If D,E,F are respectively the mid points of −−→AB,−−→AC and −−→BC in ΔABC, then −−→BE+−−→AF=
A
−−→DC
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B
12−−→BF
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C
2−−→BF
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D
3−−→BF2
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Solution
The correct option is A−−→DC Let us represent A,B and C with vectors →a,→b and →c respectively.
Here vectors representing the mid points D,E and F are →a+→b2,→a+→c2 and →b+→c2 respectively. −−→BE=→a−2→b+→c2 −−→AF=→b−2→a+→c2 −−→CD=→a+→b−2→c2 −−→BE+−−→AF=2→c−→a−→b2=−−−→CD=−−→DC