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Vector Algebra

Question
CBSEENMA12033974

Show that the sum of three vector determined by the medians of a triangle directed from the vertices is zero.

Solution
Let D. E. F be the mid-points of the sides BC, CA and AB respectively of ∆ ABC
Now,          AD with rightwards arrow on top space equals space AB with rightwards arrow on top space plus space BD with rightwards arrow on top                                    [Triangle Law of Vectors]
therefore space space space space AD with rightwards arrow on top space equals space AB with rightwards arrow on top space plus space 1 half BC with rightwards arrow on top                          ...(1)
Again,  BE with rightwards arrow on top space equals space BC with rightwards arrow on top space plus space CE with rightwards arrow on top
therefore space space space space space BE with rightwards arrow on top space plus space BC with rightwards arrow on top space plus space 1 half CA with rightwards arrow on top                         ...(2)
Also,    CF with rightwards arrow on top space equals CA with rightwards arrow on top space plus space AF with rightwards arrow on top
therefore space space space space CF with rightwards arrow on top space equals space CA with rightwards arrow on top space plus space 1 half AB with rightwards arrow on top                           ...(3)

Adding (1), (2) and (3),
AD with rightwards arrow on top space plus space BE with rightwards arrow on top space plus space CF with rightwards arrow on top space equals space AB with rightwards arrow on top space plus space 1 half BC with rightwards arrow on top space plus space BC with rightwards arrow on top space plus space 1 half CA with rightwards arrow on top space plus space CA with rightwards arrow on top space plus space 1 half AB with rightwards arrow on top
                                         equals space 3 over 2 open square brackets AB with rightwards arrow on top space plus space BC with rightwards arrow on top space plus CA with rightwards arrow on top close square brackets
equals space 3 over 2 open square brackets AC with rightwards arrow on top space plus space CA with rightwards arrow on top close square brackets space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space open square brackets because space space space AB with rightwards arrow on top plus BC with rightwards arrow on top space equals space AC with rightwards arrow on top close square brackets
equals space 3 over 2 open square brackets stack AC space with rightwards arrow on top space minus space AC with rightwards arrow on top close square brackets space equals space 3 over 2 open square brackets 0 with rightwards arrow on top close square brackets space equals space 0 with rightwards arrow on top
therefore space space space AD with rightwards arrow on top space plus space BE with rightwards arrow on top space plus space CF with rightwards arrow on top space equals space 0 with rightwards arrow on top
Hence the result.