#$&*
course Mth 277
2/8 2
If your solution to stated problem does not match the given solution, you should self-critique per instructions at
http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm
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Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
At the end of this document, after the qa problems (which provide you with questions and solutions), there is a series of Questions, Problems and Exercises.
query_09_4
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Question: Find v X w when v = sin(theta)i + cos(theta)j and w = -cos(theta)i + sin(theta)j (theta is any angle).
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Your solution:
v X w = ((sin(`theta)*sin(`theta)) - (-cos(`theta)*cos(`theta))
= sin(`theta)^2 - (-cos(`theta)^2)
= sin(`theta)^2 + cos(`theta)^2
=1
confidence rating #$&*:
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Given Solution:
@& Good, but the cross product is a vector.
v X w
= (sin(theta)i + cos(theta)j) X (-cos(theta)i + sin(theta)j )
= sin(theta) * (-cos(theta)) i X i + sin(theta) ) * (sin (theta)) i X j + cos(theta) * (-cos(theta)) j X i + cos(theta) * (sin(theta)) j X j
= 0 + sin^2(theta) k + (-cos^2(theta)) * (-k) + 0
= (sin^2(theta) + cos^2(theta) ) * k
= k
This calculation can also be expressed in determinant format as depicted in
http://vhcc2.vhcc.edu/dsmith/geninfo/qa_query_etc/multivar/qa_04_7.gif
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Self-critique (if necessary):
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Question: Find sin(theta) where theta is the angle between v = -i + j and w = -i + j + 2k.
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Your solution:
using ||v X w|| = ||v||*||w||*sin(`theta)
sin(`theta) = ||v X w||/||v||*||w||
=`sqrt(8)/ `sqrt(12)
So sin(`theta) = `sqrt(8)/ `sqrt(12)
confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary):
@& Good.
Note that sqrt(8) / sqrt(12) = 2 sqrt(2) / (2 sqrt(3)). See how the radicals were handled below. The answer given here is identical to yours, except for the form of the expression.
`v * `w = (-i + j) X (-i + j + 2k) = -2 `k + 2 `j
|| v || = sqrt( 2 ) and || w || = sqrt(6), while || v X w || = 2 sqrt(2).
||v X w|| = ||v||*||w||*sin(`theta) so
sin(`theta)
= ||v X w|| / (||v||*||w||)
=2 sqrt(2) / (sqrt(2) * sqrt(6) )
= 2 / sqrt(6)
= sqrt(6) / 3.
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Question: Find a unit vector which is orthogonal to both v = 2i - j and w = 2j - k.
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Your solution:
The cross product v X w is orth to both v and w so
v X w = i + 2j + 4k, unit vector is
i + 2j + 4k/`sqrt(21)
= <1/`sqrt(21), 2/`sqrt(21), 4/`sqrt(21)>
Confidence rating: 3
@& I've included an approximation, but the exact solution is identical to yours. I didn't mess with the radical, since nothing simplifies.
The cross product of two vectors is orthogonal to both, so
`v X `w = `i + 2 `j + 4 `k is orthogonal to both `v and `w.
A vector divided by its magnitude is a unit vector so
`v X `w / || `v X `w |
= ( `i + 2 `j + 4 `k)/`sqrt(21)
is the desired unit vector.
A rough approximation to this vector:
.2 i + .4 j + .9 k
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Given Solution:
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Self-critique (if necessary):
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Question: Find the area of the triangle with vertices P(2,0,0), Q(1,1,-1), R(3,1,2).
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Your solution:
Using area = (1/2)||PQ X PR||
PQ = -i + j -k
PR = i + j + 2k
PQ X PR = 3i - 3j + 2k
|| PQ X PR|| = `sqrt(22)
Area = .5*`sqrt(22) = approx. 2.345
@& I get a different result for that cross product, but that doesn't mean your solution isn't the right one:
The solution given here addresses the solution from the perspective of the geometry of the triangle.
Find the area of the triangle with vertices P(2,0,0), Q(1,1,-1), R(3,1,2).
The triangle comprises have the parallelogram defined by any two of its sides.
The parallelogram defined by the sides PQ = - i + j - k and PR = i + j + 2 k has one of its sides as its base. If the angle between the sides is theta, then its altitude will be the component of the other side perpendicular to the base, so the altitude will be the product of that side and sin(theta).
If PQ is taken to be the base, the altitude is therefore || PR || sin(theta). It follows that the area of the parallelogram is
|| PQ || * || PR || sin(theta),
which as we know is the magnitude || PQ X PR ||.
Thus the area of the parallelogram is the magnitude of the cross product of the vectors forming two adjacent sides.
The area of the triangle is half the area:
area of triangle = (1/2) || PQ X PR ||
PQ X PR
= (- i + j - k) X ( i + j + 2 k)
= 3 i + j - 2 k
Thus
area of triangle = (1/2) || PQ X PR || = 1/2 || 3 i + j - 2 k || = 1/2 sqrt(14).
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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: 8) Determine if each of the following products is a vector, scalar, or not defined at all. Explain why. u X (v X w) , u dot (v dot w), (u X v) dot (w X r).
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Your solution:
u X (v X w), product is a vector because cross product only involves vectors
u dot (v dot w), product is undefined because v dot w is scalar and I’m not sure how it would be dotted with u vector
(u X v) dot (w X r) product is scalar because u X v and w X r will produce sep vectors which when dotted with each other produces a scalar.
confidence rating #$&*:
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Given Solution:
@&
`v X `w is a vector, so that
`u X (`v X `w) is the cross product of two vectors and is hence a vector,
while
`u dot (`v X `w) is the dot product of two vectors and is hence a scalar.
`w X `r is also a vector, so
(`u X `v) dot (`w X `r) is a dot product of two vectors and is hence a vector.
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Self-critique (if necessary):
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Question: Find a number t such that the vectors -i - j, i - (1/2) j + (1/2)k and -2i -2j - 2tk all lie in the same plane.
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Your solution:
t = ¼
(-i -j) + (i -.5j + .5k) + (-2i -2j - (2*(1/4)k)
= -2i -3.5j
confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary):
@&
Three vectors all lie in the same plane of one is a sum of multiples of the other two.
This condition could be expressed for the given vectors as
s * (-i - j ) + r * ( i - (1/2) j + (1/2)k ) = -2i -2j - 2 t k, for some scalars s and r.
The i components of both sides must be equal, so
-s + r = -2
The j components must also be equal, so
-s - 1/2 r = -2
And the k components must be equal, so
1/2 r = -2 t.
We solve the system for the parameters s, r and t obtaining
s = 2, r = 0, t = 0.
Thus only for t = 0 can the three vectors all lie in the same plane.
Three vectors all lie in the same plane if they are all perpendicular to a common vector. Assuming none of the vectors is parallel to any of the others (in which case, as you should verify, the result would follow easily) and two of the vectors are perpendicular to their cross product. So if the third vector is also perpendicular to the cross product of the first two, all three of the vectors will lie in the same plane.
(-i - j ) X ( i - (1/2) j + (1/2)k ) = -1/2 i + 1/2 j + 3/2 k
Since (-1/2 i + 1/2 j + 3/2 k) dot (-2i -2j - 2 t k) = 1 - 1 + 3 t = 3 t
we can set 3 t equal to zero and solve for t (the answer is obvious):
3 t = 0
t = 0.
Thus t = 0 is the only value of t for which the three vectors lie in the same plane.
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@& Check the notes I've inserted. Most are for your reference, but there are a couple of discrepancies, and I've also addressed some problems from a broader perspective.*@