Question

Show that (a, b, c), (b, c, a) and (c, a, b) are the vertices of an equilateral triangle.

Answer 1

Hint: We have to use distance formula in this problem the distance between points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is given by-
$$\mathrm d=\sqrt{\left({\mathrm x}_2-{\mathrm x}_1\right)^2+\left({\mathrm y}_2-{\mathrm y}_1\right)^2+\left({\mathrm z}_2-{\mathrm z}_1\right)^2}$$

Complete step-by-step solution -

Let the points be A(a, b, c), B(b, c, a) and C(c, a, b). An equilateral triangle has all three sides equal. To prove that they form an equilateral triangle, we can prove that the distance between these three points is equal. We can do this by using the distance formula-
$$\mathrm{AB}=\sqrt{\left(\mathrm b-\mathrm a\right)^2+\left(\mathrm c-\mathrm b\right)^2+\left(\mathrm a-\mathrm c\right)^2}\\\mathrm{BC}=\sqrt{\left(\mathrm c-\mathrm b\right)^2+\left(\mathrm a-\mathrm c\right)^2+\left(\mathrm b-\mathrm a\right)^2}\\\mathrm{CA}=\sqrt{\left(\mathrm a-\mathrm c\right)^2+\left(\mathrm b-\mathrm a\right)^2+\left(\mathrm c-\mathrm b\right)^2}$$.
Here AB=BC=AC.
We can see that all the three terms in AB, BC and CA are the same. Hence, the final distance will be the same between all the three points. This shows that they form an equilateral triangle and lie at its vertices.

Hence, proved.

Note: Instead of proving their distances equal, we can also use a longer method to prove our statement. This can be done by proving that all the three lines formed by the points are at an equal angle from each other.