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महाराष्ट्र राज्य शिक्षण मंडळएचएससी विज्ञान (सामान्य) इयत्ता १२ वी

A line passes through (3, –1, 2) and is perpendicular to lines rijkijkandrjkjkr¯=(i^+j^-k^)+λ(2i^-2j^+k^)andr¯=(2^+j^-3k^)+μ(^-2j^+2k^). Find its equation. - Mathematics and Statistics

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प्रश्न

A line passes through (3, –1, 2) and is perpendicular to lines `bar"r" = (hat"i" + hat"j" - hat"k") + lambda(2hat"i" - 2hat"j" + hat"k") and bar"r" = (2hat"i" + hat"j" - 3hat"k") + mu(hat"i" - 2hat"j" + 2hat"k")`. Find its equation.

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उत्तर

The line `bar"r" = (hat"i" + hat"j" - hat"k") + λ(2hat"i" - 2hat"j" + hat"k")` is parallel to the vector `bar"b" = 2hat"i" - 2hat"j" + hat"k"` and the line `bar"r" = (2hat"i" + hat"j" - 3hat"k") + mu(hat"i" - 2hat"j" + 2hat"k")` is parallel to the vector. `bar"c" = hat"i" - 2hat"j" + 2hat"k"`.

The vector perpendicular to the vectors `bar"b" and bar"c"` is  given by

`bar"b" xx bar"c" = |(hat"i"       hatj      hatk), (2  -2    1),(1 -2      2)|`

`= hat"i"(-4 + 2) - hat"j"(4 - 1) + hat"k"(-4 + 2)`
`= -2hat"i" - 3hat"j" - 2hat"k"`
Since the required line is perpendicular to the given lines,
it is perpendicular to both `bar"b" and bar"c"`.
∴ It is parallel to `bar"b" xx bar"c"`
The equation of the line passing through `"A"(bara)` and parallel to `bar"b" and bar"c"` is
`bar"r" = bar"a" + λ(bar"b" xx bar"c")`, where λ is a scalar.
Here, `bar"a" = 3hat"i" -  hat"j" + 2hat"k"`
∴ the equation of the required line is
`bar"r" = (3hat"i" - hat"j" + 2hat"k") + λ(-2hat"i" - 3hat"j" - 2hat"k")`
or
`bar"r" = (3hat"i" - hat"j" + 2hat"k") + mu(2hat"i" + 3hat"j" + 2hat"k")`, where μ = `-λ`.

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पाठ 6: Line and Plane - Exercise 6.1 [पृष्ठ २००]

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