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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 
Option: 1 a^{2}+b^{2}  
Option: 2 a^{2}-b^{2}
Option: 3 \sqrt{a^{2}+b^{2}}
Option: 4 \sqrt{a^{2}-b^{2}}

Answers (1)

The distance between the points 

\\\(x_1,y_1)\ and\ (x_2,y_2)\ is\\d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

\\\(x_1,y_1)\ and\ (x_2,y_2)\ is\\d=\sqrt{(0-acos\theta-bsin\theta)^2+(asin\theta-bcos\theta-0)^2}\\=\sqrt{a^2cos^2\theta+b^2sin^2\theta+2absin\thetacos\theta+a^2sin^2\theta+b^2cos^2\theta-2absin\thetacos\theta}\\=\sqrt{a^2(sin^2\theta+cos^2\theta)+b^2(sin ^2\theta+cos^2\theta)}\\=\sqrt{a^2+b^2}

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Safeer PP

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