Question

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 $△ABC$, then $GA+GB+GC$ is equal to

• A. $0$
• B. $a+b+c$
• C. $\frac{a+b+c}{3}$
• D. $\frac{a-b-c}{3}$

Answer 1

Answer: (A)

$0$

A.B, C are vertices of triangle

Position vectors of A,B,C be a,b,c

G is the centroid of $=△ABC$

$\therefore G=\frac{a+b+c}{3}$

$GA=a-\left(\frac{a+b+c}{3}\right),GB=b-\left(\frac{a+b+c}{3}\right)$

$GC=c-\left(\frac{a+b+c}{3}\right)$

$\therefore GA+GB+GC=(a+b+c)-3\left(\frac{a+b+c}{3}\right)$

$=0$