Consider the vectors, `vec(a),vec(b),vec(c),vec(d)` such that `(vec(a)xxvec(b))xx(vec(c)xxvec(d))=vec(0)" Let "P_(1)andP_(2)` be the planes determined by the pairs of vectors, `vec(a),vec(b)andvec(c),vec(d)` respectively. Then the angle between `P_(1)andP_(2)` is

Options

A. 0
B. `pi//4`
C. `pi//3`
D. `pi//2`

Answer

a

Explanation

Given that `veca. Vecb , vecc and vecd` are vectors such that
`( veca xx vecb) xx (vecc xx vecd)= vec0`
` P_(1)` is the plane determined by vectors `veca and vecb` therefore, normal vectors `vecn_(1) " to" P_(1)` will be given by
`vecn_(1) = veca xx vecb`
Similarly. `P_(2)` is the plane determined by vectors `vecc and vecd`. therefore, normal vectors `vecn_(1) " to " P_(2)` will be given by
`vecn_(2) = veca xx vecb`
Substituting the values of `vecn_(1) and vecn_(2)` in (i), we get
`vecn_(1) xx vecn_(2) = vec0`
Hence ` vecn_(1) ||vecn_(2)`
Hence, the planes will also be parallel to each other. thus angle between the planes is 0.