Question

The position vector of the vertices of a triangle $ABC$ are $\hat{i},\hat{j},\hat{k}$ then the position vector of its orthocentre is

• A. $\hat{i}+\hat{j}+\hat{k}$
• B. $2(\hat{i}+\hat{j}+\hat{k})$
• C. $\frac{1}{3}(\hat{i}+\hat{j}+\hat{k})$
• D. $\frac{1}{\sqrt{3}}(\hat{i}+\hat{j}+\hat{k})$

Answer 1

The correct option is C $\frac{1}{3}(\hat{i}+\hat{j}+\hat{k})$

We have, the position vectors of the vertices of a triangle $ABC$are $\hat{i},\hat{j},\hat{k}$.

Let $O$ be the fixed point.

$\hat{i}$ = position vector of $A=\overrightarrow{OA}$

$\hat{j}$ = position vector of $B=\overrightarrow{OB}$

$\hat{k}$ = position vector of $C=\overrightarrow{OC}$

Let $AD$ be the median ofthe triangle $ABC$.

Since, $D$ is the mis point of $BC$.

Position vector of $D=$ vector $OD=\frac{(\overrightarrow{OB}+\overrightarrow{OC})}{2}$

Now, position vector of $G$,

$=\frac{2\left(\overrightarrow{OD}\right)+1\left(\overrightarrow{OA}\right)}{3}$

$=\frac{2\left(\frac{\overrightarrow{OB}+\overrightarrow{OC}}{2}\right)+\overrightarrow{OA}}{3}$

$=\frac{\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}}{3}$

So,

$=\frac{\hat{i}+\hat{j}+\hat{k}}{3}$

Hence, the position vector of the orthocentre is $\frac{1}{3}(\hat{i}+\hat{j}+\hat{k})$.