Question

The circum radius of the triangle formed by the points $(2,-1,1),(1,-3,-5)$ and $(3,-4,-4)$ is

• A. $\frac{\sqrt{6}}{2}$
• B. $\frac{\sqrt{35}}{2}$
• C. $\frac{\sqrt{41}}{2}$
• D. $\sqrt{41}$

Answer 1

Answer: (C)

$\frac{\sqrt{41}}{2}$

Let the points $A(2,-1,1)B(1,-3,-5)$ and $C(3,-4,-4)$
Now using distance formula in $3D$, we have
$AB=\sqrt{41}$, $BC=\sqrt{6}$ and $AC=\sqrt{35}$
Since, ${BC}^2+{AC}^2$$=$${AB}^2$
Hence, it is right angle triangle and as we know that the circum radius of right angled triangle is equal to half of hypotenuse.
Thus, radius $=\frac{\sqrt{41}}{2}$