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).