Animated Solution for Mathematics - Three Dimensional Geometry: Let $L_1:\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ $L_2:\frac{x-2}{3}=\frac{y-4}{4}=\frac{z-5}{5}$ and be two lines. Then which of the following points lies on the line of the shortest distance between L₁ and $L_2$?

Conquering Skew Lines

Welcome, future engineer. Today, we are not just solving a problem; we are visualizing the architecture of 3D space. We are dealing with two skew lines, $L_1$ and $L_2$.

These lines are like two airplanes flying at different altitudes, never touching, yet passing near each other. Our mission is to find the 'bridge'—the unique line of shortest distance that connects them.

The Direction of the Bridge

First, let us extract the DNA of these lines. From the symmetric equations, we immediately see their direction vectors.

For $L_1$, the direction is ${\vec{d}}_1=2\hat{i}+3\hat{j}+4\hat{k}$. For $L_2$, it is ${\vec{d}}_2=3\hat{i}+4\hat{j}+5\hat{k}$.

Now, think geometrically. The shortest distance line must be perpendicular to both $L_1$ and $L_2$. If you have two vectors and you need a third vector perpendicular to both, the go-to tool is the cross product.

We define the direction of our bridge, $\vec{n}$, as $\vec{n}={\vec{d}}_1\times {\vec{d}}_2$. Calculating this determinant, we find:

This vector $\vec{n}$ is the compass pointing along our shortest distance line.

Trapping the Line in Planes

Here is where the magic happens. Instead of chasing parametric coordinates, let us trap this line. Imagine a plane, $P_1$, that contains the line $L_1$ and our bridge vector $\vec{n}$.

Since this plane contains both ${\vec{d}}_1$ and $\vec{n}$, its normal vector ${\vec{n}}_1$ must be ${\vec{d}}_1\times \vec{n}$. Performing the cross product:

Using the point $(1,2,3)$ from $L_1$, the equation of plane $P_1$ becomes $-11(x-1)-2(y-2)+7(z-3)=0$, which simplifies beautifully to:

We repeat this for $P_2$, the plane containing $L_2$ and $\vec{n}$. The normal ${\vec{n}}_2={\vec{d}}_2\times \vec{n}$ yields ${\vec{n}}_2=-14\hat{i}-2\hat{j}+10\hat{k}$. Scaling this down for simplicity, we use the vector $(7,1,-5)$.

Using the point $(2,4,5)$ from $L_2$, the equation of $P_2$ becomes $7(x-2)+1(y-4)-5(z-5)=0$, simplifying to:

The Final Verification

The shortest distance line is the intersection of these two planes. Any point on this line must satisfy both $11x+2y-7z+6=0$ and $7x+y-5z+7=0$.

We test our candidate point $(\frac{14}{3},-3,\frac{22}{3})$ by plugging it into $P_1$:

It satisfies $P_1$! Checking $P_2$:

It satisfies $P_2$ as well. We have found our point. Remember, in JEE Advanced, the most complex problems often yield to the most elegant geometric constructions. Keep visualizing, keep calculating, and never stop questioning the geometry.