If D,E,F are respectively the mid points of AB,AC and BC in ΔABC, then →BE+→AF is equal to.
A
→DC
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B
12→BF
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C
2→BF
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D
32→BF
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Solution
The correct option is C→DC Let O be the origin. ∴→BE+→AF=→OE−→OB+→OF−→OA =→OA+→OC2−→OB+→OB+→OC2−→OA =→OC2+→OC2+→OA2−→OA+→OB2−→OB =→OC−→OA+→OB2 =→OC−→OD=→DC