Question

The $P.V{.}^{\prime }s$ of the vertices of a $△ABC$ are $\overline{i}+\overline{j}+\overline{k},4\overline{i}+\overline{j}+\overline{k},4\overline{i}+5\overline{j}+\overline{k}$. The $P.V.$ of the circumcentre of $△ABC$ is

• A. $\frac{5}{2}\overline{i}+3\overline{j}+\overline{k}$
• B. $5\overline{i}+\frac{3}{2}\overline{j}+\overline{k}$
• C. $5\overline{i}+3\overline{j}+\frac{1}{2}\overline{k}$
• D. $\overline{i}+\overline{j}+\overline{k}$

Answer 1

The correct option is A $\frac{5}{2}\overline{i}+3\overline{j}+\overline{k}$

We have,

$A(1,1,1)$ $B(4,1,1)$ $C(4,5,1)$

$AB=3$ $BC=4$ $CA=5$

Hence,

Triangle is right angle at $B$.

$\therefore$ CIrcumcentre is mid point of $AC=\left(\frac{5}{2},3,1\right)$

$=\left(\frac{5}{2}\overline{i},3\overline{j},\overline{k}\right)$

Then,

Option $A$ is correct answer.