3D Geometry Ques 4
4. 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: 4.(a)
Solution:
- 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 $(1,0,-1)$.
$\therefore \quad \cos \theta=\frac{1+0+0}{\sqrt{2} \times \sqrt{2}}=\frac{1}{2} \Rightarrow \theta=\frac{\pi}{3}$
BETA
