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📐 Mathematics  ·  Introduction to Three Dimensional Geometry  ·  JEE

A point equidistant from all four vertices of a tetrahedron is found by:

Answer: Solving the three equations obtained by equating squared distances pairwise.

  • A Averaging the four vertices, which always gives the required point
  • B Taking the midpoint of the longest edge of the tetrahedron
  • C Projecting the centroid onto the XY-plane in every case
  • D Solving the three equations obtained by equating squared distances pairwise

Correct answer: D. Solving the three equations obtained by equating squared distances pairwise

Explanation: Equating squared distances pairwise removes the quadratic terms and leaves three linear equations in x, y and z - the circumcentre generally differs from the centroid.

zxyOP(x, y, z)first octant:x>0, y>0, z>0

Concept context

Coordinate axes and planes in space, octants, distance between two points, and the section formula

Read the full Introduction to Three Dimensional Geometry notes →