First slide

Introduction to 3-D Geometry

Question

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

Easy

A

$(\frac{7}{2}, - \frac{1}{2}, 1)$
B

$(\frac{5}{2}, 1, 0)$
C

$(2, \frac{1}{2}, 4)$
D

$(\frac{8}{3}, \frac{1}{3}, \frac{5}{3})$

Solution

Let the vertices are $A (1, 2, 3), B (3, - 1, 5), C (4, 0, - 3)$
The sides are
$\begin{matrix} A B = \sqrt{4 + 9 + 4} = \sqrt{17} \\ B C = \sqrt{1 + 1 + 64} = \sqrt{66} \\ C A = \sqrt{9 + 4 + 36} = \sqrt{49} \end{matrix}$
Observing the sides of the triangle, square of one side is sum of squares of other two sides, so that the triangle is right angled triangle.
Hence, the circum center is midpoint of hypotenuse. It is midpoint of $B C$
Therefore, the circum center is $(\frac{3 + 4}{2}, \frac{- 1 + 0}{2}, \frac{5 - 3}{2}) = (\frac{7}{2}, - \frac{1}{2}, 1)$

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