sectionNineThree

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course MTH 277

09/14/20119:38 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

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temporary disclaimer: Solutions to these problems were erroneously deleted and the problem solutions have been quickly reconstructed. These solutions are therefore not guaranteed, though the process by which they are obtained should be correct. So if you have discrepancies in arithmetic and other details, feel free to question the given solutions.

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:

12i + j + 2k

confidence rating #$&*:232; 3

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Given Solution: v dot w = 4 * 3 + 1 * 2 = 12 + 2 = 14.

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Self-critique (if necessary):

I thought we had to stick to the same signs, so I didn’t add the others.

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Self-critique rating:2

@& My solution should actually have been

4 * 3 + 1 * 0 + 0 * 2 = 12.*@

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Question: Determine whether v = 5i - 5j + 5k and w = 8i - 10j -2k are orthogonal.

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Your solution:

no

v dot w does not equal 0

confidence rating #$&*:232; 3

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Given Solution:

Two vectors are orthogonal if the angle between them is 90 deg, i.e., if and onlye if their dot product is zero.

The dot product of these vectors is 5 * 8 - 5 * (-8) + 5 * (-2) = 40 + 40 - 10 = 70.

They are not orthogonal.

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Self-critique (if necessary):OK

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Self-critique rating:OK

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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||)

V dot w = -2 +12 = 10

||v|| = sqrt(13)

||w|| = sqrt(17)

||v|| ||w|| = sqrt(30)

10/sqrt(30) roughly= 1.82574

cos(theta) roughly= 1.82574

confidence rating #$&*:232; 2

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Given Solution:

Since v dot w = || v || || w || cos(theta) we have

theta = cos^-1 ( v dot w ) || v || || w || = cos^-1 ( 10 / (sqrt(13) * sqrt( 17) ) = cos^-1 (.67) = 48 degrees, approx., or roughly.8 radians.

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Self-critique (if necessary):

Screwed up with the calculations of the denominator. Added the numbers together instead of multiplying the square roots.

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Self-critique rating: 2

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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:

1: origin (0,0,0)

2: <2, 0, 1>

confidence rating #$&*:232; 2

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Given Solution:

Suppose a i + b j + c k is orthogonal to both. Then the dot product of this vector with each of the given vectors is zero, and we have

a + 2 b - 2 c = 0

a + b - 2 c = 0

Subtracting the second equation from the first we get b = 0.

With this value of b both our first and our second equation become

a - 2 c = 0

so that

a = 2 c.

Any vector of the form 2c i + c k is therefore orthogonal to our two vectors.

Any such vector has magnitude sqrt( (2 c)^2 + c^2) = sqrt( 5 c^2) = sqrt(5) | c |.

If c is positive then | c | = c and our vector is

(2 c i + c k ) / (sqrt(5) c) = 2 sqrt(5) / 5 i + sqrt(5) / 5 k.

If c is negative then | c | = - c and our vector will be

(2 c i + c k ) / (- sqrt(5) c) = - 2 sqrt(5) / 5 i - sqrt(5) / 5 k.

Our two solution vectors are equal and opposite. Each is a unit vector perpendicular to the two given vectors.

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Self-critique (if necessary):

I clearly had no idea what I was doing here.

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Self-critique rating: 1

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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

cos(theta) = v dot w / ( || v || || w ||) = 4 / (sqrt(18) sqrt(14) ) = 4 / (12 sqrt(7) ).

The condition v orthogonal to s v - w is

v dot (s v - w ) = 0

(i - j + 4 k ) dot ( (s - 1) i + (-s + 3) j + (4 s + 2) k ) = 0

which becomes

s - 1 + s - 3 + 16 s + 8 = 0

so that

18 s = 4

and

s = 4 / 18.

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Your solution:

First part: (v dot w)/(||v|| ||w||)

[1 + (-3) + 8] = 6

||v|| = sqrt(18)

||w|| = sqrt(14)

6/(sqrt(18) * sqrt(14))

cos(theta) = 0.377964

Second Part:

W dot (v - tw) = 0

(-i + 3j + 2k) dot (i - j + 4k - t(-i + 3j + 2k)) = 0

(-i + 3j + 2k) dot ((1 - t)i + (

confidence rating #$&*:232;

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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: 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:

[(42/11) + (28/11) - (42/11)]

confidence rating #$&*:232; 3

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Given Solution: The work is F dot `ds = ( (6/11)i - (2 / 11) j + (6 / 11) k ) dot (-7 i - 14 j - 7 k ) = -42/11 + 28 / 11 - 42 /11 = 28 / 11.

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Self-critique (if necessary):OK

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Self-critique rating:OK

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#*&!

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