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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
`3/4`
`4/3`
`7/5`
1
Solution
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 `underlinebb(7/5)`.
Explanation:
Since the plane is perpendicular to the given line, its direction ratios are proportional to 3, 0, 4
So, the required equation of the plane is of the form
\[3x + 0y + 4z + d = 0 ... \left( 1 \right), \text{where d is a constant}.\]
\[\text{Since this plane passes through} (1, 1, 1),\]
\[3 + 0 + 4 + d = 0\]
\[\Rightarrow d = - 7\]
\[\text{Substituting this in (1), we get}\]
\[3x + 0y + 4z - 7 = 0 ... \left( 2 \right)\]
\[\text{Perpendicular distance of (2) from the origin}\]
\[= \frac{\left|3\left(0\right) + 0 + 4 \left(0\right) - 7 \right|}{\sqrt{3^2 + 0^2 + 4^2}}\]
\[= \frac{\left| 0 + 0 - 7 \right|}{\sqrt{25}}\]
\[= \frac{7}{5} \text{units}\]
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