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The distance between the points $ (a \cos \theta-b \sin \theta, 0) $ and $ (0, a \sin \theta-b \cos \theta) $ is_____.
Given:
Points \( (a \cos \theta-b \sin \theta, 0) \) and \( (0, a \sin \theta-b \cos \theta) \).
To do:
We have to find the distance between \( \mathrm{A}(a \cos \theta+b \sin \theta, 0) \) and \( B(0, a \sin \theta-b \cos \theta) \).
Solution:
We know that,
The distance between \( \mathrm{A}\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right) \) and \( \mathrm{B}\left(\mathrm{x}_{2}, \mathrm{y}_{2}\right) \) is given by, \( A B=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}} \)
Therefore,
$A B =\sqrt{(a \cos \theta+b \sin \theta-0)^{2}+(0-a \sin \theta+b \cos \theta)^{2}}$
$=\sqrt{(a^{2}(\sin ^{2} \theta+\cos ^{2} \theta)+b^{2}(\sin ^{2} \theta+\cos ^{2} \theta)}$
$A B=\sqrt{a^{2}+b^{2}}$ [Since \( \sin ^{2} \theta+\cos ^{2} \theta=1 \)]
The distance between the points \( (a \cos \theta-b \sin \theta, 0) \) and \( (0, a \sin \theta-b \cos \theta) \) is $\sqrt{a^{2}+b^{2}}$.
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