Answer:
Explanation:
Given
vector A has magnitude of [tex]\mid A\mid =17.6 [/tex]
and at an angle of [tex]\theta =120.5^{\circ}[/tex] CCW from x-axis
Vector B has a magnitude of [tex]\mid B\mid =+21.7[/tex]
at an angle of [tex]\theta =240.3^{\circ}[/tex] from x-axis
[tex]\vec{A}=17.6[\cos (120.5)+\sin (120.5)][/tex]
[tex]\vec{B}=21.7[\cos (240.3)+\sin (240.3)][/tex]
Now unit form of vector A and B
[tex]\hat{A}=\frac{\vec{A}}{\mid A\mid }[/tex]
[tex]\hat{A}=\frac{17.6[\cos (120.5)+\sin (120.5)]}{17.6}[/tex]
[tex]\hat{A}=\cos (120.5)+\sin (120.5)[/tex]
similarly,
[tex]\hat{B}=\frac{\vec{B}}{\mid B\mid }[/tex]
[tex]\hat{B}=\cos (240.3)+\sin (240.3)[/tex]
The unit vector forms of the given vectors is required.
The required vectors are [tex]A=-8.93\hat{i}+15.16\hat{j}[/tex] and [tex]B=-10.75\hat{i}-18.84\hat{j}[/tex]
Magnitude of vector A = [tex]|A|=17.6[/tex]
Angle vector A makes with positive x axis counter clockwise = [tex]\theta_1=120.5^{\circ}[/tex]
Magnitude of vector B = [tex]|B|=21.7[/tex]
Angle vector B makes with positive x axis counter clockwise = [tex]\theta_2=240.3^{\circ}[/tex]
The vectors need to be resolved in order to write in the unit vector forms.
The vectors are
[tex]A=|A|(\cos\theta_1\hat{i}+\sin\theta_1\hat{j})\\\Rightarrow A=17.6(\cos120.5\hat{i}+\sin120.5\hat{j})\\\Rightarrow A=-8.93\hat{i}+15.16\hat{j}[/tex]
[tex]B=|B|(\cos\theta_2\hat{i}+\sin\theta_2\hat{j})\\\Rightarrow B=21.7(\cos240.3\hat{i}+\sin240.3\hat{j})\\\Rightarrow B=-10.75\hat{i}-18.84\hat{j}[/tex]
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