question_answer

The distance between the points \[\left( a\,\cos \theta +b\,\sin \theta ,\,0 \right)\]and\[\left( 0,\,a\,\sin \theta -b\,\cos \theta  \right)\]is

A) \[{{a}^{2}}+{{b}^{2}}\]

B) \[{{a}^{2}}-{{b}^{2}}\]

C) \[\sqrt{{{a}^{2}}+{{b}^{2}}}\]

D) \[\sqrt{{{a}^{2}}-{{b}^{2}}}\]

Correct Answer: C

Solution :

The given points are \[\left( a\,\cos \theta +b\,\sin \theta ,\,0 \right)\] and \[\left( 0,\,a\,\sin \theta -b\,\cos \theta  \right)\].

Here, \[{{x}_{1}}=a\,\cos \theta +b\,\sin \theta ,\,{{y}_{1}}=0\]

\[{{x}_{2}}=0,\,{{y}_{2}}=a\,\sin \theta -b\,\cos \theta \]

\[\therefore \] Required distance

\[=\sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}}\]

\[=\sqrt{{{\left( a\,\cos \,\theta +b\,\sin \theta  \right)}^{2}}+{{\left( b\,\cos \theta -a\,\sin \theta  \right)}^{2}}}\]

\[=\sqrt{{{\cos }^{2}}\theta \left( {{a}^{2}}+{{b}^{2}} \right)+{{\sin }^{2}}\theta \left( {{a}^{2}}+{{b}^{2}} \right)}\]

\[=\sqrt{{{a}^{2}}+{{b}^{2}}}\]

\[\left[ \because \,\,{{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1 \right]\]