A triangle abc , d, e, f are points on bc, ca, ab respectively, such that bd : dc=1:2, ce:ea=1:2, and af: fb=1:2. Find the ratio of the areas of triangle def and abc.
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since a and b are unit vectors then
|a|=|b|=1
Let the angle between a and b be theta
(a-b)2=|a|2+|b|2-2|a||b| cos theta
a-b=(1+1-2 cos theta)1/2
a-b=(2(1-cos theta))1/2
a-b=(2(2 sin2theta/2))1/2
a-b/2=sin theta/2
hence proved
|a|=|b|=1
Let the angle between a and b be theta
(a-b)2=|a|2+|b|2-2|a||b| cos theta
a-b=(1+1-2 cos theta)1/2
a-b=(2(1-cos theta))1/2
a-b=(2(2 sin2theta/2))1/2
a-b/2=sin theta/2
hence proved
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