Assignment 3 Query

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course Mth 277

9/13/2011 @ 9:35 p.m.

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

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: (4 * 3 + 1 * 2) = 14.

confidence rating #$&*:y

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

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Self-critique (if necessary): On dot product questions when one vector has a j component and the other vector has a k component, why can/do we multiply j component by k component? I want to leave the above answer as 12, since 1j * 0 = 0 and 2k * 0 = 0.

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Self-critique rating:

@& You are correct. I didn't read my own problem carefully enough.*@

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

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Your solution: Orthogonal is another way of saying perpendicular, which requires a dot product of 0. ((5*8) + (-5*-10) + (5*-2)) = (40 + 50 - 10) = 80…which obviously isn’t 0.

confidence rating #$&*:y

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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): bit of a number discrepancy, but you explained that above.

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Self-critique rating:

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Question: Find the angle between v = 2i +3 k and w = -j + 4k.

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Your solution: v dot w = ||v||*||w||cos(theta). V dot w = 10, ||v|| = sqrt.13, ||w|| = sqrt. 17. Theta = cos^-1(10/ (sqrt. 13 * sqrt. 17). Theta = cos^-1(.67). Theta = 47.7 degrees.

confidence rating #$&*:y

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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):

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Self-critique rating:

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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: This one was really tricky for me…all of my work is just guess work to get a dot product = 0, but this thinking is probably wrong to begin with. For example, a vector of (-2i -j + 1k) would give us a dot product equal to zero.

confidence rating #$&*:r

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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): So, we make a vector composed of variables, do the dot product against both given vectors, add our results. We then simplify, and write a theoretical equation to find the magnitude of our equation composed of variables. The only part I’m still not 100% on is where 2ci + ck came from.

@& a = 2 c, b = 0 and c is still itself.

So a `i + b `j + c `k = 2 c `i + 0 c `j + c `k, or just 2 c `i + c `k.*@

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Self-critique rating: It makes sense now, for the most part.

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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: All of this work is pretty simple and logical, though I believe you intended to paste the above work in the solution section of this problem. But, regardless, I am completely confident I could have come to this solution on my own. For t, we use the same technique as with s. (-I + 3j + 2k) * [(1 + t) + (-1 - 3t) + (4 - 2t)] = 0. -1 - t -3 - 9t + 8 - 4t = 0. -14t = -4 t = 2/7.

confidence rating #$&*:tty confident

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Given Solution:

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Self-critique (if necessary):

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Self-critique rating:

@& You're right. I compose these responses right under the problem, so I can refer to the details of the problem. And sometimes I forget to move the solution to the intended location.

*@

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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: PQ = -7i + 14j - 7k, this is the ‘ds. F * ‘ds = W. ((6/22)I - (2/11)j + (6/11)k) * (-7i + 14j - 7k) = -42/11i - 28/11j - 42/11k = -112/11.

confidence rating #$&*:y

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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): Same method, different numbers.

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#*&!

Assignment 3 Query

@& You are correct. I didn't read my own problem carefully enough.*@

@& a = 2 c, b = 0 and c is still itself. So a `i + b `j + c `k = 2 c `i + 0 c `j + c `k, or just 2 c `i + c `k.*@

@& You're right. I compose these responses right under the problem, so I can refer to the details of the problem. And sometimes I forget to move the solution to the intended location. *@

&#This looks good. See my notes. Let me know if you have any questions. &#

@& a = 2 c, b = 0 and c is still itself.

So a `i + b `j + c `k = 2 c `i + 0 c `j + c `k, or just 2 c `i + c `k.*@

@& You're right. I compose these responses right under the problem, so I can refer to the details of the problem. And sometimes I forget to move the solution to the intended location.

*@