Correct Answer - A
Let `theta` be angle between `vecA and vecB`.
`therefore` Resultant of `vecA and vecB ` is `P = sqrt(A^(2) + B^(2) + 2ABcos theta)`
When `vecB` is reversed, then th angle between `vecA and -vecB` is `(180^(@) - theta)`.
Resultant of `vecA and -vecB` is
`Q = sqrt(A^(2) + B^(2) +2AB cos(180^(@) - theta))`
` Q = sqrt(A^(2) + B^(2) - 2AB cos theta)" "...(ii)`
Squaring and adding (i) and (ii), we get
`P^(2) + Q^(2) + 2(A^(2) + B^(2))`