Animated Solution for Mathematics - Three Dimensional Geometry: Let the shortest distance between the lines $\frac{x-3}{3}=\frac{y-\alpha }{-1}=\frac{z-3}{1}$ and $\frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-\beta }{4}$ be $3\sqrt{30}$. Then the positive value of $5\alpha +\beta$ is

Imagine you are standing in a vast, three-dimensional space. You see two lines, $L_1$ and $L_2$, that seem to pass each other like ships in the night. They are not parallel, yet they never intersect.

These are skew lines. In the world of JEE Advanced, these lines are not just abstract entities; they are geometric puzzles waiting to be solved. We are given the shortest distance between them as $3\sqrt{30}$, and our mission is to find the value of $5\alpha +\beta$.

Extracting the DNA of the Lines

Every line in 3D space is defined by a point it passes through and a direction in which it travels.

For $L_1:\frac{x-3}{3}=\frac{y-\alpha }{-1}=\frac{z-3}{1}$, we identify a point $A(3,\alpha ,3)$ and a direction vector $\vec{p}=3\hat{i}-\hat{j}+\hat{k}$.

Similarly, for $L_2:\frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-\beta }{4}$, we find a point $B(-3,-7,\beta )$ and a direction vector $\vec{q}=-3\hat{i}+2\hat{j}+4\hat{k}$. These vectors are the keys to unlocking the geometry of the space between the lines.

The Common Normal

To find the shortest distance, we need a bridge—a vector that is perpendicular to both lines. This is where the cross product shines. We calculate $\vec{n}=\vec{p}\times \vec{q}$:

This vector $\vec{n}=-6\hat{i}-15\hat{j}+3\hat{k}$ is our common normal. Its magnitude is calculated as follows:

This is a beautiful coincidence, as it matches the given shortest distance perfectly!

The Final Calculation

The shortest distance formula is $SD=\frac{|\overrightarrow{BA}\cdot \vec{n}|}{|\vec{n}|}$. We construct the vector $\overrightarrow{BA}$:

Substituting these into our formula, we obtain:

Multiplying both sides by $3\sqrt{30}$, we get:

Simplifying the expression inside the absolute value, we arrive at $|-132-15\alpha -3\beta |=270$, which is equivalent to $|15\alpha +3\beta +132|=270$.

This leads to two cases: $15\alpha +3\beta +132=270$ or $15\alpha +3\beta +132=-270$. Solving the first case gives $15\alpha +3\beta =138$, which simplifies to $5\alpha +\beta =46$.

The second case yields a negative value, which we reject as per the problem context. Thus, the final result is 46.