The points D,E,F divide −−→BC,−−→CA and −−→AB of the triangle ABC in the ratio 1:4,3:2 and 3:7 respectively and the point K divides −−→AB in the ratio 1:3, then (−−→AD+−−→BE+−−→CF):−−→CK is equal to
A
1:1
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B
2:5
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C
5:2
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D
3:5
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Solution
The correct option is B2:5 Let →a,→b,→c be the position vectors of vertices →A,→B,→C of the △ABC. Then the position vectors of →D,→E,→F are respectively ⎛⎝4→b+→c5⎞⎠,(3→a+2→c5),⎛⎝7→a+3→b10⎞⎠ ∴−−→AD=⎛⎝4→b+→c5⎞⎠−→a −−→BE=(3→a+2→c5)−→b−−→CF=⎛⎝7→a+3→b10⎞⎠−→c
Now −−→AD+−−→BE+−−→CF=8→b+2→c−10→a+6→a+4→c−10→b+7→a+3→b−10→c10 =3→a+→b−4→c10 ∴−→R1=3→a+→b−4→c10 ⇒10−→R1=3→a+→b−4→c
The position vector of the point K is →b+3→a4 ∴−−→CK=→b+3→a4−→c=→b+3→a−4→c4 ∴−→R2=→b+3→a−4→c4 ∴4−→R2=→b+3→a−4→c ∴10−→R1=4−→R2 ⇒5−→R1=2−→R2