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A level Maths - Vectors

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  • Audi's Avatar
    196 posts since Jun '09
    • Relative to an origin O, the position vectors of points A and B are a and b respectively. Givent that angle AOB is 90deg, show that the position vector of the foot of the perpendicular from O to AB is

      a  +  (|a|^2 / (|a|^2 + |b|^2))  (b-a)

      Need help for this question. Thanks!

  • Mad Hat's Avatar
    91 posts since May '10
    • Let F be the foot in question. Notice that OF = OA + k(AB), where k is an unknown fraction.

      In other words, OF = a + k(ba).

      To answer the question, we simply need to find k, so let's do it.

      Since OF is perpendicular to AB, it follows that OF · AB = 0.
      So ( a + k(ba) ) · (ba) = 0.

      Simplify and re-arrange to solve for k. (Your "vector algebra" is being tested.)

  • SBS261P's Avatar
    181 posts since Dec '07
    • ^beautifully done.

      alternatively A level students are also taught the ratio theorem. you can try applying ratio theorem here.

  • Moderator
    wee_ws's Avatar
    1,521 posts since Mar '08
    • Hi,

        This type of question may be back in vogue, now that Cambridge may have exhausted its list of numerically-based questions that can be asked.

        Here is one question that involves vector product algebra as well as trigonometry:

        With respect to the origin O, point P has position vector given by p = −2i + j + k and the variable point Q has position vector given by q = (cos t)i + 2j , where 0 <= t <= pi. Show that
      |p x q| = sqrt [ 2(cos t  + 2)^2 + 12 ]. Hence find the smallest exact area of triangle OPQ.

        The approach to deal with questions involving variables is to use the USUAL manner of solving numerically-based problems.

        Thanks.

      Cheers,
      Wen Shih

      Edited by wee_ws 12 Aug `10, 8:46AM
  • Moderator
    eagle's Avatar
    23,320 posts since Aug '01
    • Originally posted by Mad Hat:

      Let F be the foot in question. Notice that OF = OA + k(AB), where k is an unknown fraction.

      In other words, OF = a + k(ba).

      To answer the question, we simply need to find k, so let's do it.

      Since OF is perpendicular to AB, it follows that OF · AB = 0.
      So ( a + k(ba) ) · (ba) = 0.

      Simplify and re-arrange to solve for k. (Your "vector algebra" is being tested.)

      Alternatively, we can make use of vector projection to solve.

      OF = OA + AF = a + |a| cos θ * (b - a) / |b - a|, where θ = angle between OA and BA

      = a + |a| [ {a.(a - b)} / {|a| |a - b| }] * (b - a) / |b - a|

      = a +  {a.a - a.b} / (|a|^2 + |b|^2) * (b - a)

      = a +  |a|^2 / (|a|^2 + |b|^2) * (b - a) since a.a = |a|^2 and a.b = 0

       

      Thanks :)

      Edited by eagle 12 Aug `10, 9:22AM
  • Moderator
    wee_ws's Avatar
    1,521 posts since Mar '08
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