Show that the points A, B and C with position vectors , `vec(a)=3hat(i)-4hat(j)-4hat(k),vec(b)=2hat(i) -hat(j)+hat(k)` and `vec (c ) hat(i) -3hat(j)-5hat(k)` respectively, from the vertices of a right angled triangle.

Explanation

Position vectors of points A,B and C respectively given as `vec(a)=3hat(i)-4hat(j)-4hat(k),vec(b)=2hat(i)-hat(j)+hat(k) and vec( c) =hat(i)-3hat(j)-5hat(k)`.
Now , `vec(AB)=vec(b)-vec(a)=(PV of B-PV of A)`
`rArr vec(b)-vec(a)=[(2hat(i)-hat(j)+hat(k))-(3hat(i)-4hat(j)-4hat(k))]=-hat(i)+3hat(j)+5hat(k)`
Comparing with `vec(X) =xhat(i)+yhat(j)+zhat(k)` , we get `x=-1,y=3,z=5`
Magnitude of `AB,|vec(AB)|=sqrt(x^2+y^2+z^2)=sqrt((-1)^2+(3)^2+(5)^2)=sqrt(1+9+25)=sqrt(35)`
`therefore|AB|^2=35`.
Similarly , ` vec(BC)=c-b=(PV of C-PV of C)`
`rArr c-b=(hat(i)-3hat(j)-5hat(k))-(2hat(i)-hat(j)+k)=-hat(i)-2hat(j)-6hat(k)`.
Similarly, `|vec(BC)|=sqrt((-1)^2+(-2)^2+(-6)^2)rArr |BC|^2=41`
and `vec(CA) =a-c = (PV of A-PV of C)`.
`rArr vec(a)-vec( c) =(3hat(i)-4hat(j)-4hat(k))-(hat(i)-3hat(j)-5hat(j)-5hat(k))=2hat(i)-hat(j)+hat(k)`.
`|vec(CA)|=sqrt((2)^2+(-1)^2+(1)^2)=sqrt(4+1+1)=sqrt(6)rArr|CA|^2=6`

Now , we fin `|vec(AB)^(2)| + |vec(CA)^(2) = 35 + 6 = 41 = "|"vec(BC)"|"^(2)|`
Therefore , `Delta ABC` is a right angled triangle with right angle at A.