Question

A plane inclined at an angle of 30° with horizontal. Prove that the component of a vector A= -10k perpendicular to this plane is 5√3(here z direction is vertically upwards

Answer 1

Answer:

Explanation:

=> Here,  z direction is vertically upwards

=> A plane inclined at an angle of 30° with horizontal.

∴θ = 30°

To prove that the component of a vector A= -10k perpendicular to this plane is 5√3, suppose, the force is acting on vertically downwards, F =-10k,

Thus, 10*cosθ = 10*cos 30° = 10 * √3/2  [because cos 30°  = √3/2]

= 5√3

Thus,  the component of a vector A= -10k perpendicular to this plane is 5√3.

Answer 2

Here is the answer in the picture attached