Given:
Vector u equals PQ has initial point P (5, 16) and terminal point Q (8, 4).
Vector v equals RS has initial point R (28, 7) and terminal point S (10, 16).
Required:
Part A: We need to find u and v in linear form.
Part B : We need to find u and v in trigonometric form
Part C: We need to find 7u − 4v.
Explanation:
Part A:
Let u =ai+bj.
Subtract the x and y values for the initial point from the terminal point to find a and b.
Consider the points (5,16) and (8,4).
[tex]a=8-5=3[/tex][tex]b=4-16=-12[/tex]Substitute a =3 and b =-12 in the equation u =ai+bj.
[tex]u=3i-12j[/tex]Let v =ai+bj.
Subtract the x and y values for the initial point from the terminal point to find a and b.
Consider the points (28,7) and (10,16).
[tex]a=10-28=-18[/tex][tex]b=16-7=9[/tex]Substitute a =-18 and b =9 in the equation v =ai+bj.
[tex]v=-18i+9j[/tex]Part B:
The trigonometry form of the vector is
[tex]u=(rcos\theta,rsin\theta)[/tex]Here
[tex]r=\sqrt{a^2+b^2}\text{ and }\theta=tan^{-1}(\frac{b}{a}).[/tex]Consider the vector
[tex]u=3i-12j[/tex][tex]Substitute\text{ a=3 and b=-12 in the equation }r=\sqrt{a^2+b^2}.[/tex][tex]r=\sqrt{3^2+(-12)^2}[/tex][tex]r=\sqrt{153}=\sqrt{3^2\times17}[/tex][tex]r=3\sqrt{17}[/tex][tex]Substitute\text{ a=3 and b=-12 in the equation }\theta=tan^{-1}(\frac{b}{a})[/tex][tex]\theta=tan^{-1}(-\frac{12}{3})[/tex][tex]\theta=-75.9638[/tex][tex]\theta=-75.96\degree[/tex]The trigonometric form of the vector u is
[tex]u=(3\sqrt{17}cos(-75.96\degree),3\sqrt{17}sin(-75.96\degree))[/tex]Consider the vector
[tex]v=-18i+9j[/tex][tex]Substitute\text{ a=-18 and b=9 in the equation }r=\sqrt{a^2+b^2}.[/tex][tex]r=\sqrt{(-18)^2+9^2}[/tex][tex]r=\sqrt{405}=\sqrt{9^2\times5}[/tex][tex]r=9\sqrt{5}[/tex][tex]Substitute\text{ a=-18 and b=9 in the equation }\theta=tan^{-1}(\frac{b}{a})[/tex][tex]\theta=tan^{-1}(\frac{-18}{9})[/tex][tex]\theta=-63.4349[/tex][tex]\theta=-63.43\degree[/tex]The trigonometric form of the vector v is
[tex]v=(9\sqrt{5}cos(-63.43\degree),9\sqrt{5}sin(-63.43\degree))[/tex]Part C:
Consider the vector
[tex]u=3i-12j[/tex]Multiply both sides by 7.
[tex]7u=7\times3i-7\times12j[/tex][tex]7u=21i-84j[/tex]Consider the vector
[tex]v=-18i+9j[/tex]Multiply both sides by (-4).
[tex]-4v=(-4)-18i+(-4)9j[/tex][tex]-4v=72i-36j[/tex][tex]Add\text{ vectors }7u=21i-84j\text{ and }-4v=72\imaginaryI-36j.[/tex][tex]7u-4v=21i-84j+72\imaginaryI-36j[/tex][tex]7u-4v=21i+72\mathrm{i}-84j-36j[/tex][tex]7u-4v=93\mathrm{i}-120j[/tex]Final answer:
Part A:
[tex]u=3i-12j[/tex][tex]v=-18i+9j[/tex]Part B:
[tex]u=(3\sqrt{17}cos(-75.96\degree),3\sqrt{17}sin(-75.96\degree))[/tex][tex]v=(9\sqrt{5}cos(-63.43\degree),9\sqrt{5}sin(-63.43\degree))[/tex]Part C:
[tex]7u-4v=93\mathrm{i}-120j[/tex]
