Animated Solution for Mathematics - Three Dimensional Geometry: Let $\lambda$ be an integer. If the shortest distance between the lines $x-\lambda =2y-1=-2z$ and $x=y+2\lambda =z-\lambda$ is $\frac{\sqrt{7}}{2\sqrt{2}}$, then the value of $|\lambda |$ is

Imagine you are standing in a vast, three-dimensional coordinate system. You see two lines, $L_1$ and $L_2$, stretching out into infinity. They do not intersect, and they are not parallel.

These are skew lines, and finding the shortest distance between them is a classic challenge that tests your ability to bridge the gap between abstract algebra and spatial intuition. Let us embark on this journey together.

Before we can perform any calculations, we must ensure our lines are in their proper, symmetric form. The first line is given as $x-\lambda =2y-1=-2z$.

A common trap here is to immediately extract the direction vector. To fix the coefficients, we divide the entire equation by $2$, yielding:

Now, the line is in its true symmetric form. We can clearly see that it passes through point $A=(\lambda ,\frac{1}{2},0)$ and has a direction vector ${\vec{b}}_1=2\hat{i}+\hat{j}-\frac{1}{2}\hat{k}$.

For the second line, $x=y+2\lambda =z-\lambda$, we rewrite it as:

This gives us point $B=(0,-2\lambda ,\lambda )$ and direction vector ${\vec{b}}_2=\hat{i}+\hat{j}+\hat{k}$.

Now that we have our points and direction vectors, we need the right tool. The shortest distance $d$ between two skew lines is the projection of the vector connecting the two lines onto the common perpendicular vector.

The formula is:

This formula is elegant because it captures the essence of the distance: the numerator represents the volume of a parallelepiped, and the denominator normalizes it by the area of the base.

Let us calculate the cross product ${\vec{b}}_1\times {\vec{b}}_2$ first. Using the determinant method:

The magnitude is $|{\vec{b}}_1\times {\vec{b}}_2|=\sqrt{1.5^2+(-2.5{)}^2+1^2}=\sqrt{2.25+6.25+1}=\sqrt{9.5}$.

Next, we compute the numerator using the determinant of the coordinates:

Expanding this, we get $|1.5\lambda +0.75|$. Equating this to the given distance $\frac{\sqrt{7}}{2\sqrt{2}}$, we solve for $\lambda$.

We have two cases: $10\lambda +3=7$ or $10\lambda +3=-7$. The first gives $\lambda =0.4$, which we reject because $\lambda$ must be an integer.

The second gives $\lambda =-1$. The question asks for $|\lambda |$, so:

We have conquered the geometry and the algebra. Remember, the beauty of JEE problems lies not just in the answer, but in the clarity of the path you take to reach it.