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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 \[\delta\theta^2=\delta l^2+\delta m^2+\delta n^2\]

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1 Answer

  • $ 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

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