sry mate, i havent got a scanner... but ill try to draw you one.
if A is a point (1,1) and B is a point (1,3), and C is point (2,2), if the midpoint of AB is M... prove that AB is perpendicular to CM.
Ok... so the idea is to prove that somethings perpendicular... and we know that the easiest way to do that is to equate the scalar dot product to 0. (because we know that a.b=|a||b|cos(theta)... and cos(theta) = 0 when perpendicular, and the magnitudes wont be 0.
So... we can establish that its asking us... CM.AB = 0.
So first we find AB... which is AO + OB (easier way to do this is OB-OA, because we know that AO = -OA)
So OB-OA = 1i + 3j - (1i + 1j) = 2j = AB
Now we need to find CM. CM is a little harder because we need to use addition of vectors. we know that AM is .5AB, because its the midpoint... and we know that CA + AM will equal CM. So.. we find CA... (OA-OC), then we find AM (.5AB) and add em together... this will give us
(1i+1j) - (2i + 2j)= -1i - 1j = CA
.5( 2j) = 1j = AM
CA+AM = -1i = CM.
okay... now CM.AB is going to be (-1i).(2j) ... and because theyre in differetn directions they equal zero, as a general rule for scalar products (ie... i.j = 0, while i.i and j.j would equal 1)
therefore, as CM.AB = -i.2j = 0, CM and AB are perpendicular.
Hope that helps. If any bit needs clarifying just ask... im not very good at explaining, generally.