The perpendicular distance from the origin to the plane containing the two lines, 3x+2=5y−2=7z+5 and 1x−1=4y−4=7z+4 is: (a) 116 (b)611 (c) 11 (d) 611

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The perpendicular distance from the origin to the plane contai

Question

The perpendicular distance from the origin to the plane containing the two lines, $\frac{x + 2}{3} = \frac{y - 2}{5} = \frac{z + 5}{7}$ and $\frac{x - 1}{1} = \frac{y - 4}{4} = \frac{z + 4}{7}$ is: (a) $116$ (b) $\frac{11}{6}$ (c) 11 (d) $611$

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Harper Discussed

The perpendicular distance from the origin to the plane containing the two lines,

\frac{x + 2}{3} = \frac{y - 2}{5} = \frac{z + 5}{7}

and

\frac{x - 1}{1} = \frac{y - 4}{4} = \frac{z + 4}{7}

is: (a)

116

(b)

\frac{11}{6}

611

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Question Text

The perpendicular distance from the origin to the plane containing the two lines,

\frac{x + 2}{3} = \frac{y - 2}{5} = \frac{z + 5}{7}

and

\frac{x - 1}{1} = \frac{y - 4}{4} = \frac{z + 4}{7}

is: (a)

116

(b)

\frac{11}{6}

611