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

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Aicharya · Accepted Answer

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).

The distance between the orthocentre and circumcentre of the triangle formed by the points \((1,2,3),(3,-1…

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