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Tamil Nadu Board of Secondary EducationHSC Science Class 12

Forces of magnitudes 52 and 102 units acting in the directions ijk3i^+4j^+5k^ and ijk10i^+6j^-8k^ respectively, act on a particle which is displaced from the point with position vector - Mathematics

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Question

Forces of magnitudes `5sqrt(2)` and `10sqrt(2)` units acting in the directions `3hat"i" + 4hat"j" + 5hat"k"` and `10hat"i" + 6hat"j" - 8hat"k"` respectively, act on a particle which is displaced from the point with position vector `4hat"i" - 3hat"j" - 2hat"k"` to the point with position vector `6hat"i" + hat"j" - 3hat"k"`. Find the work done by the forces

Sum

Solution

`vec"F"_1 = (3hat"i" + 4hat"j" + 5hat"k")/|3hat"i" + 4hat"j" + 5hat"k"|`

= `(3hat"i" + 4hat"j" + 5hat"k")/sqrt(9 + 16 + 25)`

= `(3hat"i" + 4hat"j" + 5hat"k")/sqrt(50)`

= `(3hat"i" + 4hat"j" + 5hat"k")/(5sqrt(2)`

`vec"F"_2 = (10hat"i" + 6hat"j" - 8hat"k")/sqrt(100 + 36 + 64)`

= `(10hat"i" + 6hat"j" - 8hat"k")/sqrt(200)`

= `(10hat"i" + 6hat"j" - 8hat"k")/(2sqrt(2)`

Resultant force `bar"F" = bar"F"_1 + bar"F"_2`

= `5sqrt(2)  bar"F"_1 + 10sqrt(2)  bar"F"_2`

= `3hat"i" + 4hat"j" + 5hat"k" + 10hat"i" + 5hat"j" - 8hat"k"`

= `13hat"i" + 10hat"j" - 3hat"k"`

`bar"OA" = 4hat"i" - 3hat"j" - 2hat"k"`

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

`bar"d" = bar"AB"`

= `bar"OB" - bar"OA"`

= `2hat"i" + 4hat"j" - hat"k"`

`bar"F" * bar"d" = (13hat"i" + 10hat"j" - 3hat"k")* (2hat"i" + 4hat"j" - hat"k")`

= 26 + 40 + 3

= 69 units

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Scalar Product and Vector Product
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Chapter 6: Applications of Vector Algebra - Exercise 6.1 [Page 231]

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Samacheer Kalvi Mathematics - Volume 1 and 2 [English] Class 12 TN Board
Chapter 6 Applications of Vector Algebra
Exercise 6.1 | Q 12 | Page 231

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