Q.

Let P(x, y, z) be a point in the first octant, whose projection in the xy-plane is the point Q. Let OP = γ; the angle between OQ and the positive x-axis be θ; and the angle between OP and the positive z-axis be ϕ, where O is the origin. Then the distance of P from the x-axis is          [2024]

1 γ1–sin2θcos2ϕ  
2 γ1–sin2ϕcos2θ  
3 γ1+cos2θsin2ϕ  
4 γ1+cos2ϕsin2θ  

Ans.

(2)

Projection of P in xy-plane is given by Q(x, y, 0)

Now, OP = γ

⇒ x2+y2+z2=γ ⇒ x2+y2+z2=γ2          ... (i)

Now, angle between OQ and positive x-axis is θ.

∴  (xi^+yj^)·(i^)x2+y2 =cosθ ⇒ cosθ=xx2+y2          ... (ii)

Similarly, angle between OP and positive z-axis is ϕ.

⇒ cosϕ=zx2+y2+z2 ⇒ sin2ϕ=x2+y2x2+y2+z2          ... (iii)

Now, distance of P from x-axis  =y2+z2

=γ2–x2          [Using (i)]

=γ2–cos2θ(x2+y2)         [Using (ii)]

=γ2–cos2θsin2ϕ·γ2          [Using (iii)]

=γ1–cos2θsin2ϕ.