Answer to Question #132983 in Field Theory for Aliha

Question #132983

Given three vector A=24i+33j, b=55 -12j and c=2i+43j

a) find magnitude of each vector

b) write an expression and direction of the vector difference A-B

d)in a vector diagram show A+B and A-B and also show the diagram agrees



1
Expert's answer
2020-09-16T10:11:52-0400

(a)

A=242+332=40.8|{\bf A}|=\sqrt{24^2+33^2}=40.8

B=552+122=56.3|{\bf B}|=\sqrt{55^2+12^2}=56.3

C=22+432=43.05|{\bf C}|=\sqrt{2^2+43^2}=43.05

(b)

AB=(24i+33j)(2i+43j)=22i10j{\bf A}-{\bf B}=(24{\bf i}+33{\bf j})-(2{\bf i}+43{\bf j})=22{\bf i}-10{\bf j}

θ=tan11022=24\theta=\tan^{-1}\frac{-10}{22}=-24^{\circ}

(d)

A+B=(24i+33j)+(2i+43j)=26i+76j{\bf A}+{\bf B}=(24{\bf i}+33{\bf j})+(2{\bf i}+43{\bf j})=26{\bf i}+76{\bf j}

θ=tan17626=71\theta=\tan^{-1}\frac{76}{26}=71^{\circ}

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