If A , B and C are the vertices of a triangle whose position vectors are veca, vecb and vecc respectively and G is the centroid of the Δ ABC, then GA + GB + GC is

Question

If A , B and C are the vertices of a triangle whose position vectors are veca, vecb and vecc respectively and G is the centroid of the Δ ABC, then GA + GB + GC is (A)  vec0 (B)  veca| vecb| vecc (C) ( veca+ vecb+ vecc/3) (D) ( veca- vecb- vecc/3)

Tardigrade Admin · Accepted Answer

Let A, B, C are the vertices of a Δ whose position vectors are veca, vecb and vecc respectivety. Let G be the centroid of Δ ABC ∴ Centroid of triangle ( G )=( veca| vecb| vecc/3) Consider, GA + GB + GC =( veca-( veca+ vecb+ vecc/3))+( vecb-( veca+ vecb+ vecc/3))+( vecc-( veca+ vecb+ vecc/3)) =(1/3)[2 veca- vecb- vecc+2 vecb- veca- vecc+2 vecc- veca- vecb]=0

Q. If $A, B$ and $C$ are the vertices of a triangle whose position vectors are $a, b$ and $c$ respectively and $G$ is the centroid of the $Δ A BC$ , then $G A + GB + GC$ is

$0$

$a ∣ b ∣ c$

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

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

Q. If A,B and C are the vertices of a triangle whose position vectors are a,b and c respectively and G is the centroid of the ΔABC, then GA+GB+GC is

0

a∣b∣c

3a+b+c​

3a−b−c​

Let A,B,C are the vertices of a Δ whose position vectors are a,b and c respectivety. Let G be the centroid of ΔABC ∴ Centroid of triangle (G)=3a∣b∣c​ Consider, GA+GB+GC =(a−3a+b+c​)+(b−3a+b+c​)+(c−3a+b+c​) =31​[2a−b−c+2b−a−c+2c−a−b]=0

Q. If $A, B$ and $C$ are the vertices of a triangle whose position vectors are $a, b$ and $c$ respectively and $G$ is the centroid of the $Δ A BC$ , then $G A + GB + GC$ is

$0$

$a ∣ b ∣ c$

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

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