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Find the parametric form of vector equation of the straight line passing through (−1, 2, 1) and parallel to the straight line rijktijkr→=(2i^+3j^-k^)+t(i^-2j^+k^) and hence find the shortest distance - Mathematics

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प्रश्न

Find the parametric form of vector equation of the straight line passing through (−1, 2, 1) and parallel to the straight line `vec"r" = (2hat"i" + 3hat"j" - hat"k") + "t"(hat"i" - 2hat"j" + hat"k")` and hence find the shortest distance between the lines

बेरीज

उत्तर

Given point `vec"a" = -vec"i" + 2vec"j" + vec"k"`

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

`vec"r" = vec"a" + "t"vec"b"`

`vec"r" = (-vec"i" + 2vec"j" + vec"k") + "t"(vec"i" - 2vec"j" + vec"k")`

Parallel to the straight line

`vec"r" = (2vec"i" + 3vec"j" - vec"k") + "t"(vec"i" - 2vec"j" + vec"k")`

Here `vec"c" = 2vec"i" + 3vec"j" - vec"k"`

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

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

= `-3hat"i" - 5hat"j" - 7hat"k"`

`|(vec"c" - vec"a") xx vec"b"| = sqrt(9 + 25 + 49)`

= `sqrt(83)`

`|vec"b"| = sqrt(1 + 4 + 1)`

= `sqrt(6)`

Shortest distance between parallel lines = `(|(vec"c" - vec"a") xx vec"b"|)/|vec"b"|`

= `sqrt(83)/sqrt(6)` 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 6 | पृष्ठ २५५

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