मराठी

If a plane passes through the point (1, 1, 1) and is perpendicular to the line x−13=y−10=z−14 then its perpendicular distance from the origin is ______.

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प्रश्न

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 ______.

पर्याय

  • `3/4`

  • `4/3`

  • `7/5`

  •  1

MCQ
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उत्तर

 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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पाठ 28: The Plane - MCQ [पृष्ठ ८५]
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