-->

Vector Algebra

Question
CBSEENMA12034049

Show that the line joining one vertex of a parallelogram to the mid-point of an opposite side trisects the diagonal and is trisected there at.

Solution
Let OABC be a parallelogram Table O as origin. Let straight a with rightwards arrow on top and straight b with rightwards arrow on top be position vectors of straight A with rightwards arrow on top space and space straight C with rightwards arrow on top such that OA with rightwards arrow on top space equals space straight a with rightwards arrow on top comma space OC with rightwards arrow on top space equals space straight b with rightwards arrow on top.
     Now,  OB with rightwards arrow on top space equals space OA with rightwards arrow on top space plus space AB with rightwards arrow on top space equals space OA with rightwards arrow on top space plus space OC with rightwards arrow on top
              therefore space space space space OB with rightwards arrow on top space equals straight a with rightwards arrow on top space plus space straight b with rightwards arrow on top
therefore space space space space space position space vector space of space straight B space is space open parentheses straight a with rightwards arrow on top space plus space straight b with rightwards arrow on top close parentheses.
Position vector of mid-point of D of A is fraction numerator straight a with rightwards arrow on top over denominator 2 end fraction

P.V. of a point dividing CD in the ratio
                 2 : 1 is fraction numerator 2 space open parentheses begin display style fraction numerator straight a with rightwards arrow on top over denominator 2 end fraction end style close parentheses space plus space left parenthesis 1 right parenthesis thin space left parenthesis straight b with rightwards arrow on top right parenthesis over denominator 2 plus 1 end fraction space space straight i. straight e. space space fraction numerator straight a with rightwards arrow on top space plus space straight b with rightwards arrow on top over denominator 3 end fraction
Again position vector of point divides OB in the ratio 1: 2 is
                          fraction numerator left parenthesis 1 right parenthesis thin space left parenthesis straight a with rightwards arrow on top plus straight b with rightwards arrow on top right parenthesis space plus space 2 left parenthesis stack 0 right parenthesis with rightwards arrow on top over denominator 1 plus 2 end fraction space straight i. straight e. space fraction numerator straight a with rightwards arrow on top plus straight b with rightwards arrow on top over denominator 3 end fraction
∴  position vectors of points trisecting CD and OB are same.
∴  CD trisects OB and CD is trisected there at.