find the magnitude of the resultant of two vectors A and B in terms of their magnitude and angle theta between them
Answers
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Answer: \theta = tan ^{-1} b sin \alpha / a + b cos \alpha.θ=tan
−1
bsinα/a+bcosα.
Solution:
To know the resultant vector magnitude and its angle between, simply apply the triangle law of vector addition. Assume the two vector A and B with an angle alpha whose resultant be vector C given by vector C = vector A + vector B. In addition drop a line from the point B and C meeting to form the right angled triangle, so AC ^2 = AD ^2 + CD ^2AC
2
=AD
2
+CD
2
=> AC = \sqrt{ AD ^2 + CD ^2 }AC=
AD
2
+CD
2
=> AC = \sqrt{ ( AB + BD) ^2 + CD ^2}AC=
(AB+BD)
2
+CD
2
=> AC = \sqrt{ Ab ^2 + 2 AB x BD + BD ^2 + CD ^2 }AC=
Ab
2
+2ABxBD+BD
2
+CD
2
=> From triangle BCD, cos alpha = BD / BC
=> BD = BC \times cos \alphaBD=BC×cosα
=> AC = \sqrt{a^2 + 2 ab cos \alpha + BC ^2}AC=
a
2
+2abcosα+BC
2
=> AC = \sqrt{a ^2 + b ^2 + 2 a b cos \alpha}AC=
a
2
+b
2
+2abcosα
=> c = \sqrt{a ^2 + b ^2 + 2 a b cos \alpha}c=
a
2
+b
2
+2abcosα
The resultant vector makes angle theta therefore, in triangle BCD,
sin \alpha = CD / BCsinα=CD/BC
=> CD = BC sin \alpha,CD=BCsinα,
and in triangle ADC,
tan \theta = CD / AD = BC sin \alpha / a + b cos \alphatanθ=CD/AD=BCsinα/a+bcosα
=> \theta = tan ^{-1} b sin \alpha / a + b cos \alpha.θ=tan
−1
bsinα/a+bcosα.