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If `bar(DA) =bar(a); bar(AB)=bar(b)` and `bar(CB)=kbar(a)` where `k>0` and `X,Y` are the midpoint of `DB` and `AC` respectively such that `|bar(a)|=17` and `|bar(XY)|=4`, then k= |
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Answer» `X=(veca+vecb)/2` `Y=(veca+vecb+veca(1-k))/2` `Y=veca+vecb/a-(vecaK)/2` `|XY|=4` `XY=veca+vecb/2-(vecaK)/2-(veca+vecb)/2` `=veca/2-(vecak)/2` `=veca/2(1-k)` `|XY|=4=(|veca|(1-k))/2` `4=(17(1-k))/2` `1-k=8/17` `k=9/17`. |
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