• 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

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).