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

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Solution

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 $γ \sqrt{1 - \sin^{2} θ \cos^{2} ϕ}$

2 $γ \sqrt{1 - \sin^{2} ϕ \cos^{2} θ}$

3 $γ \sqrt{1 + \cos^{2} θ \sin^{2} ϕ}$

4 $γ \sqrt{1 + \cos^{2} ϕ \sin^{2} θ}$

Ans.
(2)

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

Now, OP = $γ$

$\Rightarrow \sqrt{x^{2} + y^{2} + z^{2}} = γ \Rightarrow x^{2} + y^{2} + z^{2} = γ^{2}$ ... (i)

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

$∴ \frac{(x \hat{i} + y \hat{j}) \cdot (\hat{i})}{\sqrt{x^{2} + y^{2}}} = \cos θ \Rightarrow \cos θ = \frac{x}{\sqrt{x^{2} + y^{2}}}$ ... (ii)

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

$\Rightarrow \cos ϕ = \frac{z}{\sqrt{x^{2} + y^{2} + z^{2}}} \Rightarrow \sin^{2} ϕ = \frac{x^{2} + y^{2}}{x^{2} + y^{2} + z^{2}}$ ... (iii)

Now, distance of P from x-axis $= \sqrt{y^{2} + z^{2}}$

$= \sqrt{γ^{2} - x^{2}}$ [Using (i)]

$= \sqrt{γ^{2} - \cos^{2} θ (x^{2} + y^{2})}$ [Using (ii)]

$= \sqrt{γ^{2} - \cos^{2} θ \sin^{2} ϕ \cdot γ^{2}}$ [Using (iii)]

$= γ \sqrt{1 - \cos^{2} θ \sin^{2} ϕ}$ .

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