Let the vectors overrightarrow{AB} = 2hat{i} | vedclass

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(B) Given vectors are $\overrightarrow{AB} = 2\hat{i} + 2\hat{j} + \hat{k}$ and $\overrightarrow{AC} = 2\hat{i} + 4\hat{j} + 4\hat{k}$.Let the positio

Vedclass · Accepted Answer

$38$

Vedclass · Answer

(B) Given vectors are $\overrightarrow{AB} = 2\hat{i} + 2\hat{j} + \hat{k}$ and $\overrightarrow{AC} = 2\hat{i} + 4\hat{j} + 4\hat{k}$.Let the position vector of $A$ be $\vec{0}$. Then the position vectors of $B$ and $C$ are $\vec{B} = 2\hat{i} + 2\hat{j} + \hat{k}$ and $\vec{C} = 2\hat{i} + 4\hat{j} + 4\hat{k}$.The centroid $G$ of $	riangle ABC$ has the position vector $\overrightarrow{AG} = \frac{\vec{A} + \vec{B} + \vec{C}}{3} = \frac{\vec{0} + (2\hat{i} + 2\hat{j} + \hat{k}) + (2\hat{i} + 4\hat{j} + 4\hat{k})}{3} = \frac{4\hat{i} + 6\hat{j} + 5\hat{k}}{3}$.Now, calculate the square of the magnitude $|\overrightarrow{AG}|^2 = \left(\frac{4}{3}ight)^2 + \left(\frac{6}{3}ight)^2 + \left(\frac{5}{3}ight)^2 = \frac{16 + 36 + 25}{9} = \frac{77}{9}$.Substitute this into the expression: $\frac{27}{7}(\overrightarrow{AG})^2 + 5 = \frac{27}{7} 	imes \frac{77}{9} + 5 = 3 	imes 11 + 5 = 33 + 5 = 38$.Thus, the correct option is $B$.

Let the vectors $\overrightarrow{AB} = 2\hat{i} + 2\hat{j} + \hat{k}$ and $\overrightarrow{AC} = 2\hat{i} + 4\hat{j} + 4\hat{k}$ be two sides of a triangle $ABC$. If $G$ is the centroid of $\triangle ABC$, then $\frac{27}{7}(\overrightarrow{AG})^2 + 5 =$

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