a. Prove that the vector `vec(A)=3hat(i)-2hat(j)+hat(k)`, `vec(B)=hat(i)-3hat(j)+5hat(k),` and `vec(C )=2hat(i)+hat(j)-4hat(k)` from a right -angled triangle.
b. Determine the unit vector parallel to the cross product of vector `vec(A)=3hat(i)-5hat(j)+10hat(k)` & `=vec(B)=6hat(i)+5hat(j)+2hat(k).`

Explanation

a.The given vectors will constitute a triangle only if onr of the given vector is equal to vector sum of the remaining two vectors. In the given problem, `vec(B)+vec(C )=vec(A)`
So,the given vectors do from a triangle .This triangle will be right-angled only if the dot product of two vectors (out of the given three)is zero.
`vec(A).vec(B)=(3hat(i)-2hat(j)+hat(k)).(hat(i)-3hat(j)+5hat(k))`
`=3(hat(i).hat(i))+6(hat(j).hat(j))+5(hat(k).hat(k))=3+6+5+14`
`vec(B).vec(C)=(hat(i)-3hat(j)+5hat(k)).(2hat(i)+hat(j)-4hat(k))`
`=2(hat(i).hat(i))-3(hat(j).hat(j))-20(hat(k).hat(k))=2-3-20=-21`
`vec(C ).vec(A)=(2hat(i)+hat(j)-4hat(k)).(3hat(i)-2hat(j)+hat(k))`
`6(hat(i).hat(i))-2(hat(j).hat(j))-4(hat(k).hat(k))=6-2-4=0`
Since the dot product of `vec(C )` and `vec(A)` is zero,therefore,it implies that `vec(C )` is perpendicular to `vec(A)`.
b. The unit vector parallel to `(vec(A)xxvec(B))` is given by
`hat(n)=(vec(A)xxvec(B))/(|vec(A)xxvec(B)|)`
So, let us first determine `vec(A)xxvec(B).`
Now `vec(A)xxvec(B)=(3hat(i)-5hat(j)+10hat(k))xx(6hat(i)+5hat(j)+2hat(k))`
`=|(hat(i), hat(j), hat(k)) ,(3,-5,10), (6 ,5 ,2)|`
`=hat(i)(-10-50)+hat(j)(60-6)+hat(k)(15+30)`
`=-60hat(i)+54hat(j)+45hat(k)`
Magnitude: `|vec(A)xxvec(B)|=sqrt((-60^(2))+(54^(2))+(45^(2)))=sqrt(8541)`
So,required unit vector: `hat(n)=(-60hat(i)+54hat(j)+45hat(k))/sqrt(8541)`