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6.03

given that p = (2, 9) and q = (4, 14), find the component form and magnitude of vector pq . (1 point)

< 2, 5> , square root of twenty nine
< -2, -5> , 29
< 2, 5> , 29
< -2, -5> , square root of twenty nine
2. let u = < -3, -5> , v = < -3, 1> . find u + v. (1 point)

< -2, -8>
< -8, -2>
< 0, -6>
< -6, -4>
3. let u = < -7, -2> . find 8u. (2 points)

< 56, 16>
< -56, 16>
< -56, -16>
< 56, -16>
4. let u = < -5, -9> , v = < 6, 8> . find -8u - 2v. (2 points)

< 52, 88>
< -8, 2>
< 112, -28>
< 28, 56>
5. let u = < 4, 3> . find the unit vector in the direction of u, and write your answer in component form. (2 points)

vector with two components. first component, four divided by seven. second component, three divided by seven.
< 1, 1>
vector with two components. first component, four divided by five. second component, three divided by five.
vector with two components. first component, four divided by twenty five. second component, three divided by twenty five.
6. 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. (2 points)

These are 6 questions and 6 answers:

1) Given that P = (2, 9) and Q = (4, 14), find the component form and magnitude of vector PQ . (1 point)

<2, 5>, square root of twenty nine  < answer
<-2, -5>, 29
<2, 5>, 29
<-2, -5>, square root of twenty nine

Explanation

1) Component form of vector PQ is the subtraction of the coordinates of vector Q less the coordinates of vector P.

So it is < 4 - 2, 14 - 9> = < 2, 5>

<2, 5>

2) the magnitude is the square root of the sum of the square of each component.

magnitude = √ (2^2 + 5^2) = √(4 + 25) = √29

√29

2. Let u = <-3, -5>, v = <-3, 1>. Find u + v. (1 point)

<-2, -8>
<-8, -2>
<0, -6>
<-6, -4>           < answer

Explanation: to find the sum of two vectors sum their corresponding components:

u + v = < -3, -5> + < - 3, 1> = < -3 - 3, - 5+ 1> = < - 6, -4>

< -6, -4>

3. Let u = <-7, -2>. Find 8u. (2 points)

<56, 16>
<-56, 16>
<-56, -16>         < answer
<56, -16>

Explanation:

This is the product of a scalar times a vector.

The procedure is to multiply each component of the vector by the scalar.

So, 8u = 8 < -7, - 2> =<8*(-7) , 8*(-2)> = < -56, -16>

< -56, -16>

4. Let u = <-5, -9>, v = <6, 8>. Find -8u - 2v. (2 points)

<52, 88>
<-8, 2>
<112, -28>
<28, 56>    < answer

Explanation:

Multiply each vector by the scalar in front of it and then perform the addition or subtraction.

-8u - 2v  = - 8 < - 5, -9> - 2 <6, 8> = <40, 72> - <12, 16> = <40 - 12, 72 - 16> =

= <28, 56>

<28, 56>

5. Let u = <4, 3>. Find the unit vector in the direction of u, and write your answer in component form. (2 points)

Vector with two components. First component, four divided by seven. Second component, three divided by seven.
<1, 1>
Vector with two components. First component, four divided by five. Second component, three divided by five.  < answer
Vector with two components. First component, four divided by twenty five. Second component, three divided by twenty five.

Explanation:

Divide each component by the magnitude of the vector:

1) magnitude of u = <4, 3> = √[4^2 + 3^2] = √(16 + 9) = √25 = 5

2) unit vector: <4, 3> / 5 = <4/5, 3/5>

<4/2, 3/2>

 6. 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. (2 points)

Explanation:

Force 1, F1 = 150, angle 30°

Force 2, F2 = 75, angle 150°

F1 = 150 cos(30) i + 150 sin (30) j
F2 = 75 cos (150) i + 75 sin (150) j

F1 = 129.9 i + 75 j
F2 = - 64.95 i + 37.5 j

Resultant force, Fr = F1 + F2

Fr = [129.9 - 64.95] i +[75 + 37.5]j = 64.95 i + 112.5 j

Magnitude = √[64.95 ^2 + 112.5^2] =  129.9 pounds

Direction  = arctan[112.5/64.95] = 60°

magnitude 129.9 pounds, direction 60°

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