Question

If a variable line in two adjacent position has direction cosines $l + \delta l,{\text{ }}m + \delta m,{\text{ }}n + \delta n,$ then the small angle $\delta \theta$ between the two positions is given by
A. $\delta {\theta ^2} = 4\left( {\delta {l^2} + \delta {m^2} + \delta {n^2}} \right)$
B. $\delta {\theta ^2} = 2\left( {\delta {l^2} + \delta {m^2} + \delta {n^2}} \right)$
C. $\delta {\theta ^2} = \left( {\delta {l^2} + \delta {m^2} + \delta {n^2}} \right)$
D.None of these

Answer 1

Hint : To solve this type of problem use the condition that the sum of squares of the direction cosines is equal to 1. Use this condition for both the lines and use one of lines to result in another. And at last use the cosine formula two find the angle between then we will be able to find the required value.

Complete step-by-step answer :
Suppose the direction cosines of two lines
$\left[ {l,m,n} \right] \;$ and $\left[ {l + \delta l,m + \delta m,n + \delta n} \right] \;$ direction ratios
We know the sum of squares of direction cosines is equal to 1
∴ ${l^2} + {m^2} + {n^2} = 1\;\;{\text{ }}\;$ ….....(1)
Similarly for 2nd line we get
${\left( {l + \delta l} \right)^2} + {\left( {m + \delta m} \right)^2} + {\left( {n + \delta n} \right)^2} = 1$
i.e., $\left( {{l^2} + {m^2} + {n^2}} \right) + \left( {\delta {l^2} + \delta {m^2} + \delta {n^2}} \right) + 2l\delta l + 2m\delta m + 2n\delta n = 1$ ………(2)
on putting the above result from equation to equation 2
we get,
$\Rightarrow \delta {l^2} + \delta {m^2} + \delta {n^2} = - 2\left( {l\delta l + m\delta m + n\delta n} \right)\;$ from (1)...(2)
we know the angle between two lines can be find by taking the dot product of direction cosine vector of both the line
Now, $cos\delta \theta = l\left( {l + \delta l} \right) + m\left( {m + \delta m} \right) + n\left( {n + \delta n} \right)$
Here $\delta \theta$ is the angle between both the lines
$= {l^2} + {m^2} + {n^2} + l\delta l + m\delta m + n\delta n$
Or,
$= 1 - \dfrac{1}{2}\left[ {\delta {l^2} + \delta {m^2} + \delta {n^2}} \right] \;$ from (1) and (2)
On simplifying we get,
$\delta {l^2} + \delta {m^2} + \delta {n^2} = 2\left( {1 - cos\delta \theta } \right)$
now applying the formula of $1 - cos\delta \theta$
we get,
$= 2.2si{n^2}\dfrac{1}{2}\delta \theta$
Or ,
$= 4{\left( {\dfrac{1}{2}\delta \theta } \right)^2}$ as we know that $sin\dfrac{1}{2}\delta \theta \approx \dfrac{1}{2}\delta \theta$ .
Or,
$= \delta {\theta ^2}$
Hence $\delta {\theta ^2} = \left( {\delta {l^2} + \delta {m^2} + \delta {n^2}} \right)$
So, the correct answer is “Option C”.

Note : The angle between two lines can be found by the dot product of the vector along the lines. Those vectors are formed by the direction cosines. Here in this question cos is also calculated in the same way.