Animated Solution for Mathematics - Three Dimensional Geometry: The shortest distance between the lines $x+1=2y=-12z$ and $x=y+2=6z-6$ is

A Journey into 3D Space

Welcome, future engineer. Today, we are not just solving a problem; we are navigating the architecture of three-dimensional space. Imagine two laser beams crossing each other in a dark room. They don't touch, they don't intersect, and they aren't parallel. These are what we call 'skew lines.'

Finding the shortest distance between them is a classic JEE Advanced challenge—it tests your ability to visualize, standardize, and execute with precision.

The Art of Standardization

Before we touch any complex formulas, we must respect the geometry. The lines provided, $L_1:x+1=2y=-12z$ and $L_2:x=y+2=6z-6$, are currently in a 'disguised' state. In the world of coordinate geometry, the symmetric form of a line is sacred:

Notice the coefficients of $x,y,$ and $z$ must be $1$. In our given equations, they are not. For $L_1$, we have $2y$ and $-12z$. If we don't fix this, our direction vector will be fundamentally flawed.

We divide the entire equation by $12$ to obtain:

Now, the line speaks clearly. We can extract our anchor point ${\vec{a}}_1=(-1,0,0)$ and our direction vector ${\vec{b}}_1=12\hat{i}+6\hat{j}-\hat{k}$.

We repeat this for $L_2$. The equation $x=y+2=6z-6$ requires us to factor out the $6$ from the $z$ term. This gives us:

Our second anchor point is ${\vec{a}}_2=(0,-2,1)$ and our second direction vector is ${\vec{b}}_2=6\hat{i}+6\hat{j}+\hat{k}$.

The Common Perpendicular

Now that we have our vectors, we need to find the 'bridge' between these lines. The shortest distance is the length of the line segment that is perpendicular to both skew lines. How do we find a direction that is perpendicular to two given vectors? The cross product!

We calculate ${\vec{b}}_1\times {\vec{b}}_2$ using the determinant method:

Expanding this, we get $12\hat{i}-18\hat{j}+36\hat{k}$. To make our lives easier, we factor out a $6$, leaving us with $6(2\hat{i}-3\hat{j}+6\hat{k})$.

This vector represents the direction of the shortest path. Its magnitude is:

The Final Projection

We are almost at the finish line. The shortest distance $d$ is the projection of the vector connecting our two points, ${\vec{a}}_2-{\vec{a}}_1$, onto the common normal vector we just found.

First, the connecting vector: ${\vec{a}}_2-{\vec{a}}_1=(0-(-1))\hat{i}+(-2-0)\hat{j}+(1-0)\hat{k}=\hat{i}-2\hat{j}+\hat{k}$.

Now, we apply the formula:

Substituting our values, the numerator becomes the absolute value of the dot product of $(1,-2,1)$ and $(12,-18,36)$:

Dividing this by our magnitude of $42$, we get:

And there it is. The shortest distance is $2$ units. It is elegant, it is precise, and it is the result of methodical, step-by-step logic. Never let the complexity of 3D vectors intimidate you. Break the problem down, standardize your variables, and let the geometry guide your hand.