Key Concepts and Formulas

1. Equations of Two Lines in 3D: When two equations involving direction cosines ($l, m, n$) are given – typically one linear and one homogeneous quadratic – these equations collectively define the direction ratios (or direction cosines) of two distinct lines.
2. Angle Between Two Lines: If $(l_1, m_1, n_1)$ are the direction ratios of the first line and $(l_2, m_2, n_2)$ are the direction ratios of the second line, the angle $\theta$ between them is given by the formula: $\cos \theta = \frac{l_1 l_2 + m_1 m_2 + n_1 n_2}{\sqrt{l_1^2 + m_1^2 + n_1^2} \sqrt{l_2^2 + m_2^2 + n_2^2}}$ It is important to note that this formula works equally well for direction ratios as it does for true direction cosines. If $(l, m, n)$ were normalized direction cosines, then $l^2+m^2+n^2=1$. However, the given equations usually provide proportional values, which are then normalized by the denominators in the formula.

Step-by-Step Solution

We are given two equations for the direction cosines ($l, m, n$) of two lines:

1. $l + 3m + 5n = 0 \quad \text{(Equation 1)}$
2. $5lm - 2mn + 6nl = 0 \quad \text{(Equation 2)}$

Our goal is to find the angle $\theta$ between these two lines.

Step 1: Express one direction cosine from the linear equation in terms of the others.

The first equation (Equation 1) is a linear relationship. We can use it to express one variable in terms of the other two. This allows us to reduce the number of variables in the quadratic equation (Equation 2) during substitution. It's often easiest to isolate $l$: $l = -3m - 5n \quad \text{(Equation 3)}$

Step 2: Substitute the expression into the quadratic equation and simplify.

Substitute the expression for $l$ from Equation 3 into Equation 2. This will transform Equation 2 into a homogeneous quadratic equation involving only $m$ and $n$. $5(-3m - 5n)m - 2mn + 6n(-3m - 5n) = 0$ Now, carefully expand and combine like terms: $(-15m^2 - 25mn) - 2mn + (-18mn - 30n^2) = 0$ $-15m^2 - 25mn - 2mn - 18mn - 30n^2 = 0$ Combine all terms containing $mn$: $-15m^2 + (-25 - 2 - 18)mn - 30n^2 = 0$ $-15m^2 - 45mn - 30n^2 = 0$ To simplify, divide the entire equation by a common factor, in this case, $-15$: $\frac{-15m^2}{-15} + \frac{-45mn}{-15} + \frac{-30n^2}{-15} = 0$ $m^2 + 3mn + 2n^2 = 0 \quad \text{(Equation 4)}$ This is a homogeneous quadratic equation in $m$ and $n$.

Step 3: Factorize the quadratic equation to find relationships between direction cosines.

Equation 4, $m^2 + 3mn + 2n^2 = 0$, can be factored into two linear expressions. We look for two numbers that multiply to $2$ (coefficient of $n^2$) and add to $3$ (coefficient of $mn$). These numbers are $1$ and $2$. $m^2 + mn + 2mn + 2n^2 = 0$ Factor by grouping: $m(m + n) + 2n(m + n) = 0$ $(m + n)(m + 2n) = 0$ This factorization yields two separate linear relationships, each corresponding to one of the two lines:

1. $m + n = 0 \implies m = -n$
2. $m + 2n = 0 \implies m = -2n$

Step 4: Determine the direction ratios for each line.

We will use Equation 3 ($l = -3m - 5n$) along with each of the relationships found in Step 3 to find the direction ratios $(l, m, n)$ for each line.

Case 1: For the first line (using $m = -n$) Substitute $m = -n$ into Equation 3: $l = -3(-n) - 5n$ $l = 3n - 5n$ $l = -2n$ So, for the first line, the relationships are $l = -2n$, $m = -n$, and $n = n$. We can choose a convenient value for $n$ (e.g., $n=1$) to find a set of direction ratios: $(l_1, m_1, n_1) = (-2, -1, 1)$

Case 2: For the second line (using $m = -2n$) Substitute $m = -2n$ into Equation 3: $l = -3(-2n) - 5n$ $l = 6n - 5n$ $l = n$ So, for the second line, the relationships are $l = n$, $m = -2n$, and $n = n$. Choosing $n=1$: $(l_2, m_2, n_2) = (1, -2, 1)$

Step 5: Calculate the angle between the two lines.

Now we have the direction ratios for both lines:

• $(l_1, m_1, n_1) = (-2, -1, 1)$
• $(l_2, m_2, n_2) = (1, -2, 1)$

We use the angle formula: $\cos \theta = \frac{l_1 l_2 + m_1 m_2 + n_1 n_2}{\sqrt{l_1^2 + m_1^2 + n_1^2} \sqrt{l_2^2 + m_2^2 + n_2^2}}$

First, calculate the numerator (dot product of direction ratios): $l_1 l_2 + m_1 m_2 + n_1 n_2 = (-2)(1) + (-1)(-2) + (1)(1)$ $= -2 + 2 + 1 = 1$

Next, calculate the magnitudes of the direction ratio vectors: $\sqrt{l_1^2 + m_1^2 + n_1^2} = \sqrt{(-2)^2 + (-1)^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6}$ $\sqrt{l_2^2 + m_2^2 + n_2^2} = \sqrt{1^2 + (-2)^2 + 1^2} = \sqrt{1 + 4 + 1} = \sqrt{6}$

Substitute these values into the formula for $\cos \theta$: $\cos \theta = \frac{1}{\sqrt{6} \cdot \sqrt{6}}$ $\cos \theta = \frac{1}{6}$

Therefore, the angle $\theta$ is: $\theta = {\cos ^{ - 1}}\left( {\frac{1}{6}} \right)$

Common Mistakes & Tips

• Algebraic Precision: Pay close attention to signs, especially during substitution and expansion. A small sign error can lead to incorrect direction ratios.
• Factoring Homogeneous Quadratics: Ensure you are proficient in factoring expressions like $m^2 + 3mn + 2n^2$. If direct factorization is difficult, you can treat it as a quadratic in $m/n$ by dividing by $n^2$ and use the quadratic formula.
• Two Lines, Not One: Remember that a linear equation and a homogeneous quadratic equation in $l, m, n$ define the direction ratios of two distinct lines, not just one. Each factor from the quadratic corresponds to one line.

Summary

To find the angle between the lines, we first used the linear equation to express one direction cosine in terms of the others. This expression was then substituted into the quadratic equation, simplifying it into a homogeneous quadratic in two variables. Factoring this quadratic yielded two linear relationships, each defining the direction ratios of one of the lines. Finally, with the direction ratios of both lines, the standard formula for the angle between two lines was applied to calculate $\cos \theta$. The calculated angle is $\cos^{-1}\left(\frac{1}{6}\right)$.

The final answer is $\boxed{\text{A}}$.