Let the direction cosines of two lines satisfy the equations : 4l + m - n = 0 and 2mn + 10nl + 3lm =

Q: Wait, why can't we just solve for $l, m, n$ directly? We have two equations and three variables, shouldn't we have a unique solution?

That is a fantastic question! Remember, direction cosines are not independent variables; they are constrained by the identity $l^2 + m^2 + n^2 = 1$. However, in this problem, we are looking for the *direction* of the lines, not the specific values of $l, m, n$ that satisfy the normalization condition. The two equations define the *ratios* of the direction cosines. By finding the ratio $\frac{l}{m}$, we are essentially finding the direction vector of the line. We don't need the absolute values, just the proportions!

Q: Why do we divide the homogeneous equation by $m^2$? Is that always allowed?

It is a standard and elegant technique! A homogeneous equation of the form $Al^2 + Blm + Cm^2 = 0$ represents a pair of lines passing through the origin in the $lm$-plane. Dividing by $m^2$ assumes $m \neq 0$. If $m$ were zero, then $40l^2 = 0$, which implies $l=0$. If $l=0$ and $m=0$, then from our linear equation $n=0$, which would mean the direction cosines are $(0,0,0)$, which is impossible for a line. Thus, $m$ cannot be zero, and dividing by $m^2$ is perfectly safe and mathematically sound.

Q: Why do we take the absolute value in the dot product formula? Does it matter for the angle?

The question specifically asks for the cosine of the *acute* angle. The dot product $\vec{d}_1 \cdot \vec{d}_2 = |\vec{d}_1||\vec{d}_2| \cos \theta$ can be negative if the angle is obtuse. By taking the absolute value, we ensure that the resulting cosine value corresponds to the acute angle between the lines. It is a crucial step to ensure your answer matches the geometric requirement!

Question

Let the direction cosines of two lines satisfy the equations : $4l + m - n = 0$ and $2mn + 10nl + 3lm = 0$. Then the cosine of the acute angle between these lines is :

Options:

(A) $\frac{20}{3\sqrt{38}}$

(B) $\frac{10}{3\sqrt{38}}$

(C) $\frac{10}{7\sqrt{38}}$

(D) $\frac{10}{\sqrt{38}}$

Sigma Percentile · Accepted Answer

### The Geometry of Constraints: A Journey into 3D Space

My dear student, welcome to a beautiful intersection of algebra and geometry. Today, we are not just solving a problem; we are peeling back the layers of how lines exist in three-dimensional space.

We are given two equations that govern the direction cosines $(l, m, n)$ of two lines. At first glance, this might look like a daunting system of equations, but I want you to see the elegance hidden within.

### Phase 1: The Algebraic Bridge

We start with the linear constraint: $4l + m - n = 0$. This is our anchor. It tells us that the direction cosines are not free to wander; they are bound by a linear relationship.

The second equation, $2mn + 10nl + 3lm = 0$, is a quadratic form. In the world of JEE Advanced, whenever you see a linear equation paired with a quadratic one, your instinct should immediately scream: **Substitution!**

Let us isolate $n$ from the linear equation. It is the path of least resistance: $n = 4l + m$.

Now, we take this expression and breathe it into the quadratic equation. Wherever we see an $n$, we replace it with $(4l + m)$. This transforms our quadratic equation into:

$$2m(4l + m) + 10l(4l + m) + 3lm = 0$$

### Phase 2: The Homogeneous Transformation

Now, do not let the expansion intimidate you. This is where the magic happens. Let us expand carefully:

$$8lm + 2m^2 + 40l^2 + 10lm + 3lm = 0$$

When we group the like terms, we arrive at a beautiful, homogeneous quadratic equation:

$$40l^2 + 21lm + 2m^2 = 0$$

Why is this homogeneous form so powerful? Because it allows us to find the **ratio** of the direction cosines. By dividing the entire equation by $m^2$ (and knowing $m 
eq 0$), we define a new variable $t = \frac{l}{m}$.

The equation becomes a simple quadratic:

$$40t^2 + 21t + 2 = 0$$

We are no longer dealing with three variables; we have reduced the complexity of the universe down to a single quadratic equation in $t$.

### Phase 3: Unlocking the Directions

Solving $40t^2 + 21t + 2 = 0$ is a classic exercise in factorization. We look for two numbers that multiply to $80$ and add to $21$. Those numbers are $16$ and $5$.

Thus, we factor it as $(8t + 1)(5t + 2) = 0$. This gives us two distinct ratios for $\frac{l}{m}$: $t_1 = -\frac{1}{8}$ and $t_2 = -\frac{2}{5}$.

For the first line, if $\frac{l}{m} = -\frac{1}{8}$, we can choose $l_1 = -1$ and $m_1 = 8$. Substituting these into our linear constraint $n = 4l + m$, we find $n_1 = 4(-1) + 8 = 4$. Thus, our first direction vector is $\vec{d}_1 = (-1, 8, 4)$.

For the second line, if $\frac{l}{m} = -\frac{2}{5}$, we choose $l_2 = -2$ and $m_2 = 5$. Again, using $n = 4l + m$, we find $n_2 = 4(-2) + 5 = -3$. Our second direction vector is $\vec{d}_2 = (-2, 5, -3)$.

### Phase 4: The Geometric Payoff

We have arrived at the final stage. We have two vectors, $\vec{d}_1 = (-1, 8, 4)$ and $\vec{d}_2 = (-2, 5, -3)$. The cosine of the angle $	heta$ between them is given by the dot product formula:

$$\cos 	heta = \frac{|\vec{d}_1 \cdot \vec{d}_2|}{||\vec{d}_1|| ||\vec{d}_2||}$$

Calculating the dot product:

$$|(-1)(-2) + (8)(5) + (4)(-3)| = |2 + 40 - 12| = 30$$

Calculating the magnitudes:

$$||\vec{d}_1|| = \sqrt{(-1)^2 + 8^2 + 4^2} = \sqrt{1 + 64 + 16} = \sqrt{81} = 9$$

$$||\vec{d}_2|| = \sqrt{(-2)^2 + 5^2 + (-3)^2} = \sqrt{4 + 25 + 9} = \sqrt{38}$$

Putting it all together:

$$\cos 	heta = \frac{30}{9\sqrt{38}}$$

Simplifying by dividing the numerator and denominator by $3$, we get the final answer:

$$\cos 	heta = \frac{10}{3\sqrt{38}}$$

Look at that result. It is clean, precise, and derived from a logical flow that started with a simple substitution. You have successfully navigated the geometry of 3D lines.

Animated Solution for Mathematics - Three Dimensional Geometry: Let the direction cosines of two lines satisfy the equations : $4 l + m - n = 0$ and $2 mn + 10 n l + 3 l m = 0$ . Then the cosine of the acute angle between these lines is :

Select Answer:

Visualized Solution

Introduction to the Problem

Strategy: Eliminating a Variable

Isolating $n$

Substitution into the Quadratic Equation

Expanding the Terms

Forming a Homogeneous Equation

Converting to a Single Variable Ratio

Solving the Quadratic Equation

Finding the Roots

Direction Ratios for Line 1

Direction Ratios for Line 2

Formula for the Angle Between Lines

Substituting the Values

Calculating the Numerator and Denominator

Final Answer

The Sigma Insight: Direction Cosines and Direction Ratios

The Geometry of Constraints

Phase 1

Phase 2

Phase 3

Phase 4

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