Question

If G and G’ be the centroids of the triangle ABC and A’B’C’ respectively, then $\overrightarrow{AA'}+\overrightarrow{BB'}+\overrightarrow{CC'}=$
(a) $\dfrac{2}{3}\overrightarrow{GG'}$
(b) $\overrightarrow{GG'}$
(c) $2\overrightarrow{GG'}$
(d) $3\overrightarrow{GG'}$

Answer 1

Write the given relative position vectors as the difference of two vectors by using the relation $\overrightarrow{XY}=\overrightarrow{Y}-\overrightarrow{X}$. Now, group the vectors representing the vertices of triangle ABC together and similarly group the vectors representing the vertices of triangle A’B’C’ together. Use the formula for the centroid of the triangle ABC in vector form as $\overrightarrow{G}=\dfrac{\overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C}}{3}$ and similarly for the centroid of the triangle A’B’C’ as $\overrightarrow{G'}=\dfrac{\overrightarrow{A'}+\overrightarrow{B'}+\overrightarrow{C'}}{3}$ to get the required relation.

Complete step by step solution:
Here we have been provided with two triangles ABC and A’B’C’ with their centroids as G and G’ respectively. We have been asked to find the value of $\overrightarrow{AA'}+\overrightarrow{BB'}+\overrightarrow{CC'}$ in terms of $\overrightarrow{GG'}$.
Now, we know that if we write a vector in the form $\overrightarrow{XY}$ it denotes the relative position of vector Y with respect to vector X. Head of this vector is towards point Y and the tail is towards point X. If we have to write them as absolute position vectors that means position vector relative to the origin then we have $\overrightarrow{XY}=\overrightarrow{Y}-\overrightarrow{X}$. Using this concept we can write the relative position vectors representing the vertices of the triangle as:
(i) $\overrightarrow{AA'}=\overrightarrow{A'}-\overrightarrow{A}$
(ii) $\overrightarrow{BB'}=\overrightarrow{B'}-\overrightarrow{B}$
(iii) $\overrightarrow{CC'}=\overrightarrow{C'}-\overrightarrow{C}$
Adding the above three relations we get,
$\Rightarrow \overrightarrow{AA'}+\overrightarrow{BB'}+\overrightarrow{CC'}=\overrightarrow{A'}-\overrightarrow{A}+\overrightarrow{B'}-\overrightarrow{B}+\overrightarrow{C'}-\overrightarrow{C}$
Grouping the vectors representing the vertices of triangle ABC together and similarly the vectors representing the vertices of triangle A’B’C’ together we get,
$\Rightarrow \overrightarrow{AA'}+\overrightarrow{BB'}+\overrightarrow{CC'}=\left( \overrightarrow{A'}+\overrightarrow{B'}+\overrightarrow{C'} \right)-\left( \overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C} \right)$
The centroid of a triangle in vector form is given as $\overrightarrow{G}=\dfrac{\overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C}}{3}$ so using this relation for the triangle ABC and similarly the relation $\overrightarrow{G'}=\dfrac{\overrightarrow{A'}+\overrightarrow{B'}+\overrightarrow{C'}}{3}$ for the triangle A’B’C’ we get,

& \Rightarrow \overrightarrow{AA'}+\overrightarrow{BB'}+\overrightarrow{CC'}=3\overrightarrow{G'}-3\overrightarrow{G} \\\ & \Rightarrow \overrightarrow{AA'}+\overrightarrow{BB'}+\overrightarrow{CC'}=3\left( \overrightarrow{G'}-\overrightarrow{G} \right) \\\ \end{aligned}$$ Converting $$\left( \overrightarrow{G'}-\overrightarrow{G} \right)$$ into relative position vector again we get, $$\therefore \overrightarrow{AA'}+\overrightarrow{BB'}+\overrightarrow{CC'}=3\overrightarrow{GG'}$$ **So, the correct answer is “Option (d)”.** **Note:** You may think from where this formula of centroid in vector form comes from. This relation is derived using the coordinate geometry in which we just place the unit vectors $\hat{i}$ and $$\hat{j}$$. We know that the coordinates of centroid is given as $x=\dfrac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}}{3}$ and $y=\dfrac{{{y}_{1}}+{{y}_{2}}+{{y}_{3}}}{3}$. If we will take $\hat{i}$ in x and similarly $$\hat{j}$$ in y and consider their sum, it will become $x\hat{i}+y\hat{j}=\dfrac{{{x}_{1}}\hat{i}+{{x}_{2}}\hat{i}+{{x}_{3}}\hat{i}}{3}+\dfrac{{{y}_{1}}\hat{j}+{{y}_{2}}\hat{j}+{{y}_{3}}\hat{j}}{3}$. This can be simplified as $x\hat{i}+y\hat{j}=\dfrac{\left( {{x}_{1}}\hat{i}+{{y}_{1}}\hat{j} \right)+\left( {{x}_{2}}\hat{i}+{{y}_{2}}\hat{j} \right)+\left( {{x}_{3}}\hat{i}+{{y}_{3}}\hat{j} \right)}{3}$. Considering G (x, y) as the centroid, $\left( {{x}_{1}},{{y}_{1}} \right),\left( {{x}_{2}},{{y}_{2}} \right),\left( {{x}_{3}},{{y}_{3}} \right)$ as the coordinates of vertices A, B, C respectively we get $\overrightarrow{G}=\dfrac{\overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C}}{3}$.