Question 1(Multiple Choice Worth 2 points)
Let u = <7, -3>, v = <-9, 5>. Find 4u - 3v.

<55, -27>
<1, 3>
<16, 12>
<-8, -6>
Question 2(Multiple Choice Worth 1 points)
Let u = <-3, -5>, v = <-3, 1>. Find u + v.

<-2, -8>
<-8, -2>
<0, -6>
<-6, -4>
Question 3(Multiple Choice Worth 2 points)
Let u = <-4, -3>. Find the unit vector in the direction of u, and write your answer in component form.

Vector with two components. First component, negative four divided by five. Second component, negative three divided by five.
<1, 1>
Vector with two components. First component, negative four divided by seven. Second component, negative three divided by seven.
Vector with two components. First component, negative four divided by twenty five. Second component, negative three divided by twenty five.
Question 4(Multiple Choice Worth 2 points)
Let u = <-7, -2>. Find 8u.

<56, 16>
<-56, 16>
<-56, -16>
<56, -16>
Question 5(Multiple Choice Worth 1 points)
Given that P = (5, 9) and Q = (13, 12), find the component form and magnitude of vector PQ.

<-8, -3>, 73
<8, 3>, 73
<-8, -3>, square root of seventy three
<8, 3>, square root of seventy three
Question 6 (Essay Worth 2 points)
Two forces with magnitudes of 150 and 75 pounds act on an object at angles of 30° and 150°, respectively. Find the direction and magnitude of the resultant force. Round to two decimal places in all intermediate steps and in your final answer.

Consider vector u as a triangle of which the short sides have lengths of 3 and 4. Using Pythagoras, the length of the vector(the hypotenuse of the triangle) can be calculated as 5. A unit vector has a length of 1 along its hypotenuse. To convert the vector to a unit vector, everything has to be divided by five.

[tex]u=<\frac{-4}{5} , \frac{-3}{5} >[/tex]

Question 4

Multiply both components of the vector by 8:

[tex]8u=<-56,-16>[/tex]

Question 5

Vector PQ has two end points, P=(5,9) and Q=(13,12)

The vector is basically a line drawn between the two points

Thus, the lengths in both the x and y direction can be found by subtracting one from the other:

PQ=<13-5,12-9>

PQ=<8,3>

The magnitude is considered as the hypotenuse of a traingle with short sides of 8 and 3. Using Pythagoras the magnitude can be solved:

[tex]L_{pq}^{2} = 8^{2} + 3^{2} \\L_{pq}^{2} = 73\\L_{pq} = \sqrt{73}[/tex]

Question 6

See attached image.

[tex]y_{2} =-150sin(30)=-75\\x_{2} =-150cos(30)=-129.9\\y_{1} =-75sin(30)=-37.5\\x_{1} =75cos(30)=64.95\\\\F_{1}+F_{2} =<-64.95, -112.5>\\\\F_{mag} =\sqrt{64.95^{2} +112.5^{2} } \\F_{mag}=132.47\\sin(\alpha )=\frac{112.5}{132.47} \\\alpha =58.13^{o} \\\\180-58.13=121.87^{o}[/tex]