Animated Solution for Mathematics - Three Dimensional Geometry: Let $P$ be the point of intersection of the lines $\frac{x-2}{1}=\frac{y-4}{5}=\frac{z-2}{1}$ and $\frac{x-3}{2}=\frac{y-2}{3}=\frac{z-3}{2}$. Then, the shortest distance of $P$ from the line $4x=2y=z$ is

Welcome, future engineer. Today, we are not just solving a problem; we are navigating the architecture of three-dimensional space. Many students look at lines in 3D and see only a mess of variables and fractions.

But I want you to see something else: I want you to see paths. Imagine two lines, $L_1$ and $L_2$, cutting through the void of 3D space. They are destined to meet at a single, precise point $P$. Our mission is to find this point and then determine how far it stands from a third, mysterious line, $L_3$.

The Meeting Point

To find where two lines intersect, we must first understand their nature. We are given:

We cannot simply guess the intersection. We need a language to describe every point on these lines. That language is parameterization.

By setting $L_1$ equal to a parameter $\lambda$, we can express any point on $L_1$ as $P(\lambda +2,5\lambda +4,\lambda +2)$. Similarly, for $L_2$, we introduce a different parameter, $\mu$, giving us $P(2\mu +3,3\mu +2,2\mu +3)$.

Why different parameters? Because the lines are independent entities. They don't 'know' about each other until they collide at point $P$.

At that collision, the $x$, $y$, and $z$ coordinates must be identical. This gives us a system of equations:

1. $\lambda +2=2\mu +3\Rightarrow \lambda -2\mu =1$

2. $5\lambda +4=3\mu +2\Rightarrow 5\lambda -3\mu =-2$

Solving this system is a test of your algebraic precision. Substituting $\lambda =2\mu +1$ into the second equation yields:

This simplifies to $7\mu =-7$, or $\mu =-1$. Consequently, $\lambda =-1$. Plugging these back into our parameterization, we find the intersection point $P(1,-1,1)$.

The Target Line

Now, we turn our attention to the third line, $L_3$, defined by $4x=2y=z$. This looks intimidating, but it is merely a line in disguise. To reveal its true form, we divide by the common factor of 4, yielding the symmetric form:

Here, $k$ is our new parameter. Any point $Q$ on this line can be represented as $(k,2k,4k)$. This point $Q$ is our destination.

We want to find the shortest distance from our intersection point $P(1,-1,1)$ to this line. Geometrically, this distance is the length of the perpendicular segment dropped from $P$ to $Q$.

The Perpendicularity Condition

This is the heart of the problem. If $PQ$ is the shortest distance, then the vector $\overrightarrow{PQ}$ must be perpendicular to the line $L_3$. The direction vector of $L_3$ is $\vec{d}=\hat{i}+2\hat{j}+4\hat{k}$.

First, let us define the vector $\overrightarrow{PQ}$:

For $\overrightarrow{PQ}$ to be perpendicular to $\vec{d}$, their dot product must vanish into nothingness:

Expanding this, we get $k-1+4k+2+16k-4=0$. Combining the terms, we find $21k-3=0$, which leads us to the elegant result $k=\frac{1}{7}$.

We are almost there. We have the parameter $k=\frac{1}{7}$, which defines the exact point $Q$ on the line $L_3$ that is closest to $P$. Now, we simply calculate the magnitude of the vector $\overrightarrow{PQ}$:

Substituting $k=\frac{1}{7}$:

Simplifying $\sqrt{126}$ as $\sqrt{9\times 14}=3\sqrt{14}$, we arrive at our final answer:

Look at that result. It is not just a number; it is the culmination of logical steps, geometric visualization, and algebraic rigor. You have navigated the intersection, identified the perpendicular path, and calculated the distance.