Animated Solution for Mathematics - Three Dimensional Geometry: If the lines $x=ay+b,z=cy+d$ and $x=a^{\prime }z+b^{\prime },y=c^{\prime }z+d^{\prime }$ are perpendicular, then:

Unmasking the Line

Welcome, fellow traveler of the 3D coordinate system. Today, we are going to peel back the layers of a problem that often intimidates students in the JEE Advanced arena. We are given two lines in space, but they aren't presented in the friendly, familiar symmetric form we love.

Instead, they are presented as projections—equations that look like $x=ay+b$ and $z=cy+d$. These equations are essentially telling us how the line behaves in the $xy$-plane and the $yz$-plane.

To conquer this, we must perform a transformation, a mathematical "unmasking" that reveals the true direction vector of these lines. Let us embark on this journey together.

The Art of Symmetric Conversion

Imagine you are standing in a 3D room. You have two lines, $L_1$ and $L_2$. To understand their orientation, we need their direction vectors, ${\vec{v}}_1$ and ${\vec{v}}_2$.

The standard symmetric form, $\frac{x-x_1}{l}=\frac{y-y_1}{m}=\frac{z-z_1}{n}$, is our North Star. It tells us that the line passes through $(x_1,y_1,z_1)$ and moves in the direction of the vector $(l,m,n)$.

For our first line, $L_1$, we are given $x=ay+b$ and $z=cy+d$. Notice the common variable here: $y$. This is our anchor.

If we isolate $y$ in both equations, we can link them together. From $x=ay+b$, a simple rearrangement gives us:

Similarly, from $z=cy+d$, we get:

Now, we equate them. Since both expressions equal $y$, we can write:

Look at that! We have successfully transformed the equation into the symmetric form. The denominators are $a,1,c$. These are the components of our direction vector ${\vec{v}}_1=(a,1,c)$.

The Mirror Image

Now, let us apply the same logic to $L_2$. The equations are $x=a^{\prime }z+b^{\prime }$ and $y=c^{\prime }z+d^{\prime }$. This time, the common variable is $z$.

From $x=a^{\prime }z+b^{\prime }$, we isolate $z$:

From $y=c^{\prime }z+d^{\prime }$, we isolate $z$:

Equating these to $z$ (or $\frac{z-0}{1}$), we get the symmetric form for $L_2$:

Just like that, the direction vector for $L_2$ reveals itself: ${\vec{v}}_2=(a^{\prime },c^{\prime },1)$. We have successfully extracted the DNA of both lines.

The Orthogonality Condition

We have arrived at the climax of our problem. We are told that the lines are perpendicular. In the language of vectors, this is the most beautiful condition of all: the dot product of their direction vectors must be zero.

Substituting our vectors, we have:

Let us perform the dot product with precision. Multiply the corresponding components and sum them up:

Which simplifies to:

Or, rearranging to match our options:

Take a moment to appreciate what we just did. We took two seemingly abstract, confusing sets of equations and, through the systematic application of symmetric form conversion, reduced them to their core geometric essence.

We didn't need to know where the lines were in space; we only needed to know how they were oriented. This is the heart of JEE Advanced physics and mathematics.

It is not about memorizing formulas; it is about having the confidence to manipulate the given information until the underlying structure reveals itself. You have mastered the conversion, you have mastered the dot product, and you have mastered the logic.