Q12,paper 3,D 12. Let V1= 2I – J + K and V2= I + J – K, then the angle between V1& V2 and a vector perpendicular to both V1& V2 shall be :

(A) 90°and (–2I + J – 3K)

(B) 60° and (2I + J + 3K)

(C) 90°and (2I + J – 3K)

(D) 90°and (–2I – J + 3K)

Answer(d).

cosθ = (u⃗ · v⃗) / (||u⃗|| ||v ⃗ ||).

(u⃗ · v⃗) is the dot product of vectors.[2 -1 1][1 

                                                                      1

                                                                     -1]

calculated as (2-1-1) as 0 .

Length (||u⃗||)=((2)²+-1²+1²)^1/2

Length(||v ⃗ ||).=(1²+1²+(-1)²)^1/2

Cosθ=0/18^1/2 hence θ is 90degrees.

And

Vector perpendicular  to both shall be.

u×v  =i(u2v3-u3v2)-j(u1v3-v1u3)+(u1v2-u2v1)K.

=i(1-1)-j(-2-1)+(2+1)k

vector is (0i+3j+3k)

let us check whether dot product of 2i-j+k and (0i+3j+3k) are perpendicular their dot product is zero. Hence perpendicular.

Let us check (0i,3j,3k) dot product with (I,j,-k) and dot product is zero.

 We can say that (0i,3j,3k ) is perpendicular to both vectors.