Respuesta :
We are given two vectors and we are asked to determine the magnitude and direction of their sum. To do that we will rewrite both vectors in coordinate form. To do this we will use the following formula:
[tex]A=(\lvert A\rvert\cos \theta,\lvert A\rvert\sin \theta)[/tex]Where:
[tex]\begin{gathered} A,\text{ any vector} \\ \lvert A\rvert,\text{ magnitude of the vector} \\ \theta,\text{ direction of the vector} \end{gathered}[/tex]For the first vector we have:
[tex]A_1=(3.14\cos 30,3.14\sin 30)[/tex]Solving the operations we get:
[tex]A_1=(2.72,1.57)[/tex]Now for the second vector:
[tex]A_2=(2.71\cos 60,2.71\sin 60)[/tex]Solving the operations:
[tex]A_2=(1.36,2.35)[/tex]Now we add both vectors by adding each corresponding component, like this:
[tex]\begin{gathered} A_s=A_1+A_2 \\ A_s=(2.72,1.57)+(1.36,2.35) \end{gathered}[/tex]Adding each component:
[tex]\begin{gathered} A_s=(2.72+1.36,1.57+2.35) \\ A_s=(4.08,3.92) \end{gathered}[/tex]Now, to determine the magnitude we will use the following formula:
[tex]A=(x,y)\rightarrow\lvert A\rvert=\sqrt[]{x^2+y^2}[/tex]That means that the magnitude is determined by taking the square root of the sum of the squares of the components of the vector. Replacing we get:
[tex]\begin{gathered} \lvert A_s\rvert=\sqrt[]{(4.08)^2+(3.92)^2} \\ \lvert A_s\rvert=\sqrt[]{32.01} \\ \lvert A_s\rvert=5.66 \end{gathered}[/tex]The direction of the vector is determined using the following formula:
[tex]\theta=\arctan (\frac{y}{x})[/tex]Replacing we get:
[tex]\theta_s=arc\tan (\frac{3.92}{4.08})[/tex]Solving the operations we get:
[tex]\theta_s=43.85[/tex]Therefore, the magnitude of the sum is 5.66m and the direction is 43.85°.
