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तामिळनाडू बोर्ड ऑफ सेकेंडरी एज्युकेशनएचएससी विज्ञान इयत्ता १२

Find the altitude of a parallelepiped determined by the vectors aijkbijka→=-2i^+5j^+3k^,b→=i^+3j^-2k^ and cijkc→=-i→+j→+4k→ if the base is taken as the parallelogram determined by bb→ and cc→

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प्रश्न

Find the altitude of a parallelepiped determined by the vectors `vec"a" = - 2hat"i" + 5hat"j" + 3hat"k", vec"b" = hat"i" + 3hat"j" - 2hat"k"` and `vec"c" = - vec"i" + vec"j" + 4vec"k"` if the base is taken as the parallelogram determined by `vec"b"` and `vec"c"`

बेरीज
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उत्तर

Volume = Base Area × Height

`|[(vec"a",  vec"b",  vec"c")]| = |vec"b" xx vec"c"| xx "Height"`

`|[(vec"a",  vec"b",  vec"c")]| = |(-2, 5, 3),(1, 3, -2),(-3, 1, 4)|`

= 2(12 + 2) − 5(4 − 6) + 3(1 + 9)

= − 28 + 10 + 30

= 12

`vec"b" xx vec"c" + |(vec"i", vec"j", vec"k"),(1, 3, -2),(-3, 1, 4)|`

= `vec"i"(14) - vec"j"(- 2) + vec"k"(10)`

`vec"b" xx vec"c" = 14vec"i" + 2vec"j" + 10vec"k"`

`|vec"b" xx vec"c"| = sqrt(196 + 4 + 100_`

= `sqrt(300)`

`(sqrt(300)) xx ("Height")` = 12

Height = `12/sqrt(300)`

= `12/(10sqrt(3))`

= `6/(5sqrt(3)) xx sqrt(3)/sqrt(3)`

= `(6sqrt(3))/(5 xx 3)`

= `(2sqrt(3))/5`

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पाठ 6: Applications of Vector Algebra - Exercise 6.2 [पृष्ठ २३७]

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सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 12 TN Board
पाठ 6 Applications of Vector Algebra
Exercise 6.2 | Q 5 | पृष्ठ २३७

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