Animated Solution for Mathematics - Three Dimensional Geometry: If the square of the shortest distance between the lines $\frac{x-2}{1}=\frac{y-1}{2}=\frac{z+3}{-3}$ and $\frac{x+1}{2}=\frac{y+3}{4}=\frac{z+5}{-5}$ is $\frac{m}{n}$ where m, n are coprime numbers, then $m+n$ is equal to:

Understanding Skew Lines

Imagine you are standing in a vast, empty room. You have two thin, straight wires suspended in the air. They are not parallel, and they do not touch.

This is the essence of skew lines in 3D space. They are like two ships passing in the night, separated by a gap that we are tasked to measure. This problem is not just about plugging numbers into a formula; it is about visualizing the bridge that connects these two infinite paths.

Every line in 3D space is defined by a point it passes through and a direction in which it travels. Look at the first line:

By comparing this to the standard form $\frac{x-x_0}{a}=\frac{y-y_0}{b}=\frac{z-z_0}{c}$, we immediately extract our first anchor point $P_1(2,1,-3)$ and its direction vector ${\vec{d}}_1=\hat{i}+2\hat{j}-3\hat{k}$.

Now, look at the second line:

Similarly, we find $P_2(-1,-3,-5)$ and ${\vec{d}}_2=2\hat{i}+4\hat{j}-5\hat{k}$. We have our coordinates and our vectors; we are ready to build our bridge.

The shortest distance between these two lines is the length of the segment that is perpendicular to both. Think of this as the shortest path between two parallel planes, each containing one of our lines.

To find this, we need a vector that is perpendicular to both ${\vec{d}}_1$ and ${\vec{d}}_2$. This is where the cross product shines. We calculate $\vec{n}={\vec{d}}_1\times {\vec{d}}_2$.

Using the determinant method, we set up the matrix with $\hat{i},\hat{j},\hat{k}$ in the first row, $⟨1,2,-3⟩$ in the second, and $⟨2,4,-5⟩$ in the third. Expanding this, we get:

This simplifies beautifully to $\vec{n}=2\hat{i}-1\hat{j}+0\hat{k}$. This vector $\vec{n}$ is the normal to the planes containing our lines.

Now, we need to measure the gap. We create a vector connecting our two lines:

The shortest distance is simply the projection of this connecting vector $\overrightarrow{P_2{P}_1}$ onto the common normal $\vec{n}$. Mathematically, this is the absolute value of the dot product of $\overrightarrow{P_2{P}_1}$ and the unit normal vector.

We calculate the dot product:

The magnitude of our normal vector is $|\vec{n}|=\sqrt{2^2+(-1{)}^2+0^2}=\sqrt{5}$. Thus, the distance $d$ is:

The problem asks for the square of this distance. So, we calculate:

We are told this equals $\frac{m}{n}$ where $m$ and $n$ are coprime. Since $4$ and $5$ share no common factors, $m=4$ and $n=5$.

The final step is to find $m+n$, which is $4+5=9$. You have navigated the 3D landscape, mastered the cross product, and arrived at the solution: 9.