If the position vectors of the vertices of a triangle be `2hati+4hatj-hatk,4hati+5hatj+hatk and 3 hati+6hatj-3hatk`, then the triangle is

Options

A. right angled
B. isosceles
C. equilateral
D. none of these

Answer

A, B

Explanation

Let A,B,C be the vertices of given triangle with position vectors, `2hati+4hatj-hatk,4hati+5hatj+hatk and 3hati+6hatj-3hatk`
respectively,
then, we have
`OA=2hati+4hatj-hatk,OB=4hati+5hatj+hatk`
and `OC=3hati+6hatj-3hatk`
clearly, `AB=OB-OA-2hati+hatj+2hatk`
`BC=-hati+hatj-4hatk`
and `AC=hati+2hatj-2hatk`
Now, `AB=|AB|=sqrt(2^(2)+1^(2)+2^(2))=3`
`BC=|BC|=sqrt((-1)^(2)+(1)^(2)+(-4)^(2))=3sqrt(2)`
and `AC=|AC|=sqrt(1^(2)+2^(2)+(-2)^(3))=3`
`because AB=AC and BC^(2)=AB^(2)+AC^(2)`
`therefore`The triangle is isosceles and right angled.