Question

Are the points A(3, 6, 9), B(10, 20, 30) and C(25, –41, 5), the vertices of a right-angled triangle?

Answer 1

Let A(3,6,9), B(10,20,30) and C( 25,$-$41,5) are vertices of $△ABC$

AB = $\sqrt{{\left(10-3\right)}^2+{\left(20-6\right)}^2+{\left(30-9\right)}^2}$
$$\begin{gathered} =\sqrt{{\left(7\right)}^2+{\left(14\right)}^2+{\left(21\right)}^2} \\ =\sqrt{49+196+441} \\ =\sqrt{686} \\ =7\sqrt{14} \end{gathered}$$

BC = $\sqrt{{\left(25-10\right)}^2+{\left(-41-20\right)}^2+{\left(5-30\right)}^2}$
$$\begin{gathered} =\sqrt{{\left(15\right)}^2+{\left(-61\right)}^2+{\left(-25\right)}^2} \\ =\sqrt{225+3721+625} \\ =\sqrt{4571} \end{gathered}$$

CA= $\sqrt{{\left(3-25\right)}^2+{\left(6+41\right)}^2+{\left(9-5\right)}^2}$
$$\begin{gathered} =\sqrt{{\left(-22\right)}^2+{\left(47\right)}^2+{\left(-4\right)}^2} \\ =\sqrt{484+2209+16} \\ =\sqrt{2709} \\ =3\sqrt[]{301} \end{gathered}$$

$$\begin{gathered} A{B}^2+B{C}^2={\left(7\sqrt{14}\right)}^2+{\left(\sqrt{4571}\right)}^2 \\ =686+4571 \\ =5257 \\ C{A}^2=2709 \\ \therefore A{B}^2+B{C}^2\ne C{A}^2 \end{gathered}$$

A triangle $△ABC$ is right-angled at B if $C{A}^2=A{B}^2+B{C}^2$.
But, $C{A}^2$ ≠ $A{B}^2+B{C}^2$
Hence, the points are not vertices of a right-angled triangle.