Question

In a $△ABC$, $P.V^{\prime }.s$ of midpoints of $AB,AC$ are $\overline{i}-\overline{j}+\overline{k}$, $\overline{i}+2\overline{k}$. The $P.V.$ of centroid of $△ABC$ is $2\overline{i}+3\overline{j}+4\overline{k}$. $P.V.$ of $A$ is

• A. $\overrightarrow{a}=\left(-2,-11,-6\right)$
• B. $\overrightarrow{a}=\left(-2,-1,+6\right)$
• C. $\overrightarrow{a}=\left(-8,-11,-2\right)$
• D. $\overrightarrow{a}=\left(-5,-13,-6\right)$

Answer 1

The correct option is A $\overrightarrow{a}=\left(-2,-11,-6\right)$

We have,

$\frac{\overrightarrow{a}+\overrightarrow{b}}{2}=\left(1,-1,1\right)\frac{\overrightarrow{a}+\overrightarrow{c}}{2}=\left(1,0,2\right)$

$\overrightarrow{a}+\overrightarrow{b}=\left(2,-2,2\right)\overrightarrow{a}+\overrightarrow{c}=\left(2,0,4\right)$

$\frac{\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}}{3}=\left(2,3,4\right)2\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\left(4,-2,6\right)$

$\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\left(6,9,12\right)$

$\overrightarrow{a}+\left(6,9,12\right)=\left(4,-2,6\right)$

$\overrightarrow{a}=\left(-2,-11,-6\right)$

Then,

Option $A$ is correct answer.