Question

The angle between lines whose direction cosines satisfy the equation $l + m + n = 0$ and $l^2 - m^2 - n^2 = 0$, is ______.

• A. $\pi/2$
• B. $\pi/3$
• C. $\pi/4$
• D. $\pi/6$

Hint:

When dealing with homogeneous equations for direction cosines, directly substitute the linear equation into the quadratic one. It will always factor neatly into two roots, instantly giving you the direction vectors of the two lines!

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We are given a system of equations defining the direction cosines ($l, m, n$) of two distinct 3D lines. We must find the vectors representing these lines and compute the angle between them.

Step 2: Detailed Explanation:
The given equations are:
1) $l + m + n = 0 \implies n = -(l + m)$
2) $l^2 - m^2 - n^2 = 0$
Substitute the expression for $n$ from Eq 1 into Eq 2:
$l^2 - m^2 - [-(l + m)]^2 = 0$
$l^2 - m^2 - (l^2 + 2lm + m^2) = 0$
$l^2 - m^2 - l^2 - 2lm - m^2 = 0$
$-2m^2 - 2lm = 0$
Divide the entire equation by $-2m$ (assuming $m$ is non-zero, otherwise we will test $m=0$ as a case):
$m(m + l) = 0$
This yields two distinct mathematical cases for the lines:
Case 1: $m = 0$
If $m = 0$, substitute into $n = -(l + m)$:
$n = -l$
The direction ratios $(l, m, n)$ are proportional to $(l, 0, -l)$.
Dividing by $l$, the direction vector is $\vec{d}_1 = \langle 1, 0, -1 \rangle$.
Case 2: $m = -l$
If $m = -l$, substitute into $n = -(l + m)$:
$n = -(l - l) = 0$
The direction ratios $(l, m, n)$ are proportional to $(l, -l, 0)$.
Dividing by $l$, the direction vector is $\vec{d}_2 = \langle 1, -1, 0 \rangle$.
Now, calculate the angle $\theta$ between the two lines using the dot product formula:
$\cos \theta = \frac{|\vec{d}_1 \cdot \vec{d}_2|}{|\vec{d}_1| |\vec{d}_2|}$
Dot product: $(1)(1) + (0)(-1) + (-1)(0) = 1$.
Magnitudes: $|\vec{d}_1| = \sqrt{1^2 + 0^2 + (-1)^2} = \sqrt{2}$
$|\vec{d}_2| = \sqrt{1^2 + (-1)^2 + 0^2} = \sqrt{2}$
Substitute these values:
$\cos \theta = \frac{1}{\sqrt{2} \times \sqrt{2}} = \frac{1}{2}$
The principal angle whose cosine is $\frac{1}{2}$ is $60^\circ$ or $\pi/3$ radians.

Step 3: Final Answer:
The angle is $\pi/3$, matching option (b).