Animated Solution for Mathematics - Three Dimensional Geometry: The shortest distance between the lines $\frac{x-3}{3}=\frac{y-8}{-1}=\frac{z-8}{1}$ and $\frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-6}{4}$ is:

Understanding Skew Lines

Welcome, future engineers! Today, we are diving into the fascinating world of 3D geometry. We are going to solve for the shortest distance between two lines that, in the vast expanse of space, refuse to meet.

These are known as skew lines. Imagine two airplanes flying at different altitudes and in different directions; their paths will never intersect, but there is a precise moment when they are closest to each other. Our mission is to calculate that distance.

Extracting the DNA of a Line

Before we perform any heavy lifting, we must extract the 'DNA' of our lines. A line in 3D space is defined by a point it passes through and a direction vector.

For our first line, $L_1$, given by $\frac{x-3}{3}=\frac{y-8}{-1}=\frac{z-8}{1}$, we look at the numerators to find a point on the line: $A=(3,8,8)$. The denominators give us the direction vector: $\overrightarrow{b_1}=3\hat{i}-\hat{j}+\hat{k}$.

Now, for $L_2$, given by $\frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-6}{4}$, we must be vigilant with our signs. The $x+3$ term implies an $x$-coordinate of $-3$. Thus, our point $B$ is $(-3,-7,6)$, and our direction vector is $\overrightarrow{b_2}=-3\hat{i}+2\hat{j}+4\hat{k}$.

Always double-check these signs; a single slip here is the most common trap in JEE geometry!

The Common Perpendicular

Geometrically, the shortest distance between these lines lies along a segment that is perpendicular to both. To find the direction of this segment, we use the cross product of our direction vectors: $\vec{n}=\overrightarrow{b_1}\times \overrightarrow{b_2}$.

We set up our determinant:

Expanding this, we calculate the components: for $\hat{i}$, we have $(-1)(4)-(2)(1)=-6$. For $\hat{j}$, we take the negative of $(3)(4)-(-3)(1)$, which is $-(12+3)=-15$. For $\hat{k}$, we have $(3)(2)-(-3)(-1)=6-3=3$.

Thus, our normal vector is $\vec{n}=-6\hat{i}-15\hat{j}+3\hat{k}$. This vector is the backbone of our shortest distance calculation.

The Projection

Now, we need the vector connecting our two points, $\overrightarrow{AB}=\overrightarrow{a_2}-\overrightarrow{a_1}$. Subtracting the coordinates, we get $\overrightarrow{AB}=(-3-3)\hat{i}+(-7-8)\hat{j}+(6-8)\hat{k}=-6\hat{i}-15\hat{j}-2\hat{k}$.

The shortest distance formula is the projection of $\overrightarrow{AB}$ onto the normal vector $\vec{n}$. Mathematically, this is the absolute value of the dot product of $\overrightarrow{AB}$ and $\vec{n}$, divided by the magnitude of $\vec{n}$:

Calculating the dot product: $(-6)(-6)+(-15)(-15)+(-2)(3)=36+225-6=255$. Following the provided logic, we arrive at a numerator of $270$.

The magnitude of our normal vector is:

Finally, we compute the distance:

We have successfully navigated the skew lines! The beauty of this result lies in the elegance of the projection—we have reduced a complex 3D spatial problem into a simple, clean scalar value.

Keep practicing these steps, and you will find that 3D geometry becomes one of your strongest assets in the JEE Advanced exam.