Question

6 math questions, answer all please for all points

Answer 1

Answer:

See below for answers and explanations

Step-by-step explanation:

Problem 1

Recall that the projection of a vector $u$ onto $v$ is $\displaystyle proj_vu=\biggr(\frac{u\cdot v}{||v||^2}\biggr)v$.

Identify the vectors:

$u=\langle-10,-7\rangle$

$v=\langle-8,4\rangle$

Compute the dot product:

$u\cdot v=(-10*-8)+(-7*4)=80+(-28)=52$

Find the square of the magnitude of vector v:

$||v||^2=\sqrt{(-8)^2+(4)^2}^2=64+16=80$

Find the projection of vector u onto v:

$\displaystyle proj_vu=\biggr(\frac{u\cdot v}{||v||^2}\biggr)v\\\\proj_vu=\biggr(\frac{52}{80}\biggr)\langle-8,4\rangle\\\\proj_vu=\biggr\langle\frac{-416}{80} ,\frac{208}{80}\biggr\rangle\\\\proj_vu=\biggr\langle\frac{-26}{5} ,\frac{13}{5}\biggr\rangle\\\\proj_vu=\langle-5.2,2.6\rangle$

Thus, B is the correct answer

Problem 2

Treat the football and wind as vectors:

Football: $u=\langle42\cos172^\circ,42\sin172^\circ\rangle$

Wind: $v=\langle13\cos345^\circ,13\sin345^\circ\rangle$

Add the vectors: $u+v=\langle42\cos172^\circ+13\cos345^\circ,42\sin172^\circ+13\sin345^\circ\rangle\approx\langle-29.034,2.481\rangle$

Find the magnitude of the resultant vector:

$||u+v||=\sqrt{(-29.034)^2+(2.481)^2}\approx29.14$

Find the direction of the resultant vector:

$\displaystyle \theta=tan^{-1}\biggr(\frac{2.841}{-29.034}\biggr)\approx -5^\circ$

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 $180^\circ-5^\circ=175^\circ$

Thus, C is the correct answer

Problem 3

We identify the initial point to be $R(-2,12)$ and the terminal point to be $S(-7,6)$. The vector in component form can be found by subtracting the initial point from the terminal point:

$v=\langle-7-(-2),6-12\rangle=\langle-7+2,-6\rangle=\langle-5,-6\rangle$

Next, we find the magnitude of the vector:

$||v||=\sqrt{(-5)^2+(-6)^2}=\sqrt{25+36}=\sqrt{61}\approx7.81$

And finally, we find the direction of the vector:

$\displaystyle \theta=tan^{-1}\biggr(\frac{6}{5}\biggr)\approx50.194^\circ$

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 $180^\circ+50.194^\circ=230.194^\circ$.

Thus, A is the correct answer

Problem 4

Add the vectors:

$v_1+v_2=\langle-60,3\rangle+\langle4,14\rangle=\langle-60+4,3+14\rangle=\langle-56,17\rangle$

Determine the magnitude of the vector:

$||v_1+v_2||=\sqrt{(-56)^2+(17)^2}=\sqrt{3136+289}=\sqrt{3425}\approx58.524$

Find the direction of the vector:

$\displaystyle\theta=tan^{-1}\biggr(\frac{17}{-56} \biggr)\approx-17^\circ$

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 $180^\circ-17^\circ=163^\circ$

Thus, A is the correct answer

Problem 5

A vector in trigonometric form is represented as $w=||w||(\cos\theta i+\sin\theta i)$ where $||w||$ is the magnitude of vector $w$ and $\theta$ is the direction of vector $w$.

Magnitude: $||w||=\sqrt{(-16)^2+(-63)^2}=\sqrt{256+3969}=\sqrt{4225}=65$

Direction: $\displaystyle \theta=tan^{-1}\biggr(\frac{-63}{-16}\biggr)\approx75.75^\circ$

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 $180^\circ+75.75^\circ=255.75^\circ$.

This means that our vector in trigonometric form is $w=65(\cos255.75^\circ i+\sin255.75^\circ j)$

Thus, C is the correct answer

Problem 6

Write the vectors in trigonometric form:

$u=\langle40\cos30^\circ,40\sin30^\circ\rangle\\v=\langle50\cos140^\circ,50\sin140^\circ\rangle$

Add the vectors:

$u+v=\langle40\cos30^\circ+50\cos140^\circ,40\sin30^\circ+50\sin140^\circ\rangle\approx\langle-3.661,52.139\rangle$

Find the magnitude of the resultant vector:

$||u+v||=\sqrt{3.661^2+52.139^2}\approx52.268$

Find the direction of the resultant vector:

$\displaystyle\theta=tan^{-1}\biggr(\frac{52.139}{-3.661} \biggr)\approx-86^\circ$

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 $180^\circ-86^\circ=94^\circ$.

Thus, B is the correct answer