CAL3_ass_03

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course Mth 277

645 pm 9/28/11

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:

v= <4,1,0>

w= <3,0,2>

v dot w = 4*3 + 1*0 + 0*2 =12

confidence rating #$&*:

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

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Self-critique (if necessary):

The given solution is wrong. In a dot product, you multiply all the 'i*'i, 'j*'j and 'k*'k values then add them together. The given solution says to multiply the 'i's and add to 'j*'k. incorrect??????????????????????

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Self-critique rating: 3

@& You are correct. *@

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

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Your solution:

<5,-5,5> dot <8,-10,-2> = 40 +50 + 10 = 80. 80=/ 0, therefore vectors are not orthogonal

confidence rating #$&*:

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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):

Am I doing my math algebra wrong!!??????????????? -8*-5 = 40, but the given vector in the problem set was -5*-10, which equals 50.

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Self-critique rating:3

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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 = sqrt(4+9)*sqrt(1+16)*cos(theta)

12/[sqrt(13)*sqrt(17)] = cos(theta)

theta = 36.2°

confidence rating #$&*:

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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):

The given solution is incorrect. v dot w = <2,0,3> dot <0, -1, 4> = 0+0+12. The given solution says v dot w is 10. This is incorrect ?????????????????????

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Self-critique rating:3

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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:

Find vectors that when you use dot product with v and w vectors, they equal 0. Two vectors are <4, 0, 2> and <-4, 0, -2>. the unit vectors would be < 4/sqrt(6), 0, 2/sqrt(6)> and < -4sqrt(6), 0, -2/sqrt(6)>

confidence rating #$&*:

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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):

My two vectors follow the standard for

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

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Self-critique rating:3

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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:

v dot w = sqrt(1+1+16)*sqrt(1+9+4)*cos(theta)

cos(theta) = 4/[sqrt(14)*sqrt(18)

theta = 75.4°

v dot sv-w = 0

<1, -1, 4> dot [ - <-1, 3, 2>] = 0

<1, -1, 4> dot = 0

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

18s = 4

s = 2/9

w dot v-tw = 0

<-1, 3, 2> dot [<1, -1, 4> - <-t, 3t, 2t>] = 0

<-1, 3, 2> dot <1+t, -1-3t, 4-2t> = 0

-1-t + -3-9t + 8-4t = 0

-14t = -4

t= 2/7

confidence rating #$&*:

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Given Solution:

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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Self-critique (if necessary): OK

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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:

PQ = <-7, -14, -7>

F dot PQ = -42/11 + 28/11 -42/11 = -56/11

confidence rating #$&*:

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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):

the given solution is half of what I got. The only reason a force could be negative is if it went in the opposite direction.

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Self-critique rating:2

Self-critique (if necessary):

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#*&!

@& Good.

I did make a number of arithmetic errors in the document, and should correct most of them. I do intentionally leave a few arithmetic errors in some solutions problems, which alerts me to students who are feeding me back my own solutions without working them through. However that's done only on a handful of document in a course, and this early document isn't the place to be doing it*@