

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(b−a).
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(b−a) ) · (b−a) = 0.Simplify and rearrange to solve for k. (Your "vector algebra" is being tested.)



Hi,
This type of question may be back in vogue, now that Cambridge may have exhausted its list of numericallybased 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 numericallybased problems.
Thanks.
Cheers,
Wen ShihEdited by wee_ws 12 Aug `10, 8:46AM



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(b−a).
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(b−a) ) · (b−a) = 0.Simplify and rearrange 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
