Position vectors of points A, B and C are, respectively given as
`" "veca=3hati-4hatj-4hatk, vecb=2hati-hatj+hatk and vecc=hati-3hatj-5hatk`
`therefore" "vec(AB)=vecb-veca=(2-1)hati+(-1+4)hatj+(1+4)hatk=-hati+3hatj+5hatk`
`" "vec(BC)=vecc-vecb=(1-2)hati+(-3+1)hatj+(-5-1)hatk=-hati-2hatj+6hatk`
`" "vec(CA)=veca-vecc=(3-1)hati+(-4+3)hatj+(-4+5)hatk=2hati-hatj+hatk`
`therefore" "|vec(AB)|^(2)=(-1)^(2)+3^(2)+5^(2)=35`
`" "|vec(BC)|^(2)=(-1)^(2)+(-2)^(2)+(-6)^(2)=41`
`" "|vec(CA)|^(2)=2^(2)+(-1)^(2)+1^(2)=6`
`therefore" "|vec(AB)|^(2)+|vec(CA)|^(2)=36+6=41=|vec(BC)|^(2)`
Hence, ABC is a right-angled triangle.
