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$\text{a}+\text{b}$
- C$\text{a}^2-\text{b}^2$
- ✓$\sqrt{\text{a}^2+\text{b}^2}$
✓
Answer
Correct option: D.
$\sqrt{\text{a}^2+\text{b}^2}$
We have to find the distance between $\text{A}(\text{a}\cos\theta+\text{b}\sin\theta,0)$ and $\text{B}(0,\text{a}\sin\theta-\text{b}\cos\theta).$
In general, the distance between $A(x_1, y_1)$ and $B(x_2, y_2)$ is given by,
$\text{AB}=\sqrt{(\text{x}_2-\text{x}_1)^2+(\text{y}_2-\text{y}_1)^2}$
So, $=\sqrt{(\text{a}\cos\theta+\text{b}\sin\theta)^2+(-\text{a}\sin\theta+\text{b}\cos\theta)^2}$
$=\sqrt{\text{a}^2\cos^2\theta+\text{b}^2\sin^2\theta+2\text{ab}\sin\theta\cos\theta+\text{a}^2\sin^2\theta+\text{b}^2\cos^2\theta-2\text{ab}\sin\theta\cos\theta}$
$\text{AB}=\sqrt{(\text{a}\cos\theta+\text{b}\sin\theta-0)^2+(0-\text{a}\sin\theta+\text{b}\cos\theta)^2}$
$=\sqrt{\text{a}^2(\sin^2\theta+\cos^2\theta)+\text{b}^2(\sin^2\theta+\cos^2\theta)}$
But according to the trigonometric identity,
$\sin^2\theta+\cos^2\theta=1$
Therefore, $\text{AB}=\sqrt{\text{a}^2+\text{b}^2}$
In general, the distance between $A(x_1, y_1)$ and $B(x_2, y_2)$ is given by,
$\text{AB}=\sqrt{(\text{x}_2-\text{x}_1)^2+(\text{y}_2-\text{y}_1)^2}$
So, $=\sqrt{(\text{a}\cos\theta+\text{b}\sin\theta)^2+(-\text{a}\sin\theta+\text{b}\cos\theta)^2}$
$=\sqrt{\text{a}^2\cos^2\theta+\text{b}^2\sin^2\theta+2\text{ab}\sin\theta\cos\theta+\text{a}^2\sin^2\theta+\text{b}^2\cos^2\theta-2\text{ab}\sin\theta\cos\theta}$
$\text{AB}=\sqrt{(\text{a}\cos\theta+\text{b}\sin\theta-0)^2+(0-\text{a}\sin\theta+\text{b}\cos\theta)^2}$
$=\sqrt{\text{a}^2(\sin^2\theta+\cos^2\theta)+\text{b}^2(\sin^2\theta+\cos^2\theta)}$
But according to the trigonometric identity,
$\sin^2\theta+\cos^2\theta=1$
Therefore, $\text{AB}=\sqrt{\text{a}^2+\text{b}^2}$
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