Answer:
See below for answers and explanations
Step-by-step explanation:
Problem 1
Recall that the projection of a vector [tex]u[/tex] onto [tex]v[/tex] is [tex]\displaystyle proj_vu=\biggr(\frac{u\cdot v}{||v||^2}\biggr)v[/tex].
Identify the vectors:
[tex]u=\langle-10,-7\rangle[/tex]
[tex]v=\langle-8,4\rangle[/tex]
Compute the dot product:
[tex]u\cdot v=(-10*-8)+(-7*4)=80+(-28)=52[/tex]
Find the square of the magnitude of vector v:
[tex]||v||^2=\sqrt{(-8)^2+(4)^2}^2=64+16=80[/tex]
Find the projection of vector u onto v:
[tex]\displaystyle proj_vu=\biggr(\frac{u\cdot v}{||v||^2}\biggr)v\\\\proj_vu=\biggr(\frac{52}{80}\biggr)\langle-8,4\rangle\\\\proj_vu=\biggr\langle\frac{-416}{80} ,\frac{208}{80}\biggr\rangle\\\\proj_vu=\biggr\langle\frac{-26}{5} ,\frac{13}{5}\biggr\rangle\\\\proj_vu=\langle-5.2,2.6\rangle[/tex]
Thus, B is the correct answer
Problem 2
Treat the football and wind as vectors:
Football: [tex]u=\langle42\cos172^\circ,42\sin172^\circ\rangle[/tex]
Wind: [tex]v=\langle13\cos345^\circ,13\sin345^\circ\rangle[/tex]
Add the vectors: [tex]u+v=\langle42\cos172^\circ+13\cos345^\circ,42\sin172^\circ+13\sin345^\circ\rangle\approx\langle-29.034,2.481\rangle[/tex]
Find the magnitude of the resultant vector:
[tex]||u+v||=\sqrt{(-29.034)^2+(2.481)^2}\approx29.14[/tex]
Find the direction of the resultant vector:
[tex]\displaystyle \theta=tan^{-1}\biggr(\frac{2.841}{-29.034}\biggr)\approx -5^\circ[/tex]
Because our resultant vector is in Quadrant II, the true direction angle is 6° clockwise from the negative axis. This means that our true direction angle is [tex]180^\circ-5^\circ=175^\circ[/tex]
Thus, C is the correct answer
Problem 3
We identify the initial point to be [tex]R(-2,12)[/tex] and the terminal point to be [tex]S(-7,6)[/tex]. The vector in component form can be found by subtracting the initial point from the terminal point:
[tex]v=\langle-7-(-2),6-12\rangle=\langle-7+2,-6\rangle=\langle-5,-6\rangle[/tex]
Next, we find the magnitude of the vector:
[tex]||v||=\sqrt{(-5)^2+(-6)^2}=\sqrt{25+36}=\sqrt{61}\approx7.81[/tex]
And finally, we find the direction of the vector:
[tex]\displaystyle \theta=tan^{-1}\biggr(\frac{6}{5}\biggr)\approx50.194^\circ[/tex]
Keep in mind that since our vector is in Quadrant III, our direction angle also needs to be in Quadrant III, so the true direction angle is [tex]180^\circ+50.194^\circ=230.194^\circ[/tex].
Thus, A is the correct answer
Problem 4
Add the vectors:
[tex]v_1+v_2=\langle-60,3\rangle+\langle4,14\rangle=\langle-60+4,3+14\rangle=\langle-56,17\rangle[/tex]
Determine the magnitude of the vector:
[tex]||v_1+v_2||=\sqrt{(-56)^2+(17)^2}=\sqrt{3136+289}=\sqrt{3425}\approx58.524[/tex]
Find the direction of the vector:
[tex]\displaystyle\theta=tan^{-1}\biggr(\frac{17}{-56} \biggr)\approx-17^\circ[/tex]
Because our vector is in Quadrant II, then the direction angle we found is a reference angle, telling us the true direction angle is 17° clockwise from the negative x-axis, so the true direction angle is [tex]180^\circ-17^\circ=163^\circ[/tex]
Thus, A is the correct answer
Problem 5
A vector in trigonometric form is represented as [tex]w=||w||(\cos\theta i+\sin\theta i)[/tex] where [tex]||w||[/tex] is the magnitude of vector [tex]w[/tex] and [tex]\theta[/tex] is the direction of vector [tex]w[/tex].
Magnitude: [tex]||w||=\sqrt{(-16)^2+(-63)^2}=\sqrt{256+3969}=\sqrt{4225}=65[/tex]
Direction: [tex]\displaystyle \theta=tan^{-1}\biggr(\frac{-63}{-16}\biggr)\approx75.75^\circ[/tex]
As our vector is in Quadrant III, our true direction angle will be 75.75° counterclockwise from the negative x-axis, so our true direction angle will be [tex]180^\circ+75.75^\circ=255.75^\circ[/tex].
This means that our vector in trigonometric form is [tex]w=65(\cos255.75^\circ i+\sin255.75^\circ j)[/tex]
Thus, C is the correct answer
Problem 6
Write the vectors in trigonometric form:
[tex]u=\langle40\cos30^\circ,40\sin30^\circ\rangle\\v=\langle50\cos140^\circ,50\sin140^\circ\rangle[/tex]
Add the vectors:
[tex]u+v=\langle40\cos30^\circ+50\cos140^\circ,40\sin30^\circ+50\sin140^\circ\rangle\approx\langle-3.661,52.139\rangle[/tex]
Find the magnitude of the resultant vector:
[tex]||u+v||=\sqrt{3.661^2+52.139^2}\approx52.268[/tex]
Find the direction of the resultant vector:
[tex]\displaystyle\theta=tan^{-1}\biggr(\frac{52.139}{-3.661} \biggr)\approx-86^\circ[/tex]
Because our resultant vector is in Quadrant II, then our true direction angle will be 86° clockwise from the negative x-axis. So, our true direction angle is [tex]180^\circ-86^\circ=94^\circ[/tex].
Thus, B is the correct answer
