If A-B-C are the vertices of a triangle whose position vectors are a- b-c and G is the centroid of the -ABC- then GA-GB-GC is equal to

OA=a, OB=b, OC=c ∴ Centroid of triangle OG=a+b+c3 Now GA+GB+GC =(OA OG)+(OB OG)+(OC OG) =(a a+b+c3)+(b a+b+c3)+(c a+b+c3) =13(3a a b c+3b a b c+3c a b c) =1 ...

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Standard VIII

Mathematics

The Centroid

If A,B,C are ...

Question

If $A, B, C$ are the vertices of a triangle whose position vectors are $\to a, \to b, \to c$ and $G$ is the centroid of the $Δ A B C,$ then $- - \to G A + - - \to G B + - - \to G C$ is equal to

\to 0

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\to a + \to b + \to c

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\frac{\to a + \to b + \to c}{3}

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\frac{\to a - \to b - \to c}{3}

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Solution

The correct option is A $\to 0$
$- - \to O A = \to a, - - \to O B = \to b, - - \to O C = \to c$
$∴$ Centroid of triangle
$- - \to O G = \frac{\to a + \to b + \to c}{3}$
Now $- - \to G A + - - \to G B + - - \to G C = (- - \to O A - - - \to O G) + (- - \to O B - - - \to O G) + (- - \to O C - - - \to O G) = ⎛ ⎝ \to a - \frac{\to a + \to b + \to c}{3} ⎞ ⎠ + ⎛ ⎝ \to b - \frac{\to a + \to b + \to c}{3} ⎞ ⎠ + ⎛ ⎝ \to c - \frac{\to a + \to b + \to c}{3} ⎞ ⎠ = \frac{1}{3} (3 \to a - \to a - \to b - \to c + 3 \to b - \to a - \to b - \to c + 3 \to c - \to a - \to b - \to c) = \frac{1}{3} (\to 0) = \to 0$

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The Centroid

Standard VIII Mathematics

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If A,B,C are the vertices of a triangle whose position vectors are →a, →b,→c and G is the centroid of the ΔABC, then −−→GA+−−→GB+−−→GC is equal to

If $A, B, C$ are the vertices of a triangle whose position vectors are $\to a, \to b, \to c$ and $G$ is the centroid of the $Δ A B C,$ then $- - \to G A + - - \to G B + - - \to G C$ is equal to