If $A, B, C$ are the vertices of a triangle whose position vectors are $a, b, c$ and $G$ is the centroid of the $Δ A B C$ , then $¯ ¯¯¯¯¯¯ ¯ G A + ¯ ¯¯¯¯¯¯ ¯ G B + ¯ ¯¯¯¯¯¯ ¯ G C$ equals:

Open in App

Solution

Let us given that,

If G is the centroid of a triangle $A B C$ and let $O$ be the position vector.

Then, let

$O A = \to a$

$O B = \to b$

$O C = \to c$

$O G = \to g$

We know that, using centroid formula

$- - \to O G = \frac{- - \to O A - -- \to + O B - -- \to + O C}{3}$

$\to g = \frac{\to a + \to b + \to c}{3}$

$3 \to g = \to a + \to b + \to c$

$\to a + \to b + \to c = 3 \to g . . . . . . (1)$

Now,

According to given question,

$- - \to G A + - - \to G B + - - \to G C$

$= (- - \to O A - - - \to O G) + (- - \to O B - - - \to O G) + (- - \to O C - - - \to O G)$

$= - - \to O A + - - \to O B + - - \to O C - 3 - - \to O G$

$\to a + \to b + \to c - 3 \to g$ by equation (1)

$= 3 \to g - 3 \to g = 0$

Hence, it is complete solution.

Suggest Corrections

Similar questions

A, B, C

\to a, \to b, \to c

G

Δ A B C,

- - \to G A + - - \to G B + - - \to G C

Δ

A B C

A, B, C

a, b, c

G

A B C

G A^{2} + G B^{2} + G C^{2} = ?

\vec{a}, \vec{b}, \vec{c}

\vec{a} + \vec{b} + \vec{c}

G

△ A B C,

{G A}^{2} + {G B}^{2} + {G C}^{2}

\vec{G A} + \vec{G B} + \vec{G C} .

Join BYJU'S Learning Program