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If bar(DA) =bar(a); bar(AB)=bar(b) and b...

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=

Text Solution

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`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`
...
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