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महाराष्ट्र राज्य शिक्षण मंडळएचएससी विज्ञान (सामान्य) इयत्ता १२ वी

Solve the following : Find the cartesian equation of the plane rijkijkr¯=λ(i^+j^-k^)+μ(i^+2j^+3k^).

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प्रश्न

Solve the following :

Find the cartesian equation of the plane `bar"r" = lambda(hat"i" + hat"j" - hat"k") + mu(hat"i" + 2hat"j" + 3hat"k")`.

बेरीज
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उत्तर

The equation `bar"r" = bar"a" + lambdabar"b" + mubar"c"` represents a plane passing through a point having position vector `bar"a"` and parallel to vectors `bar"b" and bar"c"`.
Here,

`bar"a"=0hat"i"+0hat"j"+0hat"k"`
`bar"b" = hat"i" + hat"j" - hat"k"`,
`bar"c" = hat"i" + 2hat"j" + 3hat"k"`

∴ `bar"b" xx bar"c" = |(hat"i",hat"j" ,hat"k"),(1, 1, -1),(1, 2, 3)|`

= `(3 + 2)hat"i" - (3 - 1)hat"j" + (2 - 1)hat"k"`
= `5hat"i" - 4hat"j" + hat"k"`
Also,
`bar"a".(bar"b" xx bar"c")`
= `(0hat"i"+0hat"j"+0hat"k").(5hat"i"-4hat"j"+hat"k")`
= 0
The vector equation of the plane passing through A`(bara)` and parallel to `bar"b" and bar"c"` is

`bar"r".(bar"b" xx bar"c") = bar"a".(bar"b" xx bar"c")`

∴ the vector equation of the given plane is

`bar"r".(5hat"i" - 4hat"j" + hat"k")` = 0

If `bar"r" = xhat"i" + yhat"j" + zhat"k"`, then this equation becomes

`(xhat"i" + yhat"j" + zhat"k").(5hat"i" - 4hat"j" + hat"k")` = 0

∴ 5x – 4y + z = 0.
This is the cartesian equation of the required plane.

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पाठ 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
पाठ 6 Line and Plane
Miscellaneous Exercise 6 B | Q 11 | पृष्ठ २२६

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