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if the vectors a=3i+j-2k,b=-i+3j+4k,c=4i-2j-6k costitiute sides of a triangle ABC then the length of median bisecting vector c is
Let CC1 be the median CC1 = |CC1| = |1/2 (CA + CB)||1/2 (a-c+b-c)| = |1/2(a+b-2c)||1/2(3i +j -2k-i+3j+4k-2(4i-2j-6k)|=|-3i+4j+7k| = sqrt(9+16+49)=> sqrt(74)Hope this helps
as usual, vikas’ answer is wrong.note that for any triangle ABC, AB + BC + CA= 0 where bold chars represent vectors.now, if here we let a= BC= 3i + j – 2k, b= AC= – CA= –i + 3j + 4k then c= 4i – 2j – 6k= BC+CA= – AB= BAso now median length bisecting c is= |b + (c/2)|= |i + 2j + k|= sqrt(1+4+1)= sqrt(6)kindly approve :=)
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