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Vector Algebra

Question
CBSEENMA12033957

If D and E are the mid-points of sides AB and AC of a triangle ABC respectively show that
BE with rightwards arrow on top space plus space DC with rightwards arrow on top space equals space 3 over 2 BC with rightwards arrow on top.

Solution

Take A is origin. Let straight b with rightwards arrow on top comma space straight c with rightwards arrow on top be position vectors of B, C respectively so that
AD with rightwards arrow on top space equals space 1 half space straight b with rightwards arrow on top comma space space space AE with rightwards arrow on top space equals 1 half straight c with rightwards arrow on top
L.H.S.  = BE with rightwards arrow on top space plus space DC with rightwards arrow on top  
                                  = (P.V. Of E - P.V. of B) + (P.V. of C - P.V. of D)

                      equals space open parentheses 1 half straight c with rightwards arrow on top space minus straight b with rightwards arrow on top close parentheses space plus space open parentheses straight c with rightwards arrow on top minus 1 half straight b with rightwards arrow on top close parentheses space equals space 3 over 2 straight c with rightwards arrow on top space minus space 3 over 2 straight b with rightwards arrow on top
equals space 3 over 2 left parenthesis straight c with rightwards arrow on top space minus space straight b with rightwards arrow on top right parenthesis space equals space 3 over 2 BC with rightwards arrow on top space equals space straight R. straight H. straight S.