Question

Show that the points $A\left(2\hat{i}-\hat{j}+\hat{k}\right),B\left(\hat{i}-3\hat{j}-5\hat{k}\right),C\left(3\hat{i}-4\hat{j}-4\hat{k}\right)$ are the vertices of a right angled triangle.

Answer 1

Given the points $A\left(2\hat{i}-\hat{j}+\hat{k}\right),B\left(\hat{i}-3\hat{j}-5\hat{k}\right)$ and $C\left(3\hat{i}-4\hat{j}-4\hat{k}\right).$
Then,
$\overrightarrow{AB}=$Position vector of $B$ $-$ Position vector of A
$$\begin{gathered} =\hat{i}-3\hat{j}-5\hat{k}-\left(2\hat{i}-\hat{j}+\hat{k}\right) \\ =\hat{i}-3\hat{j}-5\hat{k}-2\hat{i}+\hat{j}-\hat{k} \\ =-\hat{i}-2\hat{j}-6\hat{k} \end{gathered}$$

$\overrightarrow{BC}=$Position vector of $C$ $-$ Position vector of $B$
$$\begin{gathered} =3\hat{i}-4\hat{j}-4\hat{k}-\left(\hat{i}-3\hat{j}-5\hat{k}\right) \\ =3\hat{i}-4\hat{j}-4\hat{k}-\hat{i}+3\hat{j}+5\hat{k} \\ =2\hat{i}-\hat{j}+\hat{k} \end{gathered}$$

$\overrightarrow{CA}=$Position vector of $A$ $-$ Position vector of $C$
$$\begin{gathered} =2\hat{i}-\hat{j}+\hat{k}-\left(3\hat{i}-4\hat{j}-4\hat{k}\right) \\ =2\hat{i}-\hat{j}+\hat{k}-3\hat{i}+4\hat{j}+4\hat{k} \\ =-\hat{i}+3\hat{j}+5\hat{k} \end{gathered}$$
Clearly, $\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CA}=\overrightarrow{0}$

$$\begin{gathered} \mathrm{Now},\overrightarrow{\left|AB\right|}=\sqrt{{\left(-1\right)}^2+{\left(-2\right)}^2+{\left(-6\right)}^2}=\sqrt{1+4+36}=\sqrt{41} \\ \overrightarrow{\left|BC\right|}=\sqrt{{\left(2\right)}^2+{\left(-1\right)}^2+{\left(1\right)}^2}=\sqrt{4+1+1}=\sqrt{6} \\ \overrightarrow{\left|CA\right|}=\sqrt{{\left(-1\right)}^2+{\left(3\right)}^2+{\left(5\right)}^2}=\sqrt{1+9+25}=\sqrt{35} \\ \mathrm{Clearly},{\overrightarrow{\left|AB\right|}}^2={\overrightarrow{\left|BC\right|}}^2+{\overrightarrow{\left|CA\right|}}^2 \\ \Rightarrow A{B}^2=B{C}^2+C{A}^2 \end{gathered}$$

So, $A,B,C$ forms a right angled triangle.