#$&*
course Mth 277
2/8 2
If your solution to stated problem does not match the given solution, you should self-critique per instructions at
http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm
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Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
At the end of this document, after the qa problems (which provide you with questions and solutions), there is a series of Questions, Problems and Exercises.
query_09_3
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Question: Find v dot w when v = 4i + j and w =3i + 2k.
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Your solution:
v dot w = 4*3 + 1*0 + 0*2 = 12
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Given Solution:
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Self-critique (if necessary):
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Question: Determine whether v = 5i - 5j + 5k and w = 8i - 10j -2k are orthogonal.
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Your solution:
v dot w = 5*8 + (-5*-10) + (5*-2) = 80, so no v and w are not zero therefore not orthogonal.
confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary):
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Question: Find the angle between v = 2i +3 k and w = -j + 4k.
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Your solution:
Cos `theta = v dot w/(||v||*||w||)
12/(`sqrt(13)*`sqrt(17)) = 12/`sqrt(221), taking arcos of 12/`sqrt(221)
theta = approx. .6314
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Given Solution:
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Self-critique (if necessary):
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Question: Find two distinct unit vectors orthogonal to both v = i + 2j -2k and w = i + j - 2k.
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Your solution:
u = i + (1/2)k
u = 2i + k
both are orth to both v and w.
confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary):
@& You found your vectors by inspection, which is possible in this case but won't generally be feasible.
Note that your vectors are distinct, having different magnitudes, and that they are also parallel.
However neither of your vectors is orthogoal.
If you divide either of these vectors by its magnitude you will get a unit vector which is orthogonal to the two given vectors. However if you do this to both you will get the same vector both times. To get a second unit vector orthogonal to v and w, you could take the negative of the existing unit vector.
This problem can be solved generally, without the need to find a mutually orthogonal vector by inspection:
Let u = a i + b j + c k be a vector orthogonal to both v = i + 2j -2k and w = i + j - 2k.
Then
v dot u and w dot u are both zero.
v dot u = a + 2 b - 2 c
and
w dot u = a + b - 2 c
so our condition for mutual orthogonality is that a, b and c satisfy the equations
a + 2 b - 2 c = 0
a + b - 2 c = 0
We have two equations in three unknowns so unless the system is degenerate and incompatible we expect to find an infinite number of nontrivial solutions. This will be a good thing, because we still have a third condition, that the vector u be a unit vector.
If we subtract the second equation from the first we get
b = 0.
This leaves us with the system
a - 2 c = 0
a - 2 c = 0.
Both equations are satisfied whenever
a = 2 c.
Thus any vector of the form
u = 2 c i + 0 j + c k
will be orthogonal to the two given vectors.
In fact we can statement stronger and say that in addition, no vector which is not of this form is orthogonal to the two given vector.
Our vector must still be a unit vector. Thus
|| u || = 1.
It follows that
(2c)^2 + c^2 = 1
so that
5 c^2 = 1
and
c = +-sqrt(5) / 5.
We now have two vectors satisfying all three conditions:
u = 2 * sqrt(5) / 5 i + sqrt(5) / 5 * k
and
u = -2 * sqrt(5) / 5 i - sqrt(5) / 5 * k
We could have formulated our solution more concisely in terms of the following system of simultaneous equations:
a + 2 b - 2 c = 0
a + b - 2 c = 0
a^2 + b^2 + c^1 = 1.
It is clear that unless the system is inconsistent, at least one solution exists.
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Question: Let v = i - j + 4k and w = -i + 3j + 2k. Find cos(theta). Find s such that v is orthogonal to sv - w. Also find t such that v - tw is orthogonal to w
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Your solution:
cos(`theta) = v dot w/(||v||*||w||)
= 4/ `sqrt(252) = approx. .252
For (sv - w) dot v = 0
sv dot v - w dot v = 0
sv dot v = w dot v
s = w dot v/v dot v = 4/18 = 2/9, which checks out(done on sep paper)
For (v - tw) dot w = 0
v dot w - tw dot w = 0
v dot w = tw dot w
t = v dot w/w dot w = 4/14 = 2/7, which checks out(done on sep paper)
confidence rating #$&*:
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Given Solution:
@& Good. For your reference:
Let v = i - j + 4k and w = -i + 3j + 2k. Find cos(theta). Find s such that v is orthogonal to sv - w. Also find t such that v - tw is orthogonal to w
v dot w = -1 - 3 + 8 = 4, || v || = sqrt(18) and || w || = sqrt(14) so cos(theta) = 4 / (sqrt(18) * sqrt(14) ) = .25, approx..
v is orthogonal to sv - w if and only if
v dot (sv - w) = 0, so that
(i - j + 4k) dot ( s (i - j + 4k) - (-i + 3j + 2k ) ) = 0
giving us the single equation
(s + 1) - (-s - 3) + 4 ( 4 s - 2 ) = 0
easily solved to yield
s = 2/9.
Geometrically s v - w is found by starting at some point moving some distance along the vector v , then moving through the displacement -w. The vector s v - w is the vector from our initial to our final point and that vector will change as we change the value of s. As long as w is not parallel to v, if we move just far enough along v (just the right value of s) the vector s v - w will be perpendicular to our original direction along v.
A solution for t is found similarly.
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Self-critique (if necessary):
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Question: Find the work performed when a force F = (6/11)i - (2/11)j + (6/11)k is applied to an object moving along the line from P(3,5,-4) to Q(-4,-9,-11).
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Your solution:
W = F dot PQ
PQ = -7i - 14j - 7k
F dot PQ = (6/11)*-7 + (2/11)*-14 + (6/11)*-7
= -42/11 - 28/11 - 42/11 = -112/11 or approx. -10.1818
W = -112/11
confidence rating #$&*:
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Given Solution:
@& Correct. A bit more on the connection to the definition of work:
Work is defined to be the component of force parallel to the displacement, multiplied by the displacement.
The dot product of F and PQ is || F || || PQ || cos(theta)
F dot PQ = || F || || PQ || cos(theta),
which can be written
( || F || cos(theta) ) * || PQ || .
|| F || | cos(theta) | is the magnitude of the component of F in the direction PQ, and
|| F || cos(theta) is positive or negative according to whether this component is in the direction of PQ or opposite.
Thus
F dot PQ = || F || || PQ || cos(theta)
is the work done by the given force through the given displacement.
F = (6/11)i - (2/11)j + (6/11)k is applied to an object moving along the line from P(3,5,-4) to Q(-4,-9,-11).
PQ = -7i - 14j - 7k
F dot PQ =
((6/11)i - (2/11)j + (6/11)k ) dot (-7i - 14j - 7k )
= (6/11)*(-7) + (2/11)*(-14) + (6/11)*(-7 )
= -42/11 - 28/11 - 42/11 = -112/11 = -10.2, approx..
Thus the force does work -10.2 through this displacement.
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Self-critique (if necessary):
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Self-critique rating:
@& Again I've inserted a number of notes, some with corrections and some simly for your reference.*@