Let the three sides of a triangle ABC be given by the vectors 2i-j+k,i-3j-5k and 3i-4j-4k.

Question

Let the three sides of a triangle ABC be given by the vectors $2\hat{i}-\hat{j}+\hat{k},\hat{i}-3\hat{j}-5\hat{k}$ and $3\hat{i}-4\hat{j}-4\hat{k}$. Let $G$ be the centroid of the triangle $ABC$. Then $6(|\overset{	o \leftarrow }{AG}{|}^{2}+|\overset{	o \leftarrow }{BG}{|}^{2}+|\overset{	o \leftarrow }{CG}{|}^{2})$ is equal to \_\_\_\_\_\_\_\_\_\_.

Answer: 164

Accepted Answer

$\begin{gathered} A \ \vec{A}={\vec{a}}_{1}=\overset{	o }{0} \ \vec{B}={\vec{a}}_{1}+\vec{u}=\vec{u}=2\hat{i}-\hat{j}+\hat{k} \ \vec{C}={\vec{a}}_{1}+\vec{v}=\vec{v}=3\hat{i}-4\hat{j}-4\hat{k} \end{gathered}$

One solving

$\vec{A}=\overset{	o }{0},\vec{B}=2\hat{i}-\hat{j}+\hat{k}$ and $\vec{C}=3\hat{i}-4\hat{j}-4\hat{k}$, are the position vector of vertices $AB$ and $C$ respectively.

$\begin{gathered} \vec{G}=\frac{1}{3}(\vec{A}+\vec{B}+\vec{C})=\frac{1}{3}(\overset{	o }{0}+\vec{B}+\vec{C})=\frac{1}{3}(\vec{B}+\vec{C}) \ \Rightarrow \vec{G}=\frac{5}{3}\hat{i}-\frac{5}{3}\hat{j}-\hat{k} \ \overset{	o \leftarrow }{AG}=\vec{G}-\vec{A}=\vec{G} \ |\overset{	o \leftarrow }{AG}{|}^{2}={(\frac{5}{3})}^{2}+{(\frac{5}{3})}^{2}+(1{)}^{2}=\frac{25}{9}+\frac{25}{9}+1=\frac{50}{9}+1=\frac{59}{9} \ \overset{	o \leftarrow }{BC}=\vec{G}-\vec{B} \ \vec{B}=2\hat{i}-\hat{j}+\hat{k} \ |\overset{	o \leftarrow }{BG}{|}^{2}={(\frac{1}{3})}^{3}+{(\frac{2}{3})}^{2}+4=\frac{1}{9}+\frac{4}{9}+4=\frac{5}{9}+4=\frac{41}{9} \ \overset{	o \leftarrow }{CG}=\vec{G}-\vec{C} \ \vec{C}=3\hat{i}-4\hat{j}-4\hat{k} \ {\overset{	o \leftarrow }{CG}|}^{2}={(\frac{4}{3})}^{2}+{(\frac{7}{3})}^{2}+9=\frac{16}{9}+\frac{49}{9}+9=\frac{65}{9}+9=\frac{65}{9}+\frac{81}{9}=\frac{146}{9} \end{gathered}$

$6(|\overset{	o \leftarrow }{AG}{|}^{2}+|\overset{	o \leftarrow }{BG}{|}^{2}+|\overset{	o \leftarrow }{CG}{|}^{2})=6\cdot (\frac{59}{9}+\frac{41}{9}+\frac{146}{9})=6\cdot \frac{246}{9}=164$

Solutions · 1