Animated Solution for Mathematics - Three Dimensional Geometry: The shortest distance between the lines $\frac{x-3}{2}=\frac{y+15}{-7}=\frac{z-9}{5}$ and $\frac{x+1}{2}=\frac{y-1}{1}=\frac{z-9}{-3}$ is

A 3D Adventure

Imagine you are standing on a bridge, watching two airplanes fly through the sky at different altitudes. From your vantage point, their paths might look like they cross, but they never actually collide because one is flying higher than the other.

In the world of 3D geometry, these are called skew lines. They are not parallel, and they never intersect. Our mission is to find the absolute shortest distance between these two lines, a task that requires us to master the art of vector projection.

Extracting the DNA of the Lines

Before we dive into the math, we must ensure our lines are in the standard symmetric form:

Looking at our first line, $\frac{x-3}{2}=\frac{y+15}{-7}=\frac{z-9}{5}$, we can immediately extract a point ${\vec{a}}_1=3\hat{i}-15\hat{j}+9\hat{k}$ and a direction vector ${\vec{b}}_1=2\hat{i}-7\hat{j}+5\hat{k}$.

For the second line, $\frac{x+1}{2}=\frac{y-1}{1}=\frac{z-9}{-3}$, we find our second point ${\vec{a}}_2=-\hat{i}+\hat{j}+9\hat{k}$ and direction vector ${\vec{b}}_2=2\hat{i}+\hat{j}-3\hat{k}$. These vectors are the DNA of our lines, defining exactly where they are and where they are headed.

The Bridge and the Normal

To find the shortest distance, we need a bridge—a vector connecting our two lines. We calculate this by subtracting our points:

Now, we need the direction of the shortest path, which must be perpendicular to both lines. We find this using the cross product: $\vec{n}={\vec{b}}_1\times {\vec{b}}_2$.

Setting up the determinant, we get:

Expanding this, we get $\hat{i}(21-5)-\hat{j}(-6-10)+\hat{k}(2-(-14))$, which simplifies to $16\hat{i}+16\hat{j}+16\hat{k}$. We can factor out the 16 to get $16(\hat{i}+\hat{j}+\hat{k})$.

The Final Projection

The shortest distance is simply the projection of our bridge vector onto the normal vector. The formula is:

First, the dot product:

Next, the magnitude of the normal vector:

Finally, we divide:

Rationalizing the denominator, we get $4\sqrt{3}$. We have successfully navigated the 3D space and found the shortest distance between these two skew lines. Remember, the key is always to identify your points and directions correctly, and the rest is just elegant vector algebra!