Let overrightarrow{OA}=2 hat{i}-3 hat{j}+hat{ | vedclass

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Question

(A) The position vector of the centroid $G$ is given by $\overrightarrow{OG} = \frac{\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}}{3}$.

Vedclass · Accepted Answer

$162$

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(A) The position vector of the centroid $G$ is given by $\overrightarrow{OG} = \frac{\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}}{3}$.Substituting the given vectors:$\overrightarrow{OG} = \frac{(2\hat{i}-3\hat{j}+\hat{k}) + (\hat{i}-4\hat{j}-3\hat{k}) + (-3\hat{i}+\hat{j}+2\hat{k})}{3} = \frac{0\hat{i}-6\hat{j}+0\hat{k}}{3} = -2\hat{j}$.Thus,$|\overrightarrow{OG}|^2 = (-2)^2 = 4$.Now,calculate the vectors for the sides:$\overrightarrow{BC} = \overrightarrow{OC} - \overrightarrow{OB} = (-3-1)\hat{i} + (1-(-4))\hat{j} + (2-(-3))\hat{k} = -4\hat{i} + 5\hat{j} + 5\hat{k}$.$BC^2 = |\overrightarrow{BC}|^2 = (-4)^2 + 5^2 + 5^2 = 16 + 25 + 25 = 66$.$\overrightarrow{CA} = \overrightarrow{OA} - \overrightarrow{OC} = (2-(-3))\hat{i} + (-3-1)\hat{j} + (1-2)\hat{k} = 5\hat{i} - 4\hat{j} - \hat{k}$.$CA^2 = |\overrightarrow{CA}|^2 = 5^2 + (-4)^2 + (-1)^2 = 25 + 16 + 1 = 42$.$\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = (1-2)\hat{i} + (-4-(-3))\hat{j} + (-3-1)\hat{k} = -\hat{i} - \hat{j} - 4\hat{k}$.$AB^2 = |\overrightarrow{AB}|^2 = (-1)^2 + (-1)^2 + (-4)^2 = 1 + 1 + 16 = 18$.Finally,calculate the expression:$BC^2 + CA^2 + AB^2 + 9(OG)^2 = 66 + 42 + 18 + 9(4) = 126 + 36 = 162$.

List $I$	List $II$
$P$. Volume of parallelepiped determined by vectors $\vec{a}, \vec{b}$ and $\vec{c}$ is $2$. Then the volume of the parallelepiped determined by vectors $2(\vec{a} \times \vec{b}), 3(\vec{b} \times \vec{c})$ and $(\vec{c} \times \vec{a})$ is	$1$. $100$
$Q$. Volume of parallelepiped determined by vectors $\vec{a}, \vec{b}$ and $\vec{c}$ is $5$. Then the volume of the parallelepiped determined by vectors $3(\vec{a}+\vec{b}), (\vec{b}+\vec{c})$ and $2(\vec{c}+\vec{a})$ is	$2$. $30$
$R$. Area of a triangle with adjacent sides determined by vectors $\vec{a}$ and $\vec{b}$ is $20$. Then the area of the triangle with adjacent sides determined by vectors $(2\vec{a}+3\vec{b})$ and $(\vec{a}-\vec{b})$ is	$3$. $24$
$S$. Area of a parallelogram with adjacent sides determined by vectors $\vec{a}$ and $\vec{b}$ is $30$. Then the area of the parallelogram with adjacent sides determined by vectors $(\vec{a}+\vec{b})$ and $\vec{a}$ is	$4$. $60$

Let $\overrightarrow{OA}=2 \hat{i}-3 \hat{j}+\hat{k}$,$\overrightarrow{OB}=\hat{i}-4 \hat{j}-3 \hat{k}$,and $\overrightarrow{OC}=-3 \hat{i}+\hat{j}+2 \hat{k}$ be the position vectors of three points $A$,$B$,and $C$ respectively. If $G$ is the centroid of triangle $ABC$,then $BC^2+CA^2+AB^2+9(OG)^2=$

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