Question

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

Answer 1

Position vectors of points A, B and C are respectively given as
$a=3\hat{i}-4\hat{j}-4\hat{k},b=2\hat{i}-\hat{j}+\hat{k}andc=\hat{i}-3\hat{j}-5\hat{k}$
Now, $AB=b-a=(PVofB-PVofA)\Rightarrow b-a=\left[\right(2\hat{i}-\hat{j}+\hat{k})-(3\hat{i}-4\hat{j}-4\hat{k}\left)\right]=-\hat{i}+3\hat{j}+5\hat{k}$
Comparing with $X=x\hat{i}+y\hat{j}+z\hat{k},wegetx=-1,y=3,z=5$
Magnitude of AB, |AB| =$\sqrt{x^2+y^2+z^2}=\sqrt{(-1{)}^2+(3{)}^2+(5{)}^2}=\sqrt{1+9+25}=\sqrt{35}$
$\therefore |AB{|}^2=35$
Similarly, $BC=c-b=(PVofC-PVofB)$
$\Rightarrow c-b=(\hat{i}-3\hat{j}-5\hat{k})-(2\hat{i}-\hat{j}+\hat{k})=-\hat{i}-2\hat{j}-6\hat{k}$
Similarly, |BC|=$\sqrt{(-1{)}^2+(-2{)}^2+(-6{)}^2}\Rightarrow |BC{|}^2=41$
and $CA=a-c=(PVofA-PVofC)\Rightarrow a-c=(3\hat{i}-4\hat{j}-4\hat{k})-(\hat{i}-3\hat{j}=5\hat{k})=2\hat{i}-\hat{j}+\hat{k}|CA|=\sqrt{(2{)}^2+(-1{)}^2+(1{)}^2}=\sqrt{4+1+1}=\sqrt{6}\Rightarrow |CA{|}^2=6$
Now, we find $|AB{|}^2+|CA{|}^2$ = $35+6+41=|BC{|}^2$
Therefore, $△$ABC is a right angled triangle with right angle at A.