Question

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

• A. $\sqrt{6}$ units
• B. $\frac{\sqrt{6}}{2}$ units
• C. $2\sqrt{6}$ units
• D. $0$

Answer 1

The correct option is D $0$

Let $A(2,1,5)$, $B(3,2,3)$, and $C(4,0,4)$ be the vertices of the triangle given.

We have the distance formula, between points $P(x_1,y_1,z_1)$ , $R(x_2,y_2,z_2)$

$PR=\sqrt{(x_1-x_2{)}^2+(y_1-y_2{)}^2+(z_1-z_2{)}^2}$

To find the length of the sides,

$AB=\sqrt{(2-3{)}^2+(1-2{)}^2+(5-3{)}^2}=\sqrt{1+1+4}=\sqrt{6}$

$CB=\sqrt{(4-3{)}^2+(0-2{)}^2+(4-3{)}^2}=\sqrt{1+4+1}=\sqrt{6}$

$CA=\sqrt{(4-2{)}^2+(0-1{)}^2+(4-5{)}^2}=\sqrt{4+1+1}=\sqrt{6}$

$⟹$The triangle is equilateral.

Hence the ortho-centre, circumcentre, incentre and centroid is the same point.

The distance between ortho-centre and circumcentre is $0$.