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Model Paper 10 — MATHS STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 ScienceMATHSModel Paper 101 Mark
MCQ
Distance of the point $(\alpha, \beta, \gamma)$ from $y$-axis is
  • A
    $\sqrt{\alpha^2+\gamma^2}$
  • B
    $|\beta|+|\gamma|$
  • C
    $|\beta|$
  • D
    $\beta$
✓

Answer

(a) $\sqrt{\alpha^2+\gamma^2}$
Explanation: The foot of perpendicular from point $P(\alpha, \beta, \gamma)$ on y -axis is $Q(0, \beta, 0)$
$\therefore$ Required distance, $P Q=\sqrt{(\alpha-0)^2+(\beta-\beta)^2+(\gamma-0)^2}=\sqrt{\alpha^2+\gamma^2}$

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