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.