Question

The $P.V.s$ of the vertices of a triangle are $2\overline{i}+3\overline{j}+4\overline{k},4\overline{i}+6\overline{j}+3\overline{k},3\overline{i}+2\overline{j}+3\overline{k}$ $P.V.$ of its orthocentre is

• A. $2\overline{i}-3\overline{j}+4\overline{k}$
• B. $2\overline{i}+3\overline{j}-4\overline{k}$
• C. $2\overline{i}+3\overline{j}+4\overline{k}$
• D. $-2\overline{i}+3\overline{j}+4\overline{k}$

Answer 1

The correct option is C $2\overline{i}+3\overline{j}+4\overline{k}$

We have,

$\begin{array}{c}A\left(2\overline{i}+3\overline{j}+4\overline{k}\right)B\left(4\overline{i}+6\overline{j}+3\overline{k}\right)C\left(3\overline{i}+2\overline{j}+3\overline{k}\right)\\ AB=\sqrt{14}Ac=\sqrt{3}BC=\sqrt{17}\end{array}$

Hence,

Triangle is right angled (Pythagorean theorem)

Clearly or the centre is $A$ $\left(2\overline{i}+3\overline{j}+4\overline{k}\right)$

Then,

Option $C$ is correct answer.