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rectangular cartensian co-ordinates — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSrectangular cartensian co-ordinates1 Mark
MCQ
The distance between the points $(a\cos\, \alpha ,\,\,a\sin\, \alpha )$ and $(a\cos \,\beta ,a\sin\, \beta )$ is
  • A
    $a\cos\, \frac{{\alpha -\, \beta }}{2}$
  • B
    $2a\cos\, \frac{{\alpha -\, \beta }}{2}$
  • C
    $a\sin\,\frac{{\alpha -\, \beta }}{2}$
  • ✓
    $2a\sin\, \frac{{\alpha - \,\beta }}{2}$
✓

Answer

Correct option: D.
$2a\sin\, \frac{{\alpha - \,\beta }}{2}$
d
(d) Distance $ = \sqrt {{a^2}{{(\cos \alpha - \cos \beta )}^2} + {a^2}{{(\sin \alpha - \sin \beta )}^2}} $

$ = a\sqrt {{{\sin }^2}\alpha + {{\cos }^2}\alpha + {{\cos }^2}\beta + {{\sin }^2}\beta - 2\cos \alpha \cos \beta - 2\sin \alpha \sin \beta } $

$ = a\sqrt {2\left\{ {1 - \cos \,(\alpha - \beta )} \right\}} $

$= 2a\,\sin \,\left( {\frac{{\alpha - \beta }}{2}} \right)$

Trick : Put $a = 1,\,\,\alpha = \frac{\pi }{2},\,\beta = \frac{\pi }{6},$ then the points will be $(0, 1)$ and $\left( {\frac{{\sqrt 3 }}{2},\,\,\frac{1}{2}} \right)$.

Obviously, the distance between these two points is 1 which is given by $(d)$.

$\left\{ {\,\,2a\,\sin \frac{{\alpha  - \beta }}{2} = 2 \times 1 \times \sin \frac{{(\pi /2) - (\pi /6)}}{2} = 2 \times \frac{1}{2} = 1} \right\}$

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