Question

The angle between the lines, whose direction cosines \( l, m, n \) satisfy the equations \( l + m + n = 0 \) and \( 2l^{2} + 2m^{2} - n^{2} = 0 \), is

• A. $60^\circ$
• B. $180^\circ$
• C. $90^\circ$
• D. $30^\circ$

Hint:

Geometry Tip: When substituting linear constraints into quadratic direction cosine constraints, if a perfect square like $(l \pm m)^2 = 0$ appears, it implies proportional direction ratios, so the lines are parallel or coincident. Hence, the angle is $0^\circ$ or $180^\circ$.

Answer 1

The Correct Option is B

Concept: 3D Geometry - Direction Cosines and Angle Between Lines.

Step 1: Express $n$ in terms of $l$ and $m$. From the first given linear equation, $l+m+n=0$, we can isolate $n$. This gives us $n = -(l+m)$.

Step 2: Substitute $n$ into the second quadratic equation. Take the derived expression for $n$ and substitute it into the second equation: $2l^{2}+2m^{2}-n^{2}=0$. This becomes $2l^{2}+2m^{2}-(-l-m)^{2}=0$.

Step 3: Expand and simplify to find the relationship between $l$ and $m$. Expand the squared term: $2l^{2}+2m^{2}-(l^{2}+m^{2}+2lm)=0$. Simplifying this yields $l^{2}+m^{2}-2lm=0$, which is a perfect square. Thus, $(l-m)^{2}=0$, leading directly to $l=m$.

Step 4: Determine the proportional direction ratios. Since we established that $l=m$, substitute this back into our equation for $n$: $n = -l-l = -2l$. Therefore, the relationship between the direction cosines is $\frac{l}{1}=\frac{m}{1}=\frac{n}{-2}$. This means the direction ratios for the lines are proportional to $(1, 1, -2)$.

Step 5: Calculate the angle between the lines. Because solving the system of equations yielded only one unique set of direction ratios $(1, 1, -2)$, it means both lines possess the exact same direction cosines. When two lines have identical direction cosines, they represent parallel or coincident lines in 3D space.

Mathematically, the angle $\theta$ between two lines with direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ is given by $\cos\theta = \frac{a_1a_2 + b_1b_2 + c_1c_2}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}$. Substituting our derived identical ratios gives $\cos\theta = \frac{(1)(1) + (1)(1) + (-2)(-2)}{1^2+1^2+(-2)^2} = \frac{1+1+4}{6} = \frac{6}{6} = 1$.

Since $\cos\theta = 1$, the angle $\theta$ must be either $0^\circ$ or $180^\circ$. Looking at the provided options, $180^\circ$ is the correct match. $$ \therefore \text{The angle between the lines is } 180^\circ. $$