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The foot of the perpendicular drawn from the origin to a plane is M(1, 2, 0). Find the vector equation of the plane. - Mathematics and Statistics

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प्रश्न

The foot of the perpendicular drawn from the origin to a plane is M(1, 2, 0). Find the vector equation of the plane.

योग
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उत्तर

The vector equation of the plane passing through `A(bara)` and perpendicular to `barn` is `barr.barn = bara.barn`.

M(1, 2, 0) is the foot of the perpendicular drawn from the origin to the plane.

Then the plane is passing through M and is perpendicular to OM.

If `barm` is the position vector of M, then `barm = hati + 2hatj`.

Normal to the plane is `barn = bar(OM) = hati + 2hatj`

`barm.barn = (hati + 2hatj) . (hati + 2hatj)`

= 1(1) + 2(2)

= 1 + 4

= 5

∴ The vector equation of the required plane is `barr.(hati + 2hatj)` = 5.

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अध्याय 6: Line and Plane - Miscellaneous Exercise 6 B [पृष्ठ २२६]

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बालभारती Mathematics and Statistics 1 (Arts and Science) [English] Standard 12 Maharashtra State Board
अध्याय 6 Line and Plane
Miscellaneous Exercise 6 B | Q 8 | पृष्ठ २२६

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