The angle between the lines whose direction cosines satisfy the equations l + m + n = 0 and l² = m²

Q: Wait, why can't we just solve for $l, m, n$ as unique values?

Great question! Remember that $l, m, n$ are direction cosines. They are not unique coordinates of a point; they are ratios that define a direction. Because we have two equations and three variables, we have one degree of freedom, which is exactly what we need to define the direction of a line. We are finding the ratios, not specific values.

Q: Why do we use the absolute value in the dot product formula?

The angle between two lines is conventionally defined as the acute angle, which lies between $0$ and $\frac{\pi}{2}$. The dot product can result in a negative value depending on the direction vectors chosen, so the absolute value ensures we get the cosine of the acute angle.

Q: Is there a geometric way to see this without all this algebra?

While you can visualize the intersection of a plane and a cone, the algebraic approach is the most robust and reliable method for JEE problems. It guarantees you won't miss any solutions and provides a clear, step-by-step path to the answer.

Question

The angle between the lines whose direction cosines satisfy the equations $l + m + n = 0$ and $l^2 = m^2 + n^2$ is

Options:

(A) $\frac{\pi}{6}$

(B) $\frac{\pi}{2}$

(C) $\frac{\pi}{3}$

(D) $\frac{\pi}{4}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow explorers of the mathematical universe. Today, we are diving into a classic 3D geometry problem that often trips up even the most prepared students.

We are given two lines in space defined by two constraints on their direction cosines, $l, m,$ and $n$. The first is a linear constraint, $l + m + n = 0$, and the second is a quadratic one, $l^2 = m^2 + n^2$.

Our goal is to find the angle between these two lines. It sounds mysterious, but let's peel back the layers together.

### The Algebraic Dance

To find the angle between any two lines, we need their direction ratios. We have a system of two equations with three variables, meaning we cannot find unique values for $l, m,$ and $n$, but we can find their ratios.

Let's use the power of substitution. From our linear equation, we can isolate one variable:
$$l = -(m + n)$$

Now, we substitute this expression into our quadratic constraint, $l^2 = m^2 + n^2$:
$$( -(m + n) )^2 = m^2 + n^2$$

Expanding the left side, we get:
$$m^2 + n^2 + 2mn = m^2 + n^2$$

### The Breakthrough

Look closely at this equation. The $m^2$ and $n^2$ terms appear on both sides and cancel out perfectly, leaving us with:
$$2mn = 0 \quad \Rightarrow \quad mn = 0$$

This is the turning point of our journey. The product of two variables being zero means that either $m = 0$ or $n = 0$.

This simple result splits our problem into two distinct cases, which correspond to the two lines we are investigating.

### Finding the Direction Vectors

For **Case 1**, let $m = 0$. Substituting this into our relation $l = -(m + n)$, we get $l = -n$.

This gives us the direction ratios $(-n, 0, n)$, which simplifies to the direction vector:
$$\vec{v}_1 = \hat{i} - \hat{k}$$

For **Case 2**, let $n = 0$. Substituting this into $l = -(m + n)$, we get $l = -m$.

This gives us the direction ratios $(-m, m, 0)$, which simplifies to the direction vector:
$$\vec{v}_2 = -\hat{i} + \hat{j}$$

### The Final Angle

Now, we bring in the heavy artillery: the dot product formula. The cosine of the angle $	heta$ between two vectors is given by:
$$\cos 	heta = \frac{|\vec{v}_1 \cdot \vec{v}_2|}{|\vec{v}_1| |\vec{v}_2|}$$

Calculating the dot product, we have:
$$\vec{v}_1 \cdot \vec{v}_2 = (1)(-1) + (0)(1) + (-1)(0) = -1$$
Taking the absolute value, we get $|-1| = 1$.

The magnitudes are:
$$|\vec{v}_1| = \sqrt{1^2 + 0^2 + (-1)^2} = \sqrt{2}$$
$$|\vec{v}_2| = \sqrt{(-1)^2 + 1^2 + 0^2} = \sqrt{2}$$

Substituting these into our formula:
$$\cos 	heta = \frac{1}{\sqrt{2} \cdot \sqrt{2}} = \frac{1}{2}$$

Since $\cos 	heta = \frac{1}{2}$, we conclude that the angle is:
$$	heta = \frac{\pi}{3} 	ext{ or } 60^\circ$$

We have successfully navigated the constraints and arrived at the solution. Remember, in 3D geometry, the algebra is just a tool to describe the physical reality of the lines.

Animated Solution for Mathematics - Three Dimensional Geometry: The angle between the lines whose direction cosines satisfy the equations $l + m + n = 0$ and $l^{2} = m^{2} + n^{2}$ is

Select Answer:

Visualized Solution

Understanding the Constraints

Strategy to Find Direction Ratios

Expressing $l$ in terms of $m$ and $n$

Substitution into the Second Equation

Expanding the Equation

Simplifying the Expression

Analyzing the Two Cases

Direction Ratios for Line 1

Direction Ratios for Line 2

The Angle Formula

Calculating the Dot Product

Finding $cos θ$

Final Conclusion

The Sigma Insight: Direction Cosines and Direction Ratios

Analyzing the Setup

The Algebraic Dance

The Breakthrough

Finding the Direction Vectors

The Final Angle

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Animated Solution for Mathematics - Three Dimensional Geometry: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The angle between the lines whose direction cosines satisfy the equations l+m+n=0 and l2=m2+n2 is