Question

The circumradius of the triangle formed by the points \( (2, -1, 1) \), \( (1, -3, -5) \), and \( (3, -4, -4) \) is

• A. \( \dfrac{\sqrt{35}}{2} \)
• B. \( \dfrac{\sqrt{25}}{3} \)
• C. \( \sqrt{41} \)
• D. \( \dfrac{\sqrt{41}}{2} \)

Hint:

In 3D geometry problems involving triangle properties, always begin with vector representations and magnitudes for side lengths.

Answer 1

The Correct Option is D

To find the circumradius \( R \) of a triangle in 3D given vertices \( A, B, C \), we use the formula: \[ R = \frac{abc}{4\Delta} \] Where \( a, b, c \) are side lengths and \( \Delta \) is the area of the triangle. Step 1: Let the points be: \( A = (2, -1, 1) \), \( B = (1, -3, -5) \), \( C = (3, -4, -4) \) Step 2: Compute vectors: \[ \vec{AB} = B - A = (-1, -2, -6) \Rightarrow |\vec{AB}| = \sqrt{1^2 + 2^2 + 6^2} = \sqrt{1 + 4 + 36} = \sqrt{41} \] \[ \vec{BC} = C - B = (2, -1, 1) \Rightarrow |\vec{BC}| = \sqrt{2^2 + 1^2 + 1^2} = \sqrt{6} \] \[ \vec{CA} = A - C = (-1, 3, 5) \Rightarrow |\vec{CA}| = \sqrt{1^2 + 3^2 + 5^2} = \sqrt{1 + 9 + 25} = \sqrt{35} \] Step 3: Use Heron’s formula or vector area to find \( \Delta \), then plug into the formula. After calculation, the radius simplifies to: \[ R = \dfrac{\sqrt{41}}{2} \]