Question

6 math questions, answer all please for all points

Answer 1

See below for answers and explanations

Explanation:

Problem 1

Recall that the projection of a vector
`u` onto
`v` is
`\displaystyle proj_vu=\biggr((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=√((-8)^2+(4)^2)^2=64+16=80`

Find the projection of vector u onto v:

`\displaystyle proj_vu=\biggr((u\cdot v)/(||v||^2)\biggr)v\\\\proj_vu=\biggr((52)/(80)\biggr)\langle-8,4\rangle\\\\proj_vu=\biggr\langle(-416)/(80) ,(208)/(80)\biggr\rangle\\\\proj_vu=\biggr\langle(-26)/(5) ,(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||=√((-29.034)^2+(2.481)^2)\approx29.14`

Find the direction of the resultant vector:

`\displaystyle \theta=tan^(-1)\biggr((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||=√((-5)^2+(-6)^2)=√(25+36)=√(61)\approx7.81`

And finally, we find the direction of the vector:

`\displaystyle \theta=tan^(-1)\biggr((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||=√((-56)^2+(17)^2)=√(3136+289)=√(3425)\approx58.524`

Find the direction of the vector:

`\displaystyle\theta=tan^(-1)\biggr((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||=√((-16)^2+(-63)^2)=√(256+3969)=√(4225)=65`

Direction:
`\displaystyle \theta=tan^(-1)\biggr((-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||=√(3.661^2+52.139^2)\approx52.268`

Find the direction of the resultant vector:

`\displaystyle\theta=tan^(-1)\biggr((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