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course Mth 277

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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:

-3-4=-7

7-(-1)=8

(-7,8) -7i+8j is the component form

magnitude of the vector is sqrt((-7)^2+(8)^2)=sqrt(113)

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Given Solution:

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Self-critique (if necessary):

Solution is not given, but I fell confident in my answer.

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Self-critique rating:OK

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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:

We are given that s*<0,3>+t*<-4,3>=<2,-0.5>

So lets solve for the xcomponent to find the value of t.

s*0+t*(-4)=2 -4t=2 t=-0.5

Now lets solve for y values and plug in the value we solved for t.

3s+(-0.5)(3)=-.5 3s=1 s=1/3

confidence rating #$&*:

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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):

We solved this question using different methods, but I believe my solution to be suffiecient.

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Self-critique rating:

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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:

First, we need to find the magnitudes of u and v.

||u||=sqrt((4)^2+(-3)^2)=sqrt(25)=5

||v||=sqrt((-3)^2+(4)^2)=sqrt(25)=5

Given the expression ||u|| ||v|| w

(5)(5)(6i-3j)=150i-75j

confidence rating #$&*:

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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):

Our answers are different,but after looking at your solution, I understand my mistake.

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Self-critique rating:OK

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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:

u-2v+2w

(4i+j)-2(4i+3j)+2(-i+2j)=-6i-j

magnitude=sqrt((-6)^2+(-1)^2)=sqrt(37)

So the unit vector would be 6/(sqrt(37))-1/(sqrt(37))

since it has a length of 3, the answer is 3(6/(sqrt(37))-1/(sqrt(37)))=18/sqrt(37)i-3/sqrt(37)j

confidence rating #$&*:

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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):OK

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Self-critique rating:OK

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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:

Since the unit vector is v / || v ||

|| v ||=sqrt(cos^2(theta)+sin^2(theta)) trig identity cos^2(theta)+sin^2(theta)=1 sqrt(1)=1

So, the unit vector is v/1. Therefore, the vector v is a unit vector for any angle given.

confidence rating #$&*:

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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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