a) Consider the following triangle
圖片參考：http://imgcld.yimg.com/8/n/HA00425268/o/701101190014613873371290.jpg
By dropping a perpendicular from the vertex with angle to the base, use trigonometry to show that:
c = a cos β + b cos α
b) It can similarly be deduced that
a = c cos β + b cos γb = c cos α + a cos γ
(no derivation is required).
Regard the above as three linear equations in unknowns cos α , cosβ
and cosγ, with a; b and c considered as known. By forming an aug-
mented matrix, solve for cosγ to show that:
圖片參考：http://imgcld.yimg.com/8/n/HA00425268/o/701101190014613873371301.jpg
c) Consider now the diagram
圖片參考：http://imgcld.yimg.com/8/n/HA00425268/o/701101190014613873371302.jpg
Identify IIuII, IIvII and IIu - vII with a; b; c in the first diagram, and use the fact that
圖片參考：http://imgcld.yimg.com/8/n/HA00425268/o/701101190014613873371313.jpg
to deduce from (b) that
圖片參考：http://imgcld.yimg.com/8/n/HA00425268/o/701101190014613873371324.jpg
Basic Vector Question
2011-01-19 5:08 pm
回答 (2)
2011-01-19 10:20 pm
✔ 最佳答案
(a) Draw a perpendicular line from the vertex with angle γ to the base c.Then c will be devided in two parts. Let the left part is x and the right part is ycosβ = x/a => x = acosβcosα = y/b => y = bcosαc = x + y = acosβ + bcosα(b) c = acosβ + bcosα...(1)a = ccosβ + bcosγ...(2)b = ccosα + acosγ...(3)From (2)cosγ = (a - ccosβ)/b= {a - c[(c - bcosα)/a]}/b= [a^2 - (c^2 - cbcosα)]/(ab)= {a^2 - [c^2 - cb(b - acosγ)/c]}/(ab)= {a^2 - [c^2 - (b^2 - abcosγ)]}/(ab)= (a^2 + b^2 - c^2 + abcosγ)/(ab)So, cosγ = (a^2 + b^2 - c^2)/(2ab)(c) Let a = ||u||, b = ||v||, c = ||u-v||a^2 + b^2 - c^2 = ||u||^2 + ||v||^2 - ||u-v||^2= 2||u||.||v||ab = ||u||||v||So, cosγ = (a^2 + b^2 - c^2)/(2ab)= (2||u||.||v||)/(2||u||||v||)= ||u||.||v||/(||u||||v||)
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