Question

The position vectors of $A,B,C$ are $\overline{i}+\overline{j}+\overline{k},4\overline{i}+\overline{j}+\overline{k},4\overline{i}+5\overline{j}+\overline{k}$ . Then the position vector of the circumcentre of the triangle $ABC$ is

• A. $3\hat{i}+2\hat{j}+\hat{k}$
• B. $\frac{1}{2}(6\hat{i}+\hat{j}+\hat{k})$
• C. $\frac{1}{2}(5\hat{i}+6\hat{j}+2\hat{k})$
• D. $\frac{1}{2}(9\overline{i}+7\overline{j}+3\overline{k})$

Answer 1

Answer: (C)

$\frac{1}{2}(5\hat{i}+6\hat{j}+2\hat{k})$

Position vector of $A$ is $\hat{i}+\hat{j}+\hat{k}$, gives coordinates as $(1,1,1)$.

Position vector of $B$ is $4\hat{i}+\hat{j}+\hat{k}$, gives coordinates as $(4,1,1)$.

Position vector of $C$ is $4\hat{i}+5\hat{j}+\hat{k}$, gives coordinates as $(4,5,1)$.

$\overrightarrow{AB}=\overrightarrow{B}-\overrightarrow{A}=3\hat{i}$

$\overrightarrow{BC}=\overrightarrow{C}-\overrightarrow{B}=4\hat{j}$

$\overrightarrow{AC}=\overrightarrow{C}-\overrightarrow{A}=3\hat{i}+4\hat{j}$

Here, we can see that $△ABC$ is a right angled triangle with right angle at $B$.

Hence, circumcenter will be the midpoint of $AC$.

Mid point of $AC=D\equiv (\frac{1+4}{2},\frac{1+5}{2},\frac{1+1}{2})=(\frac{5}{2},3,1)$

So, position vector of $D$ will be

$\frac{1}{2}(5\hat{i}+6\hat{j}+2\hat{k})$

which is a circumcenter.