Query Assignment 93

#$&*

course Mth 277

9/17 1:30 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:

(4 * 3 + 1 * 2) = 14

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):ok

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Self-critique rating:3

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

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Your solution:

judging from google Orthogonal means perpendicular, which means a dot product of 0. ((5*8) + (-5*-10) + (5*-2)) = (40 + 50 - 10) = 80 which isn’t 0, which means they 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):

ok

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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 = ||v||*||w||cos(theta). v dot w = 12, ||v|| = sqrt.13, ||w|| = sqrt. 17.

Theta = cos^-1(12/ (sqrt. 13 * sqrt. 17). Theta = cos^-1(.80). Theta = 37 degrees.

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):

I see that you have v dot w is 12. but when you said v = 2i + 3k and w = - j + 4 k which translates to ( 2, 0, 3 ) dot ( 0, -1, 4 )

2*0 = 0

0*-1 = 0

3*4=12

therefore dot product should be 12? either way I understand the concept.

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Self-critique rating:3

@& I'll take your numbers over mine.

Looks good.*@

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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:

since orthogonal is perpendicular. the dot product should be 0. this being said. we can find two unit vecors with mag of 1 that gives the unit vectors a dot product of 0.

(this is a lot of typing so i'm going to sum up what i have written on paper by not typing the simple arithmatic)

ai + bj + ck = 0 ( first )

take v - w and solve that b = 0

put b into w and get that a = 2c

going back to the "" ( first ) "" equation. put in these values and get 2c i + ck ( new )

take || new || and get sqrt(5) note that when taking sqrt we have a + and - value.

for + sqrt(5) we get 2sqrt(5)/5 + sqrt(5)/5

for - sqrt(5) we get -2sqrt(5)/5 - sqrt(5)/5

these are opposite, this implies that the go cancel and go to 0. which is orthogonal

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):

Got the answer but it took me a while to figure out to make a (+) solution for sqrt(5) and ( - ) solution

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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:

cos(theta) = v dot w / ( || v || || w ||) = 4 / (sqrt(18) sqrt(14) ) = 4 / (15.87) ) = 75.4 degres.

v dot (sv-w) = 0

after much dot producting on paper i get 18s = - 4

because when you have the equation

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

you have 18s

-1 -3 is -4 then + 8 is +4.

so you are left with 18s + 4 = 0. then 18s = -4. s = -4/18

is there a sign error?

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):

I was very fustrated with the dot product of v dot sv-w because i kept getting sign errors with multiplication with my (s-1) dot 1 or something to give an example.

and I keep getting the 18s = -4 . somehow i'm nearly positive thats how it is. if not, i'm even more frustrated with sign errors

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Self-critique rating:3

@& The 4 should be -4 and the result is -4/18 (which reduces to -2/9).*@

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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:

work is f * distance in this case F dot Q-P

(6/11)i - (2/11)j + (6/11)k dot (-7, -14, - 7 k ) = -42/11 + 28 / 11 - 42 /11 = 28 / 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):

ok

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Self-critique rating:3

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#*&!

@& As you know, signs were invented solely to drive people nuts. It was centuries before anyone figured out that they're actually good for something.

Not really, of course, but sometimes it seems that way.

Looks like you're doing OK here.*@