The perpendicular distance of the point (6, 5, 8) from y-axis is

5 unit

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6 unit

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8 unit

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10 unit

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Solution

The correct option is D 10 unit
$L e t P (6, 5, 8)$
$Q be the point on y axis$
$Hence Q (0, 5, 0)$
$P Q = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2} + {(z_{2} - z_{1})}^{2}}$
$= \sqrt{6^{2} + 8^{2}}$
$= \sqrt{36 + 64}$
$= \sqrt{100}$
$= 10$

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The perpendicular distance of the point (6, 5, 8) from y-axis is

The correct option is D 10 unit LetP(6,5,8) Qbe the point on y axis HenceQ(0,5,0) PQ=√(x2−x1)2+(y2−y1)2+(z2−z1)2 =√62+82 =√36+64 =√100 =10

The correct option is D 10 unit
$L e t P (6, 5, 8)$
$Q be the point on y axis$
$Hence Q (0, 5, 0)$
$P Q = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2} + {(z_{2} - z_{1})}^{2}}$
$= \sqrt{6^{2} + 8^{2}}$
$= \sqrt{36 + 64}$
$= \sqrt{100}$
$= 10$