Question

The points (5, –4, 2), (4, –3, 1), (7, 6, 4) and (8, –7, 5) are the vertices of
(a) a rectangle
(b) a square
(c) a parallelogram
(d) none of these

Answer 1

(d) None of these

Suppose:
A(5, –4, 2)
B(4, –3, 1)
C(7, 6, 4)
D(8, –7, 5)

$$\begin{gathered} AB=\sqrt{{\left(4-5\right)}^2+{\left(-3+4\right)}^2+{\left(1-2\right)}^2} \\ =\sqrt{{\left(-1\right)}^2+{\left(1\right)}^2+{\left(-1\right)}^2} \\ =\sqrt{1+1+1}=\sqrt{3} \\ BC=\sqrt{{\left(7-4\right)}^2+{\left(6+3\right)}^2+{\left(4-1\right)}^2} \\ =\sqrt{{\left(3\right)}^2+{\left(9\right)}^2+{\left(3\right)}^2} \\ =\sqrt{9+81+9}=\sqrt{99}=3\sqrt{11} \\ CD=\sqrt{{\left(8-7\right)}^2+{\left(-7-6\right)}^2+{\left(5-4\right)}^2} \\ =\sqrt{{\left(1\right)}^2+{\left(-13\right)}^2+{\left(1\right)}^2} \\ =\sqrt{1+169+1}=\sqrt{171} \\ DA=\sqrt{{\left(8-5\right)}^2+{\left(-7+4\right)}^2+{\left(5-2\right)}^2} \\ =\sqrt{{\left(3\right)}^2+{\left(-3\right)}^2+{\left(3\right)}^2} \\ =\sqrt{9+9+9}=\sqrt{27}=3\sqrt{3} \end{gathered}$$

We see that none of the sides are equal.