If a plane passes through the point (1,1,1) and is perpendicular to the line x − 1 3 = y − 1 0 = z − 1 4 then its perpendicular distance from the origin is Options: A . &nbsp34 B . &nbsp43 C . &nbsp75 D . &nbsp1

If a plane passes through the point (1,1,1) and is perpendicular to the line x&#

Question

If a plane passes through the point (1,1,1) and is perpendicular to the line

\frac{x - 1}{3} = \frac{y - 1}{0} = \frac{z - 1}{4}

then its perpendicular distance from the origin is

Options:

A . 34

B . 43

C . 75

D . 1

Answer: Option C
:
C
The d.r of the normal to the plane is

(3, 0, 4)

The equation of the plane is 3x + 0y + 4z + d = 0 since it passes through (1, 1, 1) so; d = --7
Now distance of the plane

3 x + 4 z - 7 = 0

from (0, 0, 0) is

\frac{7}{\sqrt{3^{2} + 4^{2}}} = \frac{7}{5}

units

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