Animated Solution for Mathematics - Three Dimensional Geometry: The shortest distance between the lines $\frac{x-5}{1}=\frac{y-2}{2}=\frac{z-4}{-3}$ and $\frac{x+3}{1}=\frac{y+5}{4}=\frac{z-1}{-5}$ is

Bridging the Gap

Imagine you are standing in a vast, three-dimensional void. Two infinite lines are floating before you. They are not parallel, yet they never touch.

In the world of geometry, we call these skew lines. They are like two ships passing in the night, separated by a specific, unyielding gap. Today, we are going to calculate that gap—the shortest distance between them.

Every line in 3D space has a unique "DNA"—a point it passes through and a direction in which it travels. Our given lines are in Cartesian form:

For our first line, $\frac{x-5}{1}=\frac{y-2}{2}=\frac{z-4}{-3}$, we can immediately extract the point ${\vec{a}}_1=5\hat{i}+2\hat{j}+4\hat{k}$ and the direction vector ${\vec{b}}_1=\hat{i}+2\hat{j}-3\hat{k}$.

Now, look closely at the second line: $\frac{x+3}{1}=\frac{y+5}{4}=\frac{z-1}{-5}$. Here is where many students stumble!

The equation is $x+3$, which is $x-(-3)$. So, our point is ${\vec{a}}_2=-3\hat{i}-5\hat{j}+\hat{k}$. The direction vector is ${\vec{b}}_2=\hat{i}+4\hat{j}-5\hat{k}$. Always watch those signs; they are the silent traps of JEE problems.

To find the shortest distance $d$, we use the projection formula:

Think of this geometrically. We are finding the vector connecting the two lines, ${\vec{a}}_2-{\vec{a}}_1$, and projecting it onto the common perpendicular, which is the direction of ${\vec{b}}_1\times {\vec{b}}_2$. It is elegant, precise, and powerful.

First, let's find the connecting vector:

Next, we determine the common perpendicular direction via the cross product:

Expanding this determinant, we get:

The magnitude of this cross product is:

Finally, the dot product of our connecting vector and the cross product vector is:

Plugging these into our formula, we get:

Rationalizing the denominator by multiplying by $\frac{\sqrt{3}}{\sqrt{3}}$, we arrive at:

And there it is! The shortest distance is $6\sqrt{3}$ units. You have successfully navigated the geometry of skew lines. Keep this visualization in your toolkit—it is a fundamental pillar of 3D geometry in JEE Advanced.