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

The number of unit vectors of the form $a \hat{i} + b \hat{j} + c \hat{k}$,where $a, b, c \in W$ is

The vector in the direction of the sum of the vectors $\vec{a}=2 \hat{i}-2 \hat{j}+5 \hat{k}$ and $\vec{b}=-2 \hat{i}+5 \hat{j}-3 \hat{k}$ is

The value of $\hat{k} \cdot (\hat{i} \times \hat{j}) + \hat{i} \cdot (\hat{j} \times \hat{k})$ is . . . . . . .

If $\bar{a} = (x + 2y - 3) \bar{i} + (2x - y + 3) \bar{j}$ and $\bar{b} = (3x - 2y) \bar{i} + (x - y + 1) \bar{j}$ are two vectors such that $\bar{a} = 2 \bar{b}$, then $y - 5x =$

$ABCDEF$ is a regular hexagon whose centre is $O$. Then, $\vec{AB} + \vec{AC} + \vec{AD} + \vec{AE} + \vec{AF}$ is equal to (in $\vec{AO}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module