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तामिळनाडू बोर्ड ऑफ सेकेंडरी एज्युकेशनएचएससी विज्ञान इयत्ता १२

Show that the lines rijksijkr→=(6i^+j^+2k^)+s(i^+2j^-3k^) and rijktijkr→=(3i^+2j^-2k^)+t(2i^+4j^-5k^) are skew lines and hence find the shortest distance between them - Mathematics

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प्रश्न

Show that the lines `vec"r" = (6hat"i" + hat"j" + 2hat"k") + "s"(hat"i" + 2hat"j" - 3hat"k")` and `vec"r" = (3hat"i" + 2hat"j" - 2hat"k") + "t"(2hat"i" + 4hat"j" - 5hat"k")` are skew lines and hence find the shortest distance between them

बेरीज

उत्तर

`vec"r" = vec"a" + "s"vec"b"`,  `vec"r" = vec"c" + "s"vec"d"`

`vec"a" = 6hat"i" + hat"j" + 2hat"k"`,  `vec"b" = hat"i" + 2hat"j" - 3hat"k"`

`vec"c" = 3hat"i" + 2hat"j" - 2hat"k"`,  `vec"d" = 2hat"i" + 4hat"j" - 5hat"k"`

`vec"b"` is not a scalar multiple of `vec"d"`

∴ They are not parallel.

∴ The given lines are skew lines.

The shortest distance δ = `(|(vec"c" - vec"a")*(vec"b" xx vec"d")|)/(|vec"b" xx vec"d"|)`

`vec"b" xx vec"d" = |(hat"i", hat"j", hat"k"),(1, 2, -3),(2, 4, 5)|`

= `"i"(- 10 + 12) - hat"j"(- 5 + 6) + hat"k"(4 - 4)`

= `2hat"i" - hat"j"`

`|vec"b" xx vec"d"| = sqrt(2^2 + 1^2) = sqrt(5)`

`vec"c" - vec"a" = 3hat"i" + 2hat"j" - 2hat"k" - 6hat"i" - hat"j" - 2hat"k"`

= `3hat"i" + hat"j" - 4hat"k"`

`("c" - "a")*(vec"b" xx vec"d") = (-3hat"i" + hat"j" - 4hat"k")*(2hat"i" - hat"j")`

= – 6 – 1 + 0

= – 7

δ = `(|(vec"c" - vec"a")*(vec"b" xx vec"d")|)/(|vec"b" xx vec"d"|)`

= `|-7|/sqrt(5)`

= `7/sqrt(5)` units

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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 2 | पृष्ठ २५५

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