First slide

Direction ratios and direction cosines

Question

If a variable line in two adjacent positions has direction cosines $(l, m, n) a n d (l + δ l, m + δ m, n + δ n)$ and $δ θ$ is the angle between the two positions, then ${(δ l)}^{2} + {(δ m)}^{2} + {(δ n)}^{2} =$

Moderate

A

$\frac{{(δ θ)}^{2}}{4}$
B

$\frac{{(δ θ)}^{2}}{2}$
C

$\frac{{(δ θ)}^{2}}{l m n}$
D

${(δ θ)}^{2}$

Solution

If $θ$ is the acute angle between two lines having direction ratios $〈a_{1}, b_{1}, c_{1}〉$ and $〈a_{2}, b_{2}, c_{2}〉$
then $\cos θ = |\frac{a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2}}{\sqrt{{a_{1}}^{2} + {b_{1}}^{2} + {c_{1}}^{2}} \cdot \sqrt{{a_{2}}^{2} + {b_{2}}^{2} + {c_{2}}^{2}}}|$
It implies that
$\begin{matrix} \cos δ θ = |l (l + δ l) + m (m + δ m) + n (n + δ n)| \\ = 1 + l (δ l) + m (δ m) + n (δ n) \end{matrix}$
And
$\begin{matrix} {(l + δ l)}^{2} + {(m + δ m)}^{2} + {(n + δ n)}^{2} = 1 \\ l^{2} + m^{2} + n^{2} + 2 (l δ l) + 2 (m δ m) + 2 (n δ n) + {(δ l)}^{2} + {(δ m)}^{2} + {(δ n)}^{2} = 1 \\ 2 (l δ l + m δ m + n δ n) + {(δ l)}^{2} + {(δ m)}^{2} + {(δ n)}^{2} = 0 \end{matrix}$
It implies that
$\begin{matrix} {(δ l)}^{2} + {(δ m)}^{2} + {(δ n)}^{2} = - 2 (l δ l + m δ m + n δ n) \\ = 2 (1 - \cos δ θ) \\ = 4 \sin^{2} (\frac{δ θ}{2}) \\ ≅ 4 {(\frac{δ θ}{2})}^{2} (∵ \sin θ ≅ θ f o r s m a l l θ) \\ = {(δ θ)}^{2} \end{matrix}$

Therefore, ${(δ l)}^{2} + {(δ m)}^{2} + {(δ n)}^{2} = {(δ θ)}^{2}$

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