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महाराष्ट्र राज्य शिक्षण मंडळएचएससी विज्ञान (सामान्य) इयत्ता १२ वी

Solve the following : Find the vector equation of the plane passing through the origin and containing the line rijkijkr¯=(i^+4j^+k^)+λ(i^+2j^+k^). - Mathematics and Statistics

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प्रश्न

Solve the following :

Find the vector equation of the plane passing through the origin and containing the line `bar"r" = (hat"i" + 4hat"j" + hat"k") + lambda(hat"i" + 2hat"j" + hat"k")`.

बेरीज

उत्तर

The vector equation of the plane passing through `"A"(bara)` and perpendicular to the vector `bar"n"` is `bar"r".bar"n" = bar"a".bar"n"`             ...(1)

We can take `bar"a" = bar"0"` since the plane passes through the origin.

The point M with position vector `bar"m" = hat"i" + 4hat"j" + hat"k"`  lies on the line and hence it lies on the plane.

∴ `bar"OM" = bar"m" = hat"i" + 4hat"j" + hat"k"` lies on the plane.

The plane contains the given line which is parallel to `bar"b" = hat"i" + 2hat"j" + hat"k"`.

Let `bar"n"` be normal to the plane. Then `bar"n"` is perpendicular to `bar"OM"`  as well as `bar"b"`

∴ `bar"n" = bar"OM" xx bar"b" = |(hat"i" ,,),(1, 4, 1),(1, 2, 1)|`

= `(4 - 2)hat"i" - (1 - 1)hat"j" + (2 - 4)hat"k"`

= `2hat"i" - 2hat"k"`

∴ from (1), the vector equation of the required plane is

`bar"r".(2hat"i" - 2hat"k") = bar"0".bar"n"` = 0

i.e. `bar"r".(hat"i" - hat"k")` = 0.

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Vector and Cartesian Equations of a Line
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 6: Line and Plane - Miscellaneous Exercise 6 B [पृष्ठ २२६]

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बालभारती Mathematics and Statistics 1 (Arts and Science) [English] 12 Standard HSC Maharashtra State Board
पाठ 6 Line and Plane
Miscellaneous Exercise 6 B | Q 19 | पृष्ठ २२६

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