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Three Dimensional Geometry — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSThree Dimensional Geometry5 Marks
Question
If a variable line in two adjacent positions has direction cosines l, m, n and $\text{l}+\delta\text{l},\text{m}+\delta\text{m},\text{n}+\delta\text{n},$ show that the small angle $\delta\theta$ between the two positions is given by $\delta\theta^2=\delta\text{l}^2+\delta\text{m}^2+\delta\text{n}^2.$
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Answer

We have l, m, n and $\text{l}+\delta\text{l},\text{m}+\delta\text{m},\text{n}+\delta\text{n}$ as direction cosines of a variable line in two different positions.
$\therefore\text{l}^2+\text{m}^2+\text{n}^2=1\ ....(\text{i})$
and $(\text{l}+\delta\text{l})^2+(\text{m}+\delta\text{m})^2+(\text{n}+\delta)\ ....(\text{ii})$
$\Rightarrow\text{l}^2+\text{m}^2+\text{n}^2+\delta\text{l}^2+\delta\text{m}^2+\delta\text{n}^2+2(\text{l}\delta\text{l}+\text{m}\delta\text{m}+\text{n}\delta\text{n})=1$
$\Rightarrow\delta\text{l}^2+\delta\text{m}^2+\delta\text{n}^2=2(\text{l}\delta\text{l}+\text{m}\delta\text{m}+\text{n}\delta\text{n})$ $[\because\text{l}^2+\text{m}^2+\text{n}^2=1]$
$\Rightarrow\text{l}\delta\text{l}+\text{m}\delta\text{m}+\text{n}\delta\text{n}=\frac{-1}{2}(\delta\text{l}^2+\delta\text{m}^2+\delta\text{n}^2)\ ....(\text{iii})$
Now $\vec{\text{a}}$ and $\vec{\text{b}}$ are unit vectors along a line with direction cosines l, m, n and 
$(\text{l}+\delta\text{l}),(\text{m}+\delta\text{m}),(\text{n}+\delta\text{n}),$ respectively.
$\therefore\vec{\text{a}}=\text{l}\hat{\text{i}}+\text{m}\hat{\text{j}}+\text{n}\hat{\text{k}}$ and $\vec{\text{b}}=(\text{l}+\delta\text{l})\hat{\text{i}}+(\text{m}+\delta\text{m})\hat{\text{j}}+(\text{n}+\delta\text{n})\hat{\text{k}}$
$\Rightarrow\cos\delta\theta=\text{l}(\text{l}+\delta\text{l})+\text{m}(\text{m}+\delta\text{m})+\text{n}(\text{n}+\delta\text{n})$
$=(\text{l}^2+\text{m}^2+\text{n}^2)+(\text{l}\delta\text{l}+\text{m}\delta\text{m}+\text{n}\delta\text{n})$
$=1-\frac{1}{2}(\delta\text{l}^2+\delta\text{m}^2+\delta\text{n}^2)$
$\Rightarrow2(1-\cos\delta\theta)=(\delta\text{l}^2+\delta\text{m}^2+\delta\text{n}^2)$
$\Rightarrow2\cdot2\sin^2\frac{\delta\theta}{2}=\delta\text{l}^2+\delta\text{m}^2+\delta\text{n}^2$
$\Rightarrow4\Big(\frac{\delta\theta}{2}\Big)=​​\delta\text{l}^2+\delta\text{m}^2+\delta\text{n}^2$$\Big[\text{Since}\frac{\delta\theta}{2}\text{is small,}\sin\frac{\delta\theta}{2}=\frac{\delta\theta}{2}\Big]$
$\Rightarrow\delta\theta^2=\delta\text{l}^2+\delta\text{m}^2+\delta\text{n}^2$

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