Let the three sides of a triangle ABC be repr | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let the position vectors of vertices $A, B, C$ be $\vec{a}, \vec{b}, \vec{c}$ respectively. Without loss of generality,let $\vec{a} = \vec{0}$.Giv

Vedclass · Accepted Answer

$164$

Vedclass · Answer

(A) Let the position vectors of vertices $A, B, C$ be $\vec{a}, \vec{b}, \vec{c}$ respectively. Without loss of generality,let $\vec{a} = \vec{0}$.Given $\vec{AB} = \vec{b} - \vec{a} = 2\hat{i}-\hat{j}+\hat{k}$,so $\vec{b} = 2\hat{i}-\hat{j}+\hat{k}$.Given $\vec{CA} = \vec{a} - \vec{c} = \hat{i}-3\hat{j}-5\hat{k}$,so $\vec{c} = -(\hat{i}-3\hat{j}-5\hat{k}) = -\hat{i}+3\hat{j}+5\hat{k}$.The centroid $G$ has position vector $\vec{g} = \frac{\vec{a}+\vec{b}+\vec{c}}{3} = \frac{\vec{0} + (2\hat{i}-\hat{j}+\hat{k}) + (-\hat{i}+3\hat{j}+5\hat{k})}{3} = \frac{\hat{i}+2\hat{j}+6\hat{k}}{3}$.Now,$\overrightarrow{AG} = \vec{g} - \vec{a} = \frac{1}{3}(\hat{i}+2\hat{j}+6\hat{k})$. Thus,$|\overrightarrow{AG}|^2 = \frac{1}{9}(1^2+2^2+6^2) = \frac{41}{9}$.$\overrightarrow{BG} = \vec{g} - \vec{b} = (\frac{1}{3}-2)\hat{i} + (\frac{2}{3}+1)\hat{j} + (2-1)\hat{k} = -\frac{5}{3}\hat{i} + \frac{5}{3}\hat{j} + \hat{k}$. Thus,$|\overrightarrow{BG}|^2 = \frac{25}{9} + \frac{25}{9} + 1 = \frac{59}{9}$.$\overrightarrow{CG} = \vec{g} - \vec{c} = (\frac{1}{3}+1)\hat{i} + (\frac{2}{3}-3)\hat{j} + (2-5)\hat{k} = \frac{4}{3}\hat{i} - \frac{7}{3}\hat{j} - 3\hat{k}$. Thus,$|\overrightarrow{CG}|^2 = \frac{16}{9} + \frac{49}{9} + 9 = \frac{65+81}{9} = \frac{146}{9}$.Finally,$6(|\overrightarrow{AG}|^2+|\overrightarrow{BG}|^2+|\overrightarrow{CG}|^2) = 6(\frac{41+59+146}{9}) = 6(\frac{246}{9}) = 6 	imes \frac{82}{3} = 2 	imes 82 = 164$.

Similar Questions

$L$ and $M$ are two points with position vectors $2 \vec{a}-\vec{b}$ and $\vec{a}+2 \vec{b}$ respectively. The position vector of the point $N$ which divides the line segment $LM$ in the ratio $2:1$ externally is

If $D, E, F$ are respectively the midpoints of $AB, AC$ and $BC$ in $\Delta ABC$,then $\overrightarrow{BE} + \overrightarrow{AF} = $

If $A(1, 0, 0)$,$B(0, 1, 0)$,and $C(0, 0, 1)$ are given,and $\vec{AB} = \vec{CX}$,then the point $X$ is:

$P$ is a point on the side $BC$ of the $\Delta ABC$ and $Q$ is a point such that $\overrightarrow{PQ}$ is the resultant of $\overrightarrow{AP}, \overrightarrow{PB}, \overrightarrow{PC}$. Then $ABQC$ is a

The ratio in which $\hat{i}+2 \hat{j}+3 \hat{k}$ divides the join of $-2 \hat{i}+3 \hat{j}+5 \hat{k}$ and $7 \hat{i}-\hat{k}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module