Distance of the point ( , , ) from y-axis is

MCQ

Distance of the point $(\alpha, \beta, \gamma)$ from $y$-axis is

A
$\beta$
B
$|\beta|$
C
$|\beta|+|\gamma|$
✓
$\sqrt{\alpha^2+\gamma^2}$

✓

Answer

Correct option: D.

$\sqrt{\alpha^2+\gamma^2}$

(d) : Foot of perpendicular from $(\alpha, \beta, \gamma)$ on the $y$-axis is $(0, \beta, 0)$.
$\therefore \quad$ Distance of $(\alpha, \beta, \gamma)$ from $y$-axis $=$ distance of $(\alpha, \beta, \gamma)$ from $(0, \beta, 0)$
$
=\sqrt{(0-\alpha)^2+(\beta-\beta)^2+(0-\gamma)^2}=\sqrt{\alpha^2+\gamma^2}
$

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