3D Geometry 1 Question 1
The angle between the lines whose direction cosines satisfy the equations $l+m+n=0$ and $l^{2}=m^{2}+n^{2}$, is
(a) $\frac{\pi}{3}$
(b) $\frac{\pi}{4}$
(c) $\frac{\pi}{6}$
(d) $\frac{\pi}{2}$
(2014 Main)
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Answer:
Correct Answer: (a)
Solution:
Steps to Solve:
- Relate Direction Cosines:
From $l + m + n = 0$, express one cosine in terms of others: $l = -m - n$
Substitute into $l^2 + m^2 + n^2 = 1$ to get:
$4(m^2 + n^2) = 1$
Therefore: $m^2 + n^2 = \dfrac{1}{4}$
- Find the Dot Product:
Since both lines are unit vectors, regardless of specific direction cosines:
$(l_1, m_1, n_1) \cdot (l_2, m_2, n_2) = \dfrac{1}{2} \cdot \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{4} = \dfrac{1}{2}$
- Find the Angle:
Using the inverse cosine function:
$\theta = \cos^{-1} \left( \dfrac{1}{2} \right) = \dfrac{\pi}{3}$
- Check for Extraneous Solutions:
The solution $\theta = \frac{5\pi}{3}$ (cos = -\frac{1}{2}) leads to a contradiction with the given equations.
Therefore, the angle between the lines is:
$\theta = \boxed{\dfrac{\pi}{3}}$
Method 2
We know that, angle between two lines is
$\cos \theta =\frac{a _1 a _2+b _1 b _2+c _1 c _2}{\sqrt{a _1^{2}+b _1^{2}+c _1^{2}} \sqrt{a _2^{2}+b _2^{2}+c _2^{2}}} $
$l+m+n=0 $
$\Rightarrow \quad l =-(m+n) \Rightarrow \quad(m+n)^{2}=l^{2} $
$\Rightarrow m^{2}+n^{2}+2 m n =m^{2}+n^{2} \quad\left[\because l^{2}=m^{2}+n^{2}, \text { given }\right] $
$2 m n =0 $
$\text { When } m =0 $
$\Rightarrow l =-n$
Hence, $(l, m, n)$ is $(1,0,-1)$.
When $\quad n=0$, then $l=-m$
Hence, $(l, m, n)$ is $(1,0,-1)$.
$\therefore \quad \cos \theta=\frac{1+0+0}{\sqrt{2} \times \sqrt{2}}=\frac{1}{2} \Rightarrow \theta=\frac{\pi}{3}$
