1. Tardigrade
  2. Question
  3. Mathematics
  4. The points A (4, - 2, 1), B (7, - 4, 7), C (2, - 5, 10) and D (- 1, - 3, 4) are the vertices of a

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. The points A (4, - 2, 1), B (7, - 4, 7), C (2, - 5, 10) and D (- 1, - 3, 4) are the vertices of a

Introduction to Three Dimensional Geometry

Solution:

Here, the mid-point of [AC] is
$\left(\frac{4+2}{2} , \frac{-2-5}{2} , \frac{1+10}{2}\right) = \left(3, - \frac{7}{2} , \frac{11}{2}\right) $
and that of [BD] is
$\left(\frac{7-1}{2}, \frac{-4-3}{2}, \frac{7+4}{2}\right) = \left(3,- \frac{7}{2} , \frac{11}{2}\right) $
So, the diagonals [AC] and [BD] bisect each other
$\Rightarrow $ ABCD is a parallelogram.
As $|AB| = \sqrt{3^2 + 2^2 + 6^2} = 7 $ and
$|AD| = \sqrt{5^2 + 1^2 + 3^2} \neq |AB | , $
therefore. ABCD is not a rhombus and naturally, it cannot be a square.