Show that the points `A,B` and C having position vectors `(3hati - 4hatj - 4hatk), (2hati - hatj + hatk)`and `(hati - 3hatj - 5hatk)` respectively, from the vertices of a right-angled triangle.

Explanation

We have `vec(AB)`=(position vector of B)-(position vector of A)
`=(3hati-hatj+hatk)-(3hati-4hatj-4hatk)`
`=(-hati+3hatj+5hatk)`
`vec(BC)`=(position vector of C)-(position vector of B)
`=(hati-3hatj-5hatk)-(2hati-hatj+hatk)=(-hati-2hatj-6hatk)`
`vec(CA)`=(position vector of A)-(position vector of C)
`=(2hati-hatj+hatk)`
`:.|vec(AB)|^(2)=35`
`|vec(BC)|^(2)=41`
And `|vec(CA)|^(2)=6`
Thus `|vec(AB)|^(2)+|vec(CA)|^(2)=|vec(BC)|^(2) rArrAB^(2)+CA^(2)=BC^(2)`
Hence `Delta` ABC is right angled at A.