Question

The distance between the orthocentre and circumcentre of the triangle formed by the points $(1,2,3),(3,-1,5)$ and $(4,0,-3)$ is

• A. $\sqrt{\frac{33}{2}}$
• B. $\sqrt{\frac{31}{2}}$
• C. $\sqrt{\frac{27}{2}}$
• D. $\sqrt{\frac{23}{2}}$

Answer 1

Answer: (A)

Let $A=(1,2,3)$
$B=(3,-1,5)$
$C=(4,0,-3)$
$D{R}^{\prime }s$ of $\overline{AB}=(2,-3,2)$
$D{R}^{\prime }$ s of $\overline{BC}=(1,1,-8)$
$D{R}^{\prime }s$ of $\overline{AC}=(3,-2,-6)$
Here, we noticed that $\overline{AB}\perp \overline{AC}$
$\Rightarrow \angle A=90^{\circ }$

Orthocentre of $\Delta ABC$ = vertex A
$H=(1,2,3)$
Circumcentre of $\Delta ABC=$ mid-point of $\overline{BC}$
$S=\left(\frac{7}{2},\frac{-1}{2},1\right)$
$\therefore$ Required distance $(HS)=\sqrt{\frac{33}{2}}$
$\sqrt{{\left(\frac{7}{2}-1\right)}^2+{\left(-\frac{1}{2}-2\right)}^2+(1-3{)}^2}$
$=\sqrt{\frac{25}{4}+\frac{25}{4}+4}=\sqrt{\frac{66}{4}}$