Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  CBSE XII  >>  Math  >>  Three Dimensional Geometry
0 votes

If a variable line in two adjacent positions has direction cosines l,m,n and $l+\delta l,m+\delta m,n+\delta n$,show that the small angle $\delta\theta$ between the two positions is given by \[\delta\theta^2=\delta l^2+\delta m^2+\delta n^2\]

Can you answer this question?

1 Answer

0 votes
  • $ 1-cos2\theta= 2sin^2\theta$
  • Angle between two lines in $ \cos \theta=(\overrightarrow {d_1}.\overrightarrow {d_2})$
Step 1:
Let the direction cosines of the two lines be $d_1=(l,m,n)\:\:and\:\: d_2=(l+\delta l,m+\delta m,n+\delta n)$
We know that angle $\theta$ between two lines is given by
$ cos\theta=(\overrightarrow d_1.\overrightarrow d_2)$
It is given that the angle between the lines is $\delta\theta$
$\Rightarrow cos\delta \theta =(l,m,n).(l+\delta l,m+\delta m,n+\delta n)$
$= l(l+\delta l)+m(m+\delta m)+n(n+\delta n)$
On expanding we get,
$ \Rightarrow l^2+m^2+n^2+l\delta l+m\delta m+n\delta n=0$
Step 2:
We know that $ l^2+m^2+n^2=1$
Square of direction cosines is one.
Therefore $ \Rightarrow cos\delta \theta = l\delta l + m \delta m+ n\delta n+1$..........(i)
$ l^2+m^2+n^2=1\: and \: (l+\delta l)^2+(m+\delta m)^2+(n+\delta n)^2=1$
On expanding we get,
$ \Rightarrow 1+\delta l^2+\delta m^2+\delta n^2+2 l\delta l+2m \delta m+2n \delta n=1$
$ \Rightarrow 1+\delta l^2+\delta m^2+\delta n^2+2 ( l\delta l+m \delta m+n \delta n)=1$
Substituting for $l^2+m^2+n^2=1$
Step 3:
$1+\delta l^2+\delta m^2+\delta n^2+2(lsl+msm+n\delta n)=1$
$\Rightarrow\: \delta l^2+\delta m^2+\delta n^2+2(cos\delta\theta-1)=0$
$ \Rightarrow \delta l^2+\delta m^2+\delta n^2=2(1-\cos \delta \theta)$
But $ 1-cos2\theta= 2sin^2\theta/2$
$ \Rightarrow \delta l^2+\delta m^2+\delta n^2=2\bigg(2\sin^2 \large\frac{\delta \theta}{2}\bigg)$
$ \Rightarrow \delta l^2+\delta m^2+\delta n^2=4 \sin^2 \large\frac{\delta \theta}{2}$
If $ \theta$ is small then $ sin\theta \approx \theta$
$ \therefore \: 4\: \large\frac{\delta\theta^2}{4}$$=\delta l^2+\delta m^2+\delta n^2$
$ \Rightarrow \delta \theta = \delta l^2+\delta m^2+\delta n^2$
Hence proved
answered Jun 12, 2013 by meena.p

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App