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    				| 1. | 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)` isA. 4B. 5C. 6D. 7 | 
| Answer» Correct Answer - B `(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 | |