MCQ
The distance between the points $(\text{a}\cos\theta+\text{b}\sin\theta,0)$ and $(0,\text{a}\sin\theta-\text{b}\cos\theta),$ is :
  • A
    $\text{a}^2+\text{b}^2$
  • B
    $\text{a}^2-\text{b}^2$
  • $\sqrt{\text{a}^2+\text{b}^2}$
  • D
    $\sqrt{\text{a}^2-\text{b}^2}$

Answer

Correct option: C.
$\sqrt{\text{a}^2+\text{b}^2}$
Since by the distance formula,
The distance between the points $(\text{a}\cos\theta+\text{b}\sin\theta,0)$ and $(0,\text{a}\sin\theta-\text{b}\cos\theta)$ is,
$\text{D}=\sqrt{(0-\text{a}\cos\theta-\text{b}\sin\theta)^2+(\text{a}\sin\theta-\text{b}\cos\theta-0)}$
$=\sqrt{(-\text{a}\cos\theta-\text{b}\sin\theta)^2+(\text{a}\sin\theta-\text{b}\cos\theta)^2}$
$=((\text{x}\pm\text{y})^2=\text{x}^2+2\text{xy}+\text{y}^2)$
$=\sqrt{\text{a}^2\cos^2\theta+2\text{ab}\sin\theta\cos\theta+\text{b}^2\sin^2\theta+ \text{a}^2\sin^2\theta-2\text{ab}\sin\theta\cos\theta+\text{b}^2\cos^2\theta}$
$=\sqrt{(\text{a}^2+\text{b}^2)\sin^2\theta+(\text{a}^2+\text{b}^2)\cos^2\theta}$
$=\sqrt{\text{a}^2+\text{b}^2(\sin^2\theta+\cos^2\theta)}$
$=\sqrt{\text{a}^2+\text{b}^2}$

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