question_answer

A plane is inclined at an angle of \[30{}^\circ \] with the horizontal. The component \[\vec{A}-10\hat{k}\] perpendicular to the plane is (here z-direction is vertically upwards)

A) \[5\sqrt{2}\]                            

B) \[5\sqrt{3}\]                

C)  5                    

D)    2.5

Correct Answer: B

Solution :

With reference to figure line OB is perpendicular to the plane. So the component of \[\vec{A}\]along OB is \[10\cos {{30}^{o}}\]i.e., \[5\sqrt{3}.\] Hence, the correction option is (b).