Animated Solution for Mathematics - Three Dimensional Geometry: The distance of the point $(1,-5,9)$ from the plane $x-y+z=5$ measured along the line $x=y=z$ is :

Imagine you are standing in a vast, three-dimensional room. You are at a specific point $P(1,-5,9)$, and there is a flat plane stretching out below you, defined by the equation $x-y+z=5$.

Usually, when we talk about the distance from a point to a plane, our minds immediately jump to the shortest possible path—the perpendicular distance. But today, we are going to take a different route. The problem asks us to measure the distance along the line $x=y=z$.

This is not the shortest path; it is a specific, directed journey through space. Let's embark on this journey together.

Parametric Navigation

To find the distance, we first need to know where our path intersects the plane. We have a starting point $P(1,-5,9)$ and a direction.

The line $x=y=z$ can be written in its symmetric form as:

This tells us that the direction ratios of our path are $(1,1,1)$.

Now, let's construct the equation of the line $PQ$ that starts at $P$ and moves in this direction. Using the parametric form, we can express any point $Q$ on this line as:

Here, $\lambda$ is our scalar parameter. As $\lambda$ changes, we move along the line. Our goal is to find the specific value of $\lambda$ that lands us exactly on the plane $x-y+z=5$.

Where Paths Meet

Since the point $Q(\lambda +1,\lambda -5,\lambda +9)$ must lie on the plane, its coordinates must satisfy the plane's equation. Let's substitute these expressions into the equation $x-y+z=5$:

Now, let's carefully expand this. Be mindful of the negative sign before the $y$-coordinate:

Look at the beauty of the algebra here: the $\lambda$ and $-\lambda$ cancel each other out, leaving us with a simple linear equation:

This negative value for $\lambda$ is perfectly fine. It just means that to reach the plane, we have to travel in the opposite direction of the vector $(1,1,1)$ from our starting point $P$.

Now that we have $\lambda =-10$, we can find the exact coordinates of our intersection point $Q$:

So, our point $Q$ is $(-9,-15,-1)$.

Finally, we calculate the distance $PQ$ using the standard 3D distance formula:

Simplifying $\sqrt{300}$, we get the final result:

We have successfully navigated the 3D space, avoided the trap of the perpendicular distance formula, and arrived at our destination. Remember, in JEE Advanced, the key is often not just knowing the formulas, but understanding the geometry behind them. Keep visualizing, keep calculating, and most importantly, keep falling in love with the elegance of mathematics.