LinAlg4.1.11

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In a triangle ABC if A=2\bar{i}+4\bar{j}-\bar{k},B=4\bar{i}+5\bar{j}+\bar{k},C=3\bar{i}+6\bar{j}-3\bar{k}\, and D is the mid point of the side BC, then find the length of AD.

Given D is the mid point of BC.

Therefore,Pposition vector of D=\frac{1}{2}(4\bar{i}+5\bar{j}+\bar{k}+3\bar{i}+6\bar{j}-3\bar{k})=\frac{1}{2}(7\bar{i}+11\bar{j}-2\bar{k})\,

Therefore,\bar{AD}=[\frac{7}{2}\bar{i}+\frac{11}{2}\bar{j}-\bar{k}]-[2\bar{i}-4\bar{j}-\bar{k}]=\frac{3}{2}\bar{i}+\frac{3}{2}\bar{j}\,

Hence |AD|=\frac{3}{2}\sqrt{1+1}=\frac{3}{\sqrt{2}}\,


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