Question

If position vector of $A,B,C$ of vertices of $△ABC$ are $(\hat{i}+2\hat{j}+3\hat{k}),(-\hat{i}-\hat{j}+8\hat{k}),(-4\hat{i}+4\hat{j}+6\hat{k})$ then, what is the type of the triangle.

Answer 1

$dist\left(AB\right)=\sqrt{(1+1{)}^2+(2+1{)}^2+(8-3{)}^2}=\sqrt{38}$

$dist\left(BC\right)=\sqrt{(-4+1{)}^2+(4+1{)}^2+(6-8{)}^2}=\sqrt{29}$

$dist\left(AC\right)=\sqrt{(-4-1{)}^2+(4-2{)}^2+(6-3{)}^2}=\sqrt{38}$

Since 2 sides have same length and $A{B}^2+A{C}^2\ne B{C}^2$, the triangle is an isoceles triangle.