3D Geometry 1 Question 1

####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: 1. (a)

Solution:

  1. 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+q=0 \newline$

$ \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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