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

$\begin{array}{1 1} (A)\;\beta\qquad(B)\;|\beta|\qquad(C)\;|\beta|+|\gamma|\qquad(D)\;\sqrt{\alpha^2+\gamma^2}\end{array} $

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  • Drop perpendicular from the point to Y axis and XY plane.
  • Join the foot of the perpendicular and the point to get right angle triangle.
  • The distance required is hypotenuse after this $ \Delta$
Ans : (D)


answered Mar 7, 2013 by thanvigandhi_1
edited Apr 5, 2013 by thanvigandhi_1

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