sectionNineOne

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course MTH 277

09/14/20119:35 am

If your solution to stated problem does not match the given solution, you should self-critique per instructions at

http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm

.

Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

At the end of this document, after the qa problems (which provide you with questions and solutions), there is a series of Questions, Problems and Exercises.

query_09_1

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Question: Sketch the vector from P to Q, write it into standard component form, and find ||PQ||. P=(4,-1) Q=(-3,7).

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Your solution:

Standard component form: (-7, 8)

Magnitude = sqrt(113)

confidence rating #$&*:232; 2

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Given Solution:

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Self-critique (if necessary):

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Self-critique rating:

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Question: Let u = <-4,3> and v = <2,-1/2>. Find scalars s and t so that s * <0,3> + tu = v.

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Your solution:

s * <0,3> + t<-4,3> = <2,-1/2>

assume s = 1

t<-4,3> = <2, -3 ½ >

-4/-2 = 2

3/x = -3 ½

*light bulb *

assume t = -2

s<0,3> = <0, 3>

s = 1

t = -2

s = 1

confidence rating #$&*:232; 2

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Given Solution:

The equation

s * <0,3> + tu = v

becomes

s * <0, 3> + t * <-4, 3 > = < 2, -1 / 2 >

or

<0, 3s > + <-4 t, 3 t > = <2, -1/2 >.

and finally

<0 - 4 t, 3 s + 3 t > = < 2, -1/2 >.

Since the two vectors are equal if an only if their two components are equal, this is equivalent to the two simultaneous equations

-4 t = 2

3 s + 3 t = -1/2.

The solution of these equations is

t = -1/2, s = 1/3.

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Self-critique (if necessary):

I needed to combine the problem one step further, it never occurred to me to add the two different variables together before solving them.

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Self-critique rating:2

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Question: Let u = 4i - 3j, v = -3i + 4j , and w = 6i - 3j. Write the expression ||u|| ||v|| w in standard form.

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Your solution:

||4i - 3j|| ||3i + 4j|| (6i -3j)

sqrt(16i^2 + 9j^2) sqrt(9i^2 + 16j^2) (6i -3j)

confidence rating #$&*:232; 1

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Given Solution:

|| u || = sqrt( 4^2 + 3^2) = 5

and

|| v || = sqrt(3^2 + 4^2) = 5

so that

|| u || || v || w = 5 * 5 * w = 25 * (6 i - 3 j ) = 150 i - 75 j.

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Self-critique (if necessary):

I have really got to stop making problems more complicated than they really are . . . all I had to do was deal find the magnitude as if the variables weren’t there.

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Self-critique rating:2

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Question: Let u = 4i + j, v = 4i + 3j, w = -i + 2j. Find a vector of length 3 with the same direction as u - 2v + 2w.

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Your solution:

Direction = u/||u||

Direction = (4i + j) - 2(4i + 3j) + 2(-i + 2j)

[4 - 2(4) + 2(-1)]i + [1 -2(3) +2(2)]j

[-6]i + [-4]j

direction = -6i - 4j

-6i - 4j = vector/3

-18i - 12j = vector

confidence rating #$&*:232; 2

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Given Solution:

u - 2v + 2w = -6 i - j

so

|| u - 2 v + 2 w || = sqrt(37)

and

( u - 2 v + 2 w ) / || u - 2 v + 2 w || = -6 sqrt(37) / 37 i - sqrt(37) / 37 j

is a unit vector in the directio of u - 2 v + 2 w .

A vector of magnitude 3 in this direction is therefore

3 ( -6 sqrt(37) / 37 * i - sqrt(37) / 37 * j ) =

-18 sqrt(37) / 37 i - 3 sqrt(37) / 37 j

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Self-critique (if necessary):

First mistake, I miscalculated with the j, ending up with a scalar of 4 instead of one.

Second mistake, I did not attempt to calculate the using the Pythagorean theorem.

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Self-critique rating: 2

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Question: Show that the vector v = cos(theta)i + sin(theta)j is a unit vector for any angle theta.

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Your solution:

Cos^2x + sin^2x = 1

||v|| = cos(theta)^2i + sin(theta)^2j

||v|| = 1

any vector found will be a unit vector

confidence rating #$&*:232; 2

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Given Solution: || v || = sqrt( cos^2(theta) + sin^2(theta) ) = sqrt(1) = 1.

A vector of magnitude 1 is a unit vector.

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Self-critique (if necessary):OK

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Self-critique rating:OK

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@& Good work, and good self-critiques.*@