Two system of rectangular axes have the same origin. If a plane cuts them at distances a, b, c and a

Q: Wait, why does the perpendicular distance $p$ stay the same? Doesn't rotating the axes change the coordinates of the plane?

That is a brilliant question! You are absolutely right that the coordinates $(x, y, z)$ of any point on the plane will change when you rotate the axes. However, the perpendicular distance from the origin to the plane is a geometric property, not a coordinate-dependent one. Think of it like the height of a mountain; it doesn't matter if you measure it from the North or the South, the height remains the same. In math, we call this an invariant.

Q: Can I solve this using the normal vector approach instead of the intercept form?

Yes, you absolutely can! You could define the plane using a normal vector $\vec{n}$ and the distance $p$ as $\vec{r} \cdot \hat{n} = p$. While this is a more general and powerful method, the intercept form is specifically tailored for this problem because the question explicitly gives you the intercepts $a, b, c$. In JEE Advanced, choosing the path of least resistance is a skill in itself!

Q: Is this relationship true for any rotation of the axes?

Yes, it is! Rotation is an orthogonal transformation. Orthogonal transformations preserve the dot product and, consequently, the distance from the origin. Whether you rotate the axes by 30 degrees or 300 degrees, the perpendicular distance $p$ remains constant, and therefore, the relationship $\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} = \frac{1}{a'^2} + \frac{1}{b'^2} + \frac{1}{c'^2}$ will always hold true.

Question

Two system of rectangular axes have the same origin. If a plane cuts them at distances $a, b, c$ and $a', b', c'$ from the origin then

Options:

(A) $\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} - \frac{1}{a'^2} - \frac{1}{b'^2} - \frac{1}{c'^2} = 0$

(B) $\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} + \frac{1}{a'^2} + \frac{1}{b'^2} + \frac{1}{c'^2} = 0$

(C) $\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} - \frac{1}{a'^2} - \frac{1}{b'^2} - \frac{1}{c'^2} = 0$

(D) $\frac{1}{a^2} - \frac{1}{b^2} - \frac{1}{c^2} + \frac{1}{a'^2} + \frac{1}{b'^2} + \frac{1}{c'^2} = 0$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineers. Today, we are going to peel back the layers of a problem that tests not just your algebraic manipulation, but your geometric intuition.

Imagine a plane slicing through 3D space. We have two sets of axes, $x, y, z$ and $x', y', z'$, both sharing the same origin $O$. The plane cuts the first set at $a, b, c$ and the second at $a', b', c'$.

The question asks for the relationship between these intercepts. At first glance, this looks like a nightmare of coordinate transformations. But the secret lies in the concept of an **invariant**. The perpendicular distance from the origin to the plane is a physical reality—it does not care about your coordinate system.

### The Intercept Form: Our First Tool

In the first system, the equation of the plane is given by the intercept form:
$$\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$$

This is a beautiful, symmetric equation. It tells us exactly where the plane hits the axes. Now, let us find the perpendicular distance $p$ from the origin $(0,0,0)$ to this plane.

Using the standard formula for the distance from a point to a plane $Ax + By + Cz + D = 0$, we have $p = \frac{|D|}{\sqrt{A^2 + B^2 + C^2}}$. For our plane, this becomes:

$$p = \frac{|-1|}{\sqrt{(\frac{1}{a})^2 + (\frac{1}{b})^2 + (\frac{1}{c})^2}}$$

Squaring both sides gives us $p^2 = \frac{1}{\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}}$. If we take the reciprocal, we get the elegant relation:

$$\frac{1}{p^2} = \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}$$

### The Power of Invariance

Here is the 'Aha!' moment. Since the plane and the origin are fixed in space, the perpendicular distance $p$ is constant. It is an invariant.

It does not matter if you look at the plane from the perspective of the first coordinate system or the second. The distance $p$ remains the same. Therefore, when we repeat this process for the second coordinate system, we get the exact same perpendicular distance $p$.

The equation for the second system is $\frac{x'}{a'} + \frac{y'}{b'} + \frac{z'}{c'} = 1$. Following the same algebraic steps, we arrive at:

$$\frac{1}{p^2} = \frac{1}{a'^2} + \frac{1}{b'^2} + \frac{1}{c'^2}$$

### The Final Synthesis

Now, we have two expressions for the same quantity, $\frac{1}{p^2}$. Since they are both equal to $\frac{1}{p^2}$, they must be equal to each other:

$$\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} = \frac{1}{a'^2} + \frac{1}{b'^2} + \frac{1}{c'^2}$$

By bringing all terms to one side, we arrive at the final result:

$$\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} - \frac{1}{a'^2} - \frac{1}{b'^2} - \frac{1}{c'^2} = 0$$

This is the beauty of physics and math—finding the constant in a world of variables. You didn't need to perform complex rotation matrices or coordinate transformations. You simply needed to identify the invariant. Keep this mindset, and you will conquer the most difficult problems in **JEE Advanced**.

Animated Solution for Mathematics - Three Dimensional Geometry: Two system of rectangular axes have the same origin. If a plane cuts them at distances $a, b, c$ and $a^{'}, b^{'}, c^{'}$ from the origin then

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Visualized Solution

Two Coordinate Systems

Plane Intercepts (System 1)

Equation of Plane (System 1)

Plane Intercepts (System 2)

Equation of Plane (System 2)

The Invariant Perpendicular Distance

Distance $p$ in System 1

Squaring the Relation (System 1)

Distance $p$ in System 2

Squaring the Relation (System 2)

Equating the Invariants

Final Result

The Sigma Insight: Equation of a Plane

Analyzing the Setup

The Intercept Form

The Power of Invariance

The Final Synthesis

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