The points A = (1, 1, 1), B = (2, 4, 2), and C = (3, 2, 2)are the vertices of a triangle in space. Which of the vertices is closest to the origin?
A)A
B)B
C)C

Question

Quizplus · Accepted Answer

The distance of a point (x, y, z) from the origin (0, 0, 0) is given by the formula $\sqrt{x^2 + y^2 + z^2}$. Calculating the distances for each point: A = $\sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}$, B = $\sqrt{2^2 + 4^2 + 2^2} = \sqrt{24}$, and C = $\sqrt{3^2 + 2^2 + 2^2} = \sqrt{17}$. Therefore, point A is closest to the origin.

The Points a = (1, 1, 1), B = (2