Answer to Question #222044 in Analytic Geometry for ellie

Question #222044

Find the sum of two vectors if one has a direction of 60 degree and a magnitude of 5 and the other has a direction of 240 and a magnitude of 2?


1
Expert's answer
2021-08-04T11:17:45-0400

First we resolve each vector into its components, we do this by computing (acosθ0,bsinθ0) for the vector a, b. Therefore we have that (5cos600,5sin600)=(52,532)(2cos2400,2sin2400)=(1,3)Next,we add both vectors using component-wise addition, which yields(32,323)To get the magnitude and direction of the vectors, we compute 322+(323)2=3The direction of the vector is given bytanθ=32332=3    θ=tan13    θ=600Therefore the vector has a magnitude of 3 in direction of 600\text{First we resolve each vector into its components, we do this by computing }\\(acos\theta^0,bsin\theta^0) \text{ for the vector a, b. Therefore we have that }\\(5cos60^0, 5sin60^0)=(\frac{5}{2},\frac{5\sqrt{3}}{2}) \\(2cos240^0, 2sin240^0) = (-1,-\sqrt{3}) \\\text{Next,we add both vectors using component-wise addition, which yields}\\(\frac{3}{2},\frac{3}{2}\sqrt{3})\\\text{To get the magnitude and direction of the vectors, we compute }\\\sqrt{\frac{3}{2}^2+(\frac{3}{2}\sqrt3)^2}=3 \\\text{The direction of the vector is given by}\\tan\theta=\frac{\frac{3}{2}\sqrt{3}}{\frac{3}{2}}=\sqrt{3}\\\implies\theta=tan^{-1}\sqrt{3}\\\implies\theta=60^0 \\\text{Therefore the vector has a magnitude of 3 in direction of $60^0$}


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