Question

Show that the vectors $2\hat{i}-\hat{j}+\hat{k},\hat{i}-3\hat{j}-5\hat{k}$ and $3\hat{i}-4\hat{j}-4\hat{k}$ from the vertices of a right-angled triangle.

Answer 1

Let vectors $2\hat{i}-\hat{j}+\hat{k},\hat{i}-3\hat{j}-5\hat{k}$ and $3\hat{i}-4\hat{j}-4\hat{k}$ be position vectors of points$A,B,and \space C$ respectively.
i.e.,$\overrightarrow{OA}=2\hat{i}-\hat{j}+\hat{k},\overrightarrow{OB}=\hat{i}-3\hat{j}-5\hat{k}\space and\space \overrightarrow{OC}=3\hat{i}-4\hat{j}-4\hat{k}$
Now,vectors $\overrightarrow{AB},\overrightarrow{BC}\space and\space \overrightarrow{AC}$represents the sides of $△ABC.$
$i.e.,\overrightarrow{OA}→=2\hat{i}-\hat{j}+\hat{k},\overrightarrow{OB}=\hat{i}-3\hat{j}-5\hat{k},\space and \space \overrightarrow{OC}=3\hat{i}-4\hat{j}-4\hat{k}.$
$∴\overrightarrow{AB}=(1-2)\hat{i}+(-3+1)\hat{j}+(-5-1)\hat{k}=-\hat{i}-2\hat{j}-6\hat{k}$
$\overrightarrow{BC}=(3-1)\hat{i}+(-4+3)\hat{j}(-4+5)\hat{k}=2\hat{i}-\hat{j}+\hat{k}$
$\overrightarrow{AC}=(2-3)\hat{i}+(-1+4)\hat{j}+(1+4)\hat{k}=-\hat{i}+3\hat{j}+5\hat{k}$
$|\overrightarrow{AB}|=\sqrt{(-1)^{2}+(-2)^{2}+(-6)^{2}}=\sqrt{1+4+36}=\sqrt{41}$
$|\overrightarrow{BC}|=\sqrt{2^{2}+(-1)^{2}+1^{2}}=\sqrt{4+1+1}=\sqrt{6}$
$|\overrightarrow{AC}|=\sqrt{(-1)^{2}+3^{2}+5^{2}}=\sqrt{1+9+25}=\sqrt{35}$
$∴|\overrightarrow{BC}|2+|\overrightarrow{AC}|2=6+35=41=|\overrightarrow{AB}|^{2}$
Hence$,△ABC$ is a right-angled triangle.