Answer:
See below for answers and explanations
Step-by-step explanation:
Problem 1
[tex]u\cdot v=\langle4,-7\rangle\cdot\langle-6,9\rangle=(4*-6)+(-7*9)=-24-63=-87[/tex], so A is correct. Remember to multiply the horizontal components with each other and then add that to the product of the vertical components to get the dot product.
Problem 2
Two vectors are orthogonal if their dot product is 0, which also means that they must be perpendicular to each other, so we can check this:
[tex]u\cdot v=(6*-24)+(-9*36)=-144-324=-468[/tex]
Since the dot product of the two vectors is 0, they aren't orthogonal. Therefore, they are parallel since [tex]v=4u[/tex], making A the best choice.
Problem 3
Recall that the angle between two vectors is [tex]\theta=cos^{-1}(\frac{u\cdot v}{||u||*||v||})[/tex] where [tex]u\cdot v[/tex] is the dot product of vectors [tex]u[/tex] and [tex]v[/tex] while [tex]||u||[/tex] and [tex]||v||[/tex] are the magnitudes of vectors [tex]u[/tex] and [tex]v[/tex] respectively. If we let [tex]F_1=u[/tex] and [tex]F_2=v[/tex], we can then continue on:
[tex]u\cdot v=(190*128)+(160*-121)=24320+(-19360)=4960[/tex]
[tex]||u||=\sqrt{190^2+160^2}=\sqrt{61700}=10\sqrt{617}[/tex]
[tex]||v||=\sqrt{128^2+(-121)^2}=\sqrt{31025}=5\sqrt{1241}[/tex]
[tex]\theta=cos^{-1}(\frac{u\cdot v}{||u||*||v||})\\\\\theta=cos^{-1}(\frac{4960}{10\sqrt{617}*5\sqrt{1241}})\\\\\theta\approx83.49^\circ\approx83^\circ[/tex]
Therefore, B is the correct answer
Problem 4
We observe that [tex]u=\langle-8,-9\rangle[/tex] and [tex]v=\langle3,7\rangle[/tex], so [tex]-3(u\cdot v)=-3[(-8*3)+(-9*7)]=-3[-24+(-63)]=-3(-87)=261[/tex]. This means that B is the correct answer.
Problem 5
Recall that the projection of a vector [tex]u[/tex] onto [tex]v[/tex] is [tex]proj_vu=(\frac{u\cdot v}{||v||^2})v[/tex], thus:
[tex]proj_vu=(\frac{u\cdot v}{||v||^2})v\\\\proj_vu=(\frac{(-10*4)+(-3*8)}{\sqrt{4^2+8^2}^2})\langle4,8\rangle\\\\proj_vu=(\frac{-40+(-24)}{\sqrt{80}^2})\langle4,8\rangle\\\\proj_vu=(\frac{-64}{80})\langle4,8\rangle\\\\proj_vu=(-\frac{4}{5})\langle4,8\rangle\\\\proj_vu=\langle-\frac{16}{5},-\frac{32}{5}\rangle[/tex]
Therefore, D is the correct answer