Key Concepts and Formulas

1. Cartesian Coordinates in 3D: A point $P$ in three-dimensional space is represented by its coordinates $(x, y, z)$. The origin $O$ is at $(0,0,0)$.
2. Projection of a Point: The projection of a point $P(x,y,z)$ onto the $xy$-plane is the point $Q(x,y,0)$. This means $Q$ shares the same $x$ and $y$ coordinates as $P$, but its $z$-coordinate is zero.
3. Distance from a Point to an Axis: The distance of a point $P(x,y,z)$ from the $x$-axis is given by the formula $\sqrt{y^2+z^2}$.
4. Trigonometric Relationships in 3D (Spherical Coordinates Adaptation): For a point $P(x,y,z)$ at a distance $OP=\gamma$ from the origin, if $\phi$ is the angle between $OP$ and the positive $z$-axis, and $\theta$ is the angle between the projection $OQ$ (of $OP$ onto the $xy$-plane) and the positive $x$-axis, then the coordinates can be expressed as:
   • $x = \gamma \sin \phi \cos \theta$
   • $y = \gamma \sin \phi \sin \theta$
   • $z = \gamma \cos \phi$

---

Step-by-Step Solution

Step 1: Define Point P and its Projection Q in Cartesian Coordinates Let the coordinates of point $P$ in the first octant be $(x, y, z)$. The origin $O$ is $(0,0,0)$. We are given that $Q$ is the projection of $P(x,y,z)$ onto the $xy$-plane.

• Why this step? This sets up the fundamental coordinate system and defines the points involved, which is essential for translating the geometric problem into algebraic expressions. Therefore, the coordinates of $Q$ are $(x, y, 0)$.

Step 2: Express the $z$-coordinate of P in terms of $\gamma$ and $\phi$ We are given that $OP = \gamma$ and $\phi$ is the angle between $OP$ and the positive $z$-axis. Consider the right-angled triangle formed by the origin $O(0,0,0)$, point $P(x,y,z)$, and the point $Z_P(0,0,z)$ on the $z$-axis (which is the projection of $P$ onto the $z$-axis). In this right triangle, the hypotenuse is $OP = \gamma$, and the side adjacent to angle $\phi$ is $OZ_P = z$. Using the cosine function: $\cos \phi = \frac{\text{adjacent side}}{\text{hypotenuse}} = \frac{z}{OP}$ Substituting $OP = \gamma$: $\cos \phi = \frac{z}{\gamma}$ Solving for $z$: $\mathbf{z = \gamma \cos \phi}$

• Why this step? This directly translates the given angle $\phi$ and distance $\gamma$ into the $z$-coordinate of $P$, which is one of the components needed for the final distance calculation.

Step 3: Express the length of $OQ$ in terms of $\gamma$ and $\phi$ Now, let's consider the relationship between $OP$, $OQ$, and $z$. The points $O(0,0,0)$, $Q(x,y,0)$, and $P(x,y,z)$ form a right-angled triangle $\triangle OQP$. The right angle is at $Q$ because $PQ$ is perpendicular to the $xy$-plane (where $OQ$ lies). The length of side $PQ$ is the $z$-coordinate of $P$. By the Pythagorean theorem in $\triangle OQP$: $OP^2 = OQ^2 + PQ^2$ We know $OP = \gamma$ and $PQ = z$. $\gamma^2 = OQ^2 + z^2$ Rearranging to find $OQ^2$: $OQ^2 = \gamma^2 - z^2$ Substitute the expression for $z$ from Step 2 ($z = \gamma \cos \phi$): $OQ^2 = \gamma^2 - (\gamma \cos \phi)^2$ $OQ^2 = \gamma^2 - \gamma^2 \cos^2 \phi$ Factor out $\gamma^2$: $OQ^2 = \gamma^2 (1 - \cos^2 \phi)$ Using the trigonometric identity $\sin^2 \phi + \cos^2 \phi = 1 \implies 1 - \cos^2 \phi = \sin^2 \phi$: $OQ^2 = \gamma^2 \sin^2 \phi$ Taking the square root (since $P$ is in the first octant, $\phi$ is between $0$ and $\pi/2$, so $\sin \phi > 0$): $\mathbf{OQ = \gamma \sin \phi}$

• Why this step? The length of $OQ$ is crucial for relating the angle $\theta$ to the $x$ and $y$ coordinates in the next step. This step establishes $OQ$ in terms of the initial parameters.

Step 4: Express the $x$ and $y$-coordinates of P in terms of $\gamma$, $\phi$, and $\theta$ We are given that $\theta$ is the angle between $OQ$ and the positive $x$-axis. Point $Q(x,y,0)$ lies in the $xy$-plane. We know its distance from the origin $O$ is $OQ = \gamma \sin \phi$. In the $xy$-plane, for point $Q(x,y)$ and angle $\theta$ with the positive $x$-axis, we use basic trigonometry: $\cos \theta = \frac{x}{OQ} \implies x = OQ \cos \theta$ $\sin \theta = \frac{y}{OQ} \implies y = OQ \sin \theta$ Substitute $OQ = \gamma \sin \phi$ (from Step 3) into these equations: $\mathbf{x = \gamma \sin \phi \cos \theta}$ $\mathbf{y = \gamma \sin \phi \sin \theta}$

• Why this step? This completes the conversion of all Cartesian coordinates $(x,y,z)$ into expressions involving the given parameters $(\gamma, \theta, \phi)$, providing all necessary components for the final calculation.

Step 5: Calculate the Distance of P from the $x$-axis The problem asks for the distance of point $P(x,y,z)$ from the $x$-axis. Using the formula for distance from an axis: $\text{Distance from } x\text{-axis} = \sqrt{y^2+z^2}$ Now, substitute the expressions for $y$ (from Step 4) and $z$ (from Step 2): $\text{Distance} = \sqrt{(\gamma \sin \phi \sin \theta)^2 + (\gamma \cos \phi)^2}$ $\text{Distance} = \sqrt{\gamma^2 \sin^2 \phi \sin^2 \theta + \gamma^2 \cos^2 \phi}$ Factor out $\gamma^2$ from under the square root: $\text{Distance} = \sqrt{\gamma^2 (\sin^2 \phi \sin^2 \theta + \cos^2 \phi)}$ Take $\gamma$ out of the square root: $\text{Distance} = \gamma \sqrt{\sin^2 \phi \sin^2 \theta + \cos^2 \phi}$ To match one of the options, we use the trigonometric identity $\cos^2 \phi = 1 - \sin^2 \phi$: $\text{Distance} = \gamma \sqrt{\sin^2 \phi \sin^2 \theta + (1 - \sin^2 \phi)}$ Rearrange the terms: $\text{Distance} = \gamma \sqrt{1 + \sin^2 \phi \sin^2 \theta - \sin^2 \phi}$ Factor out $\sin^2 \phi$ from the last two terms: $\text{Distance} = \gamma \sqrt{1 + \sin^2 \phi (\sin^2 \theta - 1)}$ Now, use the identity $\sin^2 \theta - 1 = - \cos^2 \theta$: $\text{Distance} = \gamma \sqrt{1 + \sin^2 \phi (-\cos^2 \theta)}$ $\mathbf{\text{Distance} = \gamma \sqrt{1 - \sin^2 \phi \cos^2 \theta}}$

• Why this step? This is the final calculation that directly answers the question using all the coordinate expressions derived. The algebraic manipulation ensures the result matches the provided options.

---

Common Mistakes & Tips

• Confusing Angles: Pay close attention to which angle ($\theta$ or $\phi$) is defined with respect to which axis or plane. $\theta$ is in the $xy$-plane, while $\phi$ is with the $z$-axis.
• Incorrect Distance Formulas: Remember that the distance from a point $(x,y,z)$ to the $x$-axis is $\sqrt{y^2+z^2}$, not just $y$ or $z$. Similarly for other axes.
• Algebraic/Trigonometric Errors: Be careful with squaring terms, factoring, and applying trigonometric identities. Double-check each step in the simplification process.
• Visualizing the Geometry: Drawing a simple sketch of the point $P$, its projection $Q$, and the angles $\theta$ and $\phi$ can greatly help in setting up the coordinate relationships correctly. Think of the projection of $OP$ onto the $xy$-plane ($OQ$) and how it relates to $OP$ and the $z$-coordinate.

---

Summary

This problem required us to translate a geometric description of a point in 3D space into its Cartesian coordinates using given distances and angles. By systematically deriving the $z$-coordinate from $OP$ and $\phi$, then the length of $OQ$, and finally the $x$ and $y$-coordinates from $OQ$ and $\theta$, we obtained all components of $P(x,y,z)$. The final step involved using the formula for the distance of a point from the $x$-axis and simplifying the resulting trigonometric expression to match the given options.

The final answer is $\boxed{\text{(A)}}$.