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

• Correct answer
  A

  \(\sqrt{\frac{33}{2}}\)
• B

  \(\sqrt{\frac{31}{2}}\)
• C

  \(\sqrt{\frac{27}{2}}\)
• D

  \(\sqrt{\frac{23}{2}}\)

Solution

Let \(A=(1,2,3)\) \(\begin{aligned} B & =(3,-1,5) \\ C & =(4,0,-3) \\ D R^{\prime} S \text { of } \overline{A B} & =(2,-3,2) \\ D R^{\prime} \text { ' of } \overline{B C} & =(1,1,-8) \\ D R^{\prime} S \text { of } \overline{A C} & =(3,-2,-6) \end{aligned}\) Here, we noticed that \(\overline{A B} \perp \overline{A C}\) \(\Rightarrow \quad \angle A=90^{\circ}\)
Orthocentre of \(\triangle A B C=\) vertex \(A\) \(H=(1,2,3)\) Circumcentre of \(\triangle A B C=\) mid-point of \(\overline{B C}\) \(S=\left(\frac{7}{2}, \frac{-1}{2}, 1\right)\) \(\therefore\) Required distance \((H S)=\sqrt{\frac{33}{2}}\) \(\begin{aligned} & \sqrt{\left(\frac{7}{2}-1\right)^2+\left(-\frac{1}{2}-2\right)^2+(1-3)^2} \\ & =\sqrt{\frac{25}{4}+\frac{25}{4}+4}=\sqrt{\frac{66}{4}} \end{aligned}\) Hence, answer is (a).

Hence, the correct answer is option: 'A'.