Let `(hatp xx vecq) xx (hatp.vecq)vecq`
`=(x^(2)+y^(2))vecq + (14-4x-6y)vecp`
Where `hatp` and `hatq` are two non-collinear vectors `vecp` is unit vector and x,y are scalars. Then the value of `(x+y)` is

Options

A. 4
B. 5
C. 6
D. 7

Answer

B

Explanation

`(vecp xx vecq) xx hatp+(hatp.vecc)vecq=(x^(2)+y^(2))vecq+(14-4x-6y)hatp`
`rArr (hatp.hatp)vecq+(1-4x-6y)hatp`
Since `hatp` and `vecq` are non-zero non-collinear.
We can compare coefficients of vectors `hatp` and `hatq`.
`therefore 1+hatp.vecq=x^(2)+y^(2)`.............(i)
And `hatp.vecq=4x+6y-14`............(ii)
From (i) and (ii), we get
`rArr x^(2) + y^(2)-4x-6y+13=0`
`rArr (x-2)^(2)+(y-3)^(2)=0`
x=2 and y=3