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

The correct option is C $\frac{\sqrt{41}}{2}$

$A=(2,-1,1);B=(1,-3,-5);C=(3,-4,-4)$

$AB=\sqrt{(1-2{)}^2+(-3+1{)}^2+(-5-1{)}^2}=\sqrt{41}$

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

$CA=\sqrt{(2-1{)}^2+(-1+4{)}^2+(1+4{)}^2}=\sqrt{35}$

We can see $(BC{)}^2+(CA{)}^2=(AB{)}^2$

Hence it is a right angled triangle;

So, circumradius $=\frac{1}{2}$ (hypotenuse)

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