The point having position vectors `2hati+3hatj+4hatk,3hati+4hatj+2hatk and 4hati+2hatj+3hatk` are the vertices of

Options

A. right angled triangle
B. isosceles triangle
C. equilateral triangle
D. collinear

Answer

C

Explanation

Here, `OA=2hati+3hatj+4hatk`
`OB=3hati+4hatj+2hatk`
and `OC=4hati+2hatj+3hatk`
So, `AB=hati+hatj-2hatk,BC=hati-2hatj+hatk,CA=2hati-hatj-hatk`
Clearly, `|AB|=|BC|=|CA|=sqrt(6)`
So, these poins are vertices of an equilateral triangle.