Key Concepts and Formulas

In three-dimensional geometry, a straight line is most conveniently represented in its symmetric (or standard) form. This form is essential for easily identifying key properties of the line, such as its direction.

1. Standard Form of a Line: A line passing through a point $(x_1, y_1, z_1)$ and having direction ratios $(a, b, c)$ is given by the equation: $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$ It is crucial that the coefficients of $x, y,$ and $z$ in the numerators are all $+1$. If they are not, the equation must be manipulated algebraically to achieve this form. The denominators $(a, b, c)$ are the direction ratios (DRs) of the line.

2. Condition for Perpendicular Lines: If two lines, $L_1$ and $L_2$, have direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ respectively, they are perpendicular (i.e., they make a right angle with each other) if and only if the dot product of their direction ratio vectors is zero: $a_1a_2 + b_1b_2 + c_1c_2 = 0$ This problem directly utilizes this perpendicularity condition.

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Step-by-Step Solution

Step 1: Convert the equation of Line 1 ($L_1$) to standard form and identify its direction ratios.

The given equation for the first line, $L_1$, is: $\frac{2-x}{3}=\frac{3 y-2}{4 \lambda+1}=4-z$ Our goal is to transform this into the standard form $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$, where the coefficients of $x, y, z$ in the numerators are $+1$.

Let's process each term:

• First term: $\frac{2-x}{3}$ To get $x$ with a coefficient of $+1$, we factor out $-1$ from the numerator: $2-x = -(x-2)$. So, $\frac{2-x}{3} = \frac{-(x-2)}{3} = \frac{x-2}{-3}$.
• Second term: $\frac{3y-2}{4\lambda+1}$ To get $y$ with a coefficient of $+1$, we factor out $3$ from the numerator: $3y-2 = 3\left(y-\frac{2}{3}\right)$. So, $\frac{3y-2}{4\lambda+1} = \frac{3\left(y-\frac{2}{3}\right)}{4\lambda+1} = \frac{y-\frac{2}{3}}{\frac{4\lambda+1}{3}}$.
• Third term: $4-z$ This term can be written as $\frac{4-z}{1}$. Similar to the first term, we factor out $-1$: $4-z = -(z-4)$. So, $\frac{4-z}{1} = \frac{-(z-4)}{1} = \frac{z-4}{-1}$.

Combining these transformations, the standard form of $L_1$ is: $L_1: \frac{x-2}{-3} = \frac{y-\frac{2}{3}}{\frac{4\lambda+1}{3}} = \frac{z-4}{-1}$ From this, we identify the direction ratios for $L_1$ as $(a_1, b_1, c_1)$: $a_1 = -3, \quad b_1 = \frac{4\lambda+1}{3}, \quad c_1 = -1$

Step 2: Convert the equation of Line 2 ($L_2$) to standard form and identify its direction ratios.

The given equation for the second line, $L_2$, is: $\frac{x+3}{3 \mu}=\frac{1-2 y}{6}=\frac{5-z}{7}$ Again, we aim for the standard form with $+1$ coefficients for $x, y, z$ in the numerators.

Let's process each term:

• First term: $\frac{x+3}{3\mu}$ This term is already in the desired form, as $x+3 = x-(-3)$. So, it remains $\frac{x+3}{3\mu}$.
• Second term: $\frac{1-2y}{6}$ To get $y$ with a coefficient of $+1$, we factor out $-2$ from the numerator: $1-2y = -2\left(y-\frac{1}{2}\right)$. So, $\frac{1-2y}{6} = \frac{-2\left(y-\frac{1}{2}\right)}{6} = \frac{y-\frac{1}{2}}{-3}$.
• Third term: $\frac{5-z}{7}$ Similar to the third term of $L_1$, we factor out $-1$: $5-z = -(z-5)$. So, $\frac{5-z}{7} = \frac{-(z-5)}{7} = \frac{z-5}{-7}$.

Combining these transformations, the standard form of $L_2$ is: $L_2: \frac{x-(-3)}{3\mu} = \frac{y-\frac{1}{2}}{-3} = \frac{z-5}{-7}$ From this, we identify the direction ratios for $L_2$ as $(a_2, b_2, c_2)$: $a_2 = 3\mu, \quad b_2 = -3, \quad c_2 = -7$

Step 3: Apply the perpendicularity condition.

The problem states that line $L_1$ makes a right angle with line $L_2$, which means they are perpendicular. Therefore, we use the condition $a_1a_2 + b_1b_2 + c_1c_2 = 0$.

Substitute the direction ratios we found in Step 1 and Step 2 into this equation: $(-3)(3\mu) + \left(\frac{4\lambda+1}{3}\right)(-3) + (-1)(-7) = 0$

Step 4: Solve the resulting equation for $4\lambda+9\mu$.

Now, we simplify and solve the equation derived in Step 3: $-9\mu - (4\lambda+1) + 7 = 0$ Distribute the negative sign over $(4\lambda+1)$: $-9\mu - 4\lambda - 1 + 7 = 0$ Combine the constant terms: $-9\mu - 4\lambda + 6 = 0$ To find the value of $4\lambda+9\mu$, we can move the terms involving $\lambda$ and $\mu$ to the right side of the equation: $6 = 4\lambda + 9\mu$ Rearranging to the desired format: $4\lambda + 9\mu = 6$

Thus, the value of $4\lambda+9\mu$ is $6$.

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Common Mistakes & Tips

1. Strict Adherence to Standard Form: The most common mistake is failing to correctly convert the line equations to the standard symmetric form. Always ensure the coefficients of $x, y, z$ in the numerator are exactly $+1$. For example, $\frac{A-Bx}{C}$ must be rewritten as $\frac{-(Bx-A)}{C} = \frac{x-A/B}{-C/B}$.
2. Sign Errors: Be extremely careful with negative signs when factoring them out or distributing them, especially in expressions like $2-x$ or $1-2y$.
3. Fractional Denominators: When you factor out a coefficient from the numerator (e.g., $3$ from $3y-2$), remember to divide the original denominator by that same coefficient to maintain equality. For instance, $\frac{3(y-2/3)}{4\lambda+1}$ becomes $\frac{y-2/3}{(4\lambda+1)/3}$.
4. Algebraic Precision: Double-check your calculations when simplifying the dot product equation. A small arithmetic or sign error can lead to an incorrect final answer.

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Summary

This problem required us to utilize the fundamental concepts of lines in 3D geometry. The primary steps involved transforming the given line equations into their standard symmetric form to accurately extract their direction ratios. Once the direction ratios for both lines were obtained, we applied the condition for perpendicularity ($a_1a_2 + b_1b_2 + c_1c_2 = 0$) and solved the resulting linear equation to find the value of the required expression, $4\lambda+9\mu$. The careful manipulation of algebraic terms and strict adherence to the standard form were key to arriving at the correct solution.

The final calculated value for $4\lambda+9\mu$ is $6$.

The final answer is $\boxed{\text{6}}$, which corresponds to option (A).