The distance between the orthocentre and circ | vedclass

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Question

(A) Let $A = (1,2,3)$,$B = (3,-1,5)$,and $C = (4,0,-3)$.First,we find the direction ratios of the sides:Direction ratios of $\overline{AB} = (3-1, -1-

Vedclass · Accepted Answer

$\sqrt{\frac{33}{2}}$

Vedclass · Answer

(A) Let $A = (1,2,3)$,$B = (3,-1,5)$,and $C = (4,0,-3)$.First,we find the direction ratios of the sides:Direction ratios of $\overline{AB} = (3-1, -1-2, 5-3) = (2, -3, 2)$.Direction ratios of $\overline{AC} = (4-1, 0-2, -3-3) = (3, -2, -6)$.Check for orthogonality: $\vec{AB} \cdot \vec{AC} = (2)(3) + (-3)(-2) + (2)(-6) = 6 + 6 - 12 = 0$.Since the dot product is $0$,$\overline{AB} \perp \overline{AC}$,which means $\angle A = 90^{\circ}$.In a right-angled triangle,the orthocentre $H$ is the vertex where the right angle is located. Thus,$H = A = (1, 2, 3)$.The circumcentre $S$ of a right-angled triangle is the midpoint of the hypotenuse $\overline{BC}$.$S = \left( \frac{3+4}{2}, \frac{-1+0}{2}, \frac{5-3}{2} \right) = \left( \frac{7}{2}, -\frac{1}{2}, 1 \right)$.The distance $HS$ is given by the distance formula:$HS = \sqrt{\left( \frac{7}{2} - 1 \right)^2 + \left( -\frac{1}{2} - 2 \right)^2 + (1 - 3)^2}$$HS = \sqrt{\left( \frac{5}{2} \right)^2 + \left( -\frac{5}{2} \right)^2 + (-2)^2}$$HS = \sqrt{\frac{25}{4} + \frac{25}{4} + 4} = \sqrt{\frac{50}{4} + \frac{16}{4}} = \sqrt{\frac{66}{4}} = \sqrt{\frac{33}{2}}$.Thus,the correct option is $(A)$.

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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