Show that the points `A ,\ B\ a n d\ C` with position vectos ` vec a=3 hat i-4 hat j-4 hat k ,\ vec b=2 hat i- hat j+ hat k\ a n d\ vec c= hat i-3 hat j-5 hat k` represent, form the vertices of a right angled triangle.

Explanation

given: `\quad \vec{a}=3 \hat{i}-4 \hat{j}-4 \hat{k}(`point `A)`

`\quad \vec{b}=2 \hat{i}-\hat{j}+\hat{k}(` point `B) \quad`

`\vec{c}=\hat{i}-3 \hat{j}-5 \hat{k}(`point C`)`

The sum of square of two side of a triangle is equal to square of third side.
`\begin{array}{ll}\overrightarrow{A B}=\vec{b}-\vec{a}=-\hat{i}+3 \hat{j}+5 \hat{k} & |\overrightarrow{A B}|=\sqrt{(-1)^{2}+(3)^{2}+(5)^{2}}=\sqrt{3} \\ \overrightarrow{B C}=\vec{c}-\vec{b}=-\hat{i}-2 \hat{j}-6 \hat{k} & |\overrightarrow{B C}|=\sqrt{(-1)^{2}+(-2)^{2}+(-b)^{2}}=\sqrt{41} \\ \overrightarrow{C A}=\vec{a}-\vec{c}=2 \hat{i}-\hat{j}+\hat{k} & |\overrightarrow{C A}|=\sqrt{(2)^{2}+(-1)^{2}+(1)^{2}}=\sqrt{6}\end{array}`



\begin{aligned}
|\overrightarrow{A B}|^{2}+|\overrightarrow{C A}|^{2} &\left.=(\sqrt{3} 5)^{2}+(\sqrt{6})^{2}=4\right) \\
|\overrightarrow{B C}|^{2} &=41 \\
|\overrightarrow{A B} \mid)^{2}+(\mid \overrightarrow{C A})^{2} &=(|\overrightarrow{B C}|)^{2}
\end{aligned}
...