Question

Show that the points $A, B,$ and $C$ with position vectors, $\vec{a}=3\hat{i}-4\hat{j}-4\hat{k}, \vec{b}=2\hat{i}-\hat{j}+\hat{k}$ and $\vec{c}=\hat{i}-3\hat{j}-5\hat{k}$, respectively from the vertices of a right angled triangle.

Answer 1

Position vectors of $A,B$,and $C$ are respectively given as:
$\vec{a}=3\hat{i}-4\hat{j}-4\hat{k}, \vec{b}=2\hat{i}-\hat{j}+\hat{k}$,$\vec{c}=\hat{i}-3\hat{j}-5\hat{k}$
$∴\vec{AB}=\vec{b}-\vec{a}=(2-3)\hat{i}+(-1+4)\hat{j}+(1+4)\hat{k}=-\hat{i}+3\hat{j}+5\hat{k}$
$\vec{BC}=\vec{c}-\vec{b}=(1-2)\hat{i}+(-3+1)\hat{j}+(-5-1)\hat{k}=-\hat{i}-2\hat{j}-6\hat{k}$
$\vec{CA}=\vec{a}-\vec{c}=(3-1)\hat{i}+(-4+3)\hat{j}+(-4+5)\hat{k}=2\hat{i}-\hat{j}+\hat{k}$
$∴|\vec{AB}|^2=(-1)^2+3^2+5^2=1+9+25=35$
$|\vec{BC}|^2=(-1)^2+(-2)^2+(-6)^2=1+4+36=41$
$|\vec{CA}|^2=2^2+(-1)^2+1^2=4+1+1=6$
$∴|\vec{AB}|^2+|\vec{CA}|^2=36+6=41=|\vec{BC}|^2$
Hence,ABC is a right angled triangle.