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:

  1. 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: $m^2 + n^2 = \dfrac{1}{4}$

  1. Find the Dot Product:

Since both lines are unit vectors, regardless of specific direction cosines:

$(l_1, m_1, n_1) \cdot (l_2, m_2, n_2) = \dfrac{1}{2} \cdot \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{4} = \dfrac{1}{2}$

  1. Find the Angle:

Using the inverse cosine function:

$\theta = \cos^{-1} \left( \dfrac{1}{2} \right) = \dfrac{\pi}{3}$

  1. 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 $(1,0,-1)$.

$\therefore \quad \cos \theta=\frac{1+0+0}{\sqrt{2} \times \sqrt{2}}=\frac{1}{2} \Rightarrow \theta=\frac{\pi}{3}$



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