Animated Solution for Mathematics - Three Dimensional Geometry: If the line $\frac{2-x}{3}=\frac{3y-2}{4\lambda +1}=4-z$ makes a right angle with the line $\frac{x+3}{3\mu }=\frac{1-2y}{6}=\frac{5-z}{7}$, then $4\lambda +9\mu$ is equal to :

A Masterclass

Welcome, future engineers! Today, we are going to conquer a classic JEE Advanced problem that tests your precision in 3D geometry. Imagine you are standing in a vast, three-dimensional room.

You have two lines stretching out into the void, and you are told they meet at a perfect right angle. Your mission is to find the value of a specific expression involving two variables, $\lambda$ and $\mu$. It sounds simple, but the devil is in the details.

The Uniform of a Line

Before we can perform any calculations, we must ensure our lines are wearing the right 'uniform.' In 3D geometry, a line is only in its standard symmetric form when the coefficients of $x$, $y$, and $z$ are all exactly $+1$.

The standard form is given by:

Here, $(a,b,c)$ are the direction ratios. If the coefficients are anything else, the denominators are not the true direction ratios. This is the most common trap in JEE exams.

Let us look at our first line: $\frac{2-x}{3}=\frac{3y-2}{4\lambda +1}=4-z$. Notice the coefficients: $-1$ for $x$, $+3$ for $y$, and $-1$ for $z$. We must fix this.

The Standardization Process

Let us standardize Line 1. For the $x$-term, $\frac{2-x}{3}$, we multiply the numerator and denominator by $-1$ to get $\frac{x-2}{-3}$. For the $z$-term, $4-z$, we do the same to get $\frac{z-4}{-1}$.

Now, the $y$-term is slightly trickier: $\frac{3y-2}{4\lambda +1}$. To make the coefficient of $y$ equal to $1$, we divide both the numerator and the denominator by $3$. This gives us:

Now, our Line 1 is perfectly standardized:

We can now confidently extract the direction ratios: $\overrightarrow{v_1}=\left(-3,\frac{4\lambda +1}{3},-1\right)$.

We apply the same logic to Line 2: $\frac{x+3}{3\mu }=\frac{1-2y}{6}=\frac{5-z}{7}$. The $x$-term is fine. For the $y$-term, we divide by $-2$ to get $\frac{y-\frac{1}{2}}{-3}$. For the $z$-term, we multiply by $-1$ to get $\frac{z-5}{-7}$.

Thus, the direction ratios for Line 2 are $\overrightarrow{v_2}=(3\mu ,-3,-7)$.

The Language of Perpendicularity

Now that we have our direction vectors, we invoke the most powerful tool in our arsenal: the dot product. When two lines are perpendicular, the dot product of their direction vectors must be zero.

This is our master equation: $a_1{a}_2+b_1{b}_2+c_1{c}_2=0$. Substituting our values, we get:

Let us simplify this step by step. The first term is $-9\mu$. The second term is beautiful—the $3$ in the denominator cancels with the $-3$, leaving us with $-(4\lambda +1)$, which is $-4\lambda -1$.

The third term is $(-1)(-7)=+7$. Our equation becomes:

Combining the constants, we get $-9\mu -4\lambda +6=0$. Rearranging this, we find the final relationship:

And there you have it! By carefully standardizing the lines and applying the dot product condition, we have arrived at the answer: $6$.

This problem is a reminder that in JEE Advanced, success isn't just about knowing the formulas; it's about paying attention to the structure of the equations. Keep practicing, stay curious, and never let a coefficient trick you again!