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Find the parametric form of vector equation and Cartesian equations of straight line passing through (5, 2, 8) and is perpendicular to the straight lines - Mathematics

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प्रश्न

Find the parametric form of vector equation and Cartesian equations of straight line passing through (5, 2, 8) and is perpendicular to the straight lines `vec"r" = (hat"i" + hat"j" - hat"k") + "s"(2hat"i" - 2hat"j" + hat"k")` and `vec"r" = (2hat"i" - hat"j" - 3hat"k") + "t"(hat"i" + 2hat"j" + 2hat"k")`

बेरीज

उत्तर

GIven point `vec"a" = 5hat"i" + 2hat"j" + 8hat"k"`

`vec"b" = 2hat"i" + hat"j" - 2hat"k"`

`vec"d" = hat"i" + 2hat"j" + 2hat"k"`

`vec"b" xx vec"d" = |(hat"i", hat"j", hat"k"),(2, -2, 1),(1, 2, 2)|`

= `- 6hat"i" - 3hat"j" + 6hat"k"`

∴ This’ vector is perpendicular to both the given straight lines.

∴ The required straight line is `vec"r" = vec"a" + "t"(vec"b" xx vec"d")`

`vec"r" = (5hat"i" + 2hat"j" + 8hat"k") + "t"(- 6hat"i" - 3hat"j" + 6hat"k")`

 OR

`vec"r" = (5hat"ii" + 2hat"j" + 8hat"") + "t"(2hat"i" + hat"j" - 2hat"k"), "t" ∈ "R"`

`(x - 5)/(-6) = (y - 2)/(-3) = (z - 8)/6`

`(x - 5)/2 = (y - 2)/1 = (z - 8)/(-2)`

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Application of Vectors to 3-dimensional Geometry
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पाठ 6: Applications of Vector Algebra - Exercise 6.5 [पृष्ठ २५५]

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सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 12 TN Board
पाठ 6 Applications of Vector Algebra
Exercise 6.5 | Q 1 | पृष्ठ २५५

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