Let P(x, y, z) be a point in the first octant, whose projection in the xy-plane is the point Q. Let

Q: Wait, why is the distance from the $x$-axis just $\sqrt{y^2 + z^2}$? Shouldn't it involve $x$?

That is a fantastic question! It is a common trap to think that the distance from an axis must involve the coordinate along that axis. However, think about the definition of distance from a line. The distance from a point $P(x, y, z)$ to the $x$-axis is the length of the perpendicular segment dropped from $P$ to the $x$-axis. The closest point on the $x$-axis to $P$ is $M(x, 0, 0)$. Using the 3D distance formula, the distance is $\sqrt{(x-x)^2 + (y-0)^2 + (z-0)^2}$, which simplifies to $\sqrt{y^2 + z^2}$. The $x$-coordinate essentially 'cancels out' because you are moving parallel to the $x$-axis to reach the foot of the perpendicular.

Q: I got stuck at $\sqrt{\gamma^2 \sin^2 \phi \sin^2 \theta + \gamma^2 \cos^2 \phi}$. How do I know which identity to use to match the options?

When you are stuck in a trigonometric maze, always look at the 'destination'—the options provided. The options contain $\cos^2 \theta$. Your current expression contains $\sin^2 \theta$. This is a massive hint! It tells you that you must convert the $\sin^2 \theta$ term into something involving $\cos^2 \theta$. Using the identity $\sin^2 \theta = 1 - \cos^2 \theta$ is the natural path forward. Trust the structure of the options; they are your roadmap.

Q: Is this just spherical coordinates? Can I just use the standard formulas?

You have a sharp eye! Yes, this is exactly the definition of spherical coordinates. In standard spherical coordinates, $x = \gamma \sin \phi \cos \theta$, $y = \gamma \sin \phi \sin \theta$, and $z = \gamma \cos \phi$. If you recognize this immediately, you can jump straight to the substitution step. However, for JEE Advanced, it is always safer to derive the components geometrically as we did, just to ensure you haven't swapped $\phi$ and $\theta$ or confused the sine and cosine terms. Never lose the geometric intuition for the sake of a formula.

Question

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

Options:

(A) $\gamma\sqrt{1 - \sin^2 \phi \cos^2 	heta}$

(B) $\gamma\sqrt{1 - \sin^2 	heta \cos^2 \phi}$

(C) $\gamma\sqrt{1 + \cos^2 \phi \sin^2 	heta}$

(D) $\gamma\sqrt{1 + \cos^2 	heta \sin^2 \phi}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a point $P(x, y, z)$ suspended in the first octant of a 3D coordinate system. We are given the distance from the origin $O(0, 0, 0)$ to $P$ as $\gamma$.

The angle between the vector $\vec{OP}$ and the positive $z$-axis is defined as $\phi$. Additionally, the projection of $OP$ onto the $xy$-plane, denoted as $OQ$, makes an angle $	heta$ with the positive $x$-axis.

### The Vertical Slice

Consider the right-angled triangle formed by the origin $O$, the point $P$, and the projection point $Q(x, y, 0)$ on the $xy$-plane. The hypotenuse is $OP = \gamma$.

Using basic trigonometry, the $z$-coordinate is the adjacent side to angle $\phi$:
$$z = \gamma \cos \phi$$

The length of the projection $OQ$ is the side opposite to $\phi$:
$$OQ = \gamma \sin \phi$$

### The Floor Plan

Now, focus on the $xy$-plane. The vector $OQ$ has length $\gamma \sin \phi$ and makes an angle $	heta$ with the $x$-axis. The $y$-coordinate of $P$ is the vertical component of this projection:
$$y = OQ \sin 	heta = (\gamma \sin \phi) \sin 	heta$$

Similarly, the $x$-coordinate is the horizontal component of this projection:
$$x = OQ \cos 	heta = (\gamma \sin \phi) \cos 	heta$$

### The Distance Calculation

We seek the distance of $P$ from the $x$-axis. Let $M$ be the foot of the perpendicular from $P$ to the $x$-axis, located at $(x, 0, 0)$. The distance $PM$ is given by:
$$PM = \sqrt{(x-x)^2 + (y-0)^2 + (z-0)^2} = \sqrt{y^2 + z^2}$$

Substituting our expressions for $y$ and $z$:
$$PM = \sqrt{(\gamma \sin \phi \sin 	heta)^2 + (\gamma \cos \phi)^2}$$

### The Algebraic Dance

Expanding the squares, we obtain:
$$PM = \sqrt{\gamma^2 \sin^2 \phi \sin^2 	heta + \gamma^2 \cos^2 \phi}$$

Factoring out $\gamma^2$:
$$PM = \gamma \sqrt{\sin^2 \phi \sin^2 	heta + \cos^2 \phi}$$

To simplify, we use the identity $\sin^2 	heta = 1 - \cos^2 	heta$:
$$PM = \gamma \sqrt{\sin^2 \phi (1 - \cos^2 	heta) + \cos^2 \phi}$$

Expanding the terms inside the square root:
$$PM = \gamma \sqrt{\sin^2 \phi - \sin^2 \phi \cos^2 	heta + \cos^2 \phi}$$

Grouping the trigonometric terms $\sin^2 \phi + \cos^2 \phi = 1$, we arrive at the final result:
**$$PM = \gamma \sqrt{1 - \sin^2 \phi \cos^2 	heta}$$**

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Visualized Solution

Visualizing the 3D Setup

Projection on the $x y$ -plane

Angle in the $x y$ -plane

Finding the $z$ -coordinate

Finding the Projection Length $O Q$

Finding the $y$ -coordinate

Distance from the $x$ -axis

Distance Formula

Substituting $y$ and $z$

Expanding the Squares

Trigonometric Substitution

Expanding the Bracket

Final Simplification

The Sigma Insight: Direction Cosines and Direction Ratios

Analyzing the Setup

The Vertical Slice

The Floor Plan

The Distance Calculation

The Algebraic Dance

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Visualized Solution

Visualizing the 3D Setup

Projection on the .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } xy-plane

Angle in the .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } xy-plane

Finding the .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } z-coordinate

Finding the Projection Length .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } OQ

Finding the .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y-coordinate

Distance from the .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x-axis