#$&*
course Mth 277
2/8 2
If your solution to stated problem does not match the given solution, you should self-critique per instructions at
http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm
.
Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
At the end of this document, after the qa problems (which provide you with questions and solutions), there is a series of Questions, Problems and Exercises.
query_09_1
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Question: Sketch the vector from P to Q, write it into standard component form, and find ||PQ||. P=(4,-1) Q=(-3,7).
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Your solution:
PQ = <-7, 8>
||PQ|| = `sqrt(-7^2 + 8^2) = approx. 10.63
confidence rating #$&*:
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Given Solution:
@& Good.
For reference:
The vector PQ has initial point (4, -1) and terminal point (-3, 7). Its components are therefore
-3 - 4 = -7 and 7 - (-1) = 8, so
PQ = <-7, 8>, which can also be written in the form -7 i + 8 j.
This vector describes the displacement from the initial to the terminal point.
Its magnitude, by the Pythagorean Theorem, is
|| PQ || = sqrt( (-7)^2 + 8^2) = sqrt(103)
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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:
For s = 1/3 and t = -1/2 we have
s<0, 3> + t<-4, 3> = <2, -1/2>, plugging in values
(1/3)*<0, 3> + (-1/2)*<-4, 3> = <0, 1> + <2, -1.5>
= <0+2, 1+(-1.5)> = <2, -.5> = v, which checks
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution:
@& In your line
(1/3)*<0, 3> + (-1/2)*<-4, 3> = <0, 1> + <2, -1.5>
you reversed 1/3 and -1/2, but that was just a typo.
For reference:
The equation
s * <0,3> + tu = v
becomes
s<0, 3> + t<-4, 3> = <2, -1/2>.
Multiplying the components of each by the scalars we get
<0, 3 s> + <-4 s, 3 t> = <2, -1/2>
so that
<0 - 4 s, 3s + 3 t > = <2, -1/2>
This is a statement of equality between two vectors. Each component of the vector on the left must therefore be equal to the same component of the vector on the right so
0- 4 s = 2
and
3 s + 3 t = -1/2.
Solving these two equations simultaneously for s and t we obtain
s = -1/2, t = 1/3.
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Self-critique (if necessary):
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Question: Let u = 4i - 3j, v = -3i + 4j , and w = 6i - 3j. Write the expression ||u|| ||v|| w in standard form.
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Your solution:
||u|| = `sqrt(4^2 + (-3)^2) = 5
||v|| = `sqrt((-3)^2 + 4^2) = 5
||u||*||v||*w = 5*5*(6i – 3j) = 25*(6i – 3j)
= (150i – 75j)
@& Good
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confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary):
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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 = <-6, -1>
k/||k|| = <-6, -1>/3 = (1/3)<-6, -1> = <-2, -(1/3)>
Thus <-2, -(1/3)> satisfies requirements.
Confidence rating: 3
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Given Solution:
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Self-critique (if necessary):
@& You appear to have used `k for the vector `u - 2`v + 2`w . However || k || = sqrt(37), not 1/3.
`u - 2`v + 2`w
= (4 `i + `j) - 2(4 `i + 3 `j) + 2(- `i + 2 `j)
= -6 `i - `j = <-6, -1>
( `u - 2 `v + 2 `w)/|| `u - 2 `v + 2 `w||
= (-6 `i - `j) / || (-6 `i - `j ||
= (-6 `i - `j) / sqrt(37)
= -6/sqrt(37) `i - 1 / sqrt(37) `j
This expression may be expressed with rationalized denominator or approximated, as appropriate.
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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:
||v|| = `sqrt(cos(theta)^2 + sin(theta)^2) = `sqrt(1) = 1
Unit vector = v/||v|| which equals
/1 =
= cos(theta)i + sin(theta)j
Confidence rating: 3
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Given Solution:
@& Good.
For reference:
|| `v || = sqrt( cos^2(theta) + sin^2(theta) ) = sqrt(1) = 1,
where we have used the Pythagorean identity
cos^2(theta) + sin^2(theta) = 1.
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#*&!
@& Very good. I've inserted reference solutions, one of which includes a correction. Let me know if you have questions.*@