had a second question.

    No. The cross product will not be zero, only the dot product.

    Call the point P (x, y, z). Thus the vector OP is (x, y, z). That point must satisfy two conditions:
    (1) The vector OP must be perpendicular to the line, thus its dot product with a vector parallel to the line must be zero. (Find a vector parallel to the line--any one will do.)
    (2) The point P must be on line, thus satisfy the same parametric equations from before. (Solve for t.)
     
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