Query9_2

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course Mth 277

Question: Find u + v, u - v, (5/2)u, and 2u + 3v for the following vectors: u = <1,2,-3>, v = < -1,-2,3>.YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

u+v=<0,0,0>

u-v=<2,4,-6>

2u+3v=<-1,-2,3>

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

No solution given, but I believe my answer to be correct.

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Self-critique rating:3

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Question: Find the standard form equation of the sphere with center (-1,2,4) and radius 2.

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Your solution:

sqrt((x+1)^2+(y-2)^2+(z-4)^2)=2

(x+1)^2+(y-2)^2+(z-4)^2=4

x^2+2x+1+y^2-4y+4+z^2-8z+16=4

x^2+2x+y^2-4y+z^2-8z+17=0 is the standard form

confidence rating #$&*:

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Given Solution:

A point (x, y, z) is on the given sphere if its distance from (-1, 2, 4) is 2, so that

sqrt( (x - (-1))^2 + (y - 2)^2 + (z - 4)^2 ) = 2

and

(x + 1)^2 + (y - 2)^2 + (z - 4)^2 = 4.

This is the equation of the sphere in one form.

Expanding the squares we obtain

x^2 + 2 x + 1 + y^2 - 4 y + 4 + z^2 - 8 x + 16 = 4

which we rearrange to the standard form

x^2 + 2 x + y^2 - 4 y + z^2 - 8 z + 13 = 0.

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Self-critique (if necessary):

I got 17 instead of the 13 seen in your solution. I reworked my problem several times and still get +17. Maybe you can see where I made my mistake.

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Self-critique rating:3

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Your result is correct. I believe I overlooked the 4.

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Question: Find the center and radius of the sphere with equation x^2 + y^2 + z^2 - 2x - 6y + 12z - 17 = 0.

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Your solution:

x^2+y^2+z^2-2x-6y+12z=17

(x^2-2x)+(y^2-6y)+(z^2+12z)=17

(x^2-2x+1)-1+(y^2-6y+9)-9+(z^2+12z+36)-36=17

(x-1)^2+(y-3)^2+(z+6)^2=63

center(1,3,-6) radius=sqrt(63)

confidence rating #$&*:

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Given Solution:

Completing the squares we obtain

(x^2 - 2 x + 1 - 1) + (y^2 - 6 y + 9 - 9) + (z^2 + 12 z + 36 - 36) = 17

which can be written as

(x - 1)^2 - 1 + (y - 3)^2 - 9 + (z + 6)^2 - 36 = 17

and finally as

(x - 1)^2 + (y - 3)^2 + (z + 6)^2 = 63

This sphere is centered at (1, 3, -6) and has radius sqrt(63) = 3 sqrt(7).

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Self-critique (if necessary):OK

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Self-critique rating:OK

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Question: Find the standard representation and length of PQ when P = (-3,1,4) and Q = (2,-4,-3).

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Your solution:

2-(-3)=5

-4-1=-5

-3-4=-7

5i-5j-7k

sqrt((5)^2+(-5)^2+(-7)^2)=sqrt(99)

confidence rating #$&*:

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Given Solution:

PQ = (2 - (-3) ) i + (-4 - 1) j + (-3 - 4) k = 5 i - 5 j - 7 k.

|| PQ || = sqrt( 5^2 + 5^2 + 7^2) = sqrt(99).

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Self-critique (if necessary):OK

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Self-critique rating:OK

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Question: Find a unit vector in the direction of v = <-1, sqrt(3), 4>.

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Your solution:

v=<-1,sqrt(3),4>

magnitude of v=sqrt((-1)^2+(sqrt(3))^2+(4)^2)=sqrt(1+3+16)=sqrt(20)

unit vector=<-1/sqrt(20),sqrt(3)/sqrt(20),4/sqrt(20)>

confidence rating #$&*:

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Given Solution:

|| v || = sqrt( 1^2 + sqrt(3) ^ 2 + 4^2 ) = sqrt( 26 )

so a unit vector in the direction of v is

v / || v ||= < -1, sqrt(3), 4 > / sqrt(26) = <-sqrt(26) / 26, sqrt(78) / 26, 4 sqrt(26) / 26)> .

4 sqrt(26) / 26 is 2 sqrt(26) / 13.

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Self-critique (if necessary):

I am somewhat confused what you did. First, I do not know how you got sqrt(26) for the magnitude. If I am correct, I think you squared 3 rather than squaring sqrt(3). Then, I thought the unit vector was the vector divided by the magnitude so your other work confuses me.

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Self-critique rating:OK

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You are correct.

I thought I had linked to the corrected query set, but I'll have to check that out to make sure. I know this error should have been corrected.

There are a few other known errors in the Chapter 9 problems, if indeed I have linked to the wrong set, mostly a couple of arithmetic errors in 9.3. I can't fix that right now but should be able to do so soon.

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The form of the given solution is written without radicals in the denominator, which is standard form.

For example

4 / sqrt(20) = 4 sqrt(20) / (sqrt(20) sqrt(20) ) = 4 sqrt(20) / 20 = sqrt(20) / 5 = 2 sqrt(5) / 5.

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Question: Sketch and describe the cylindrical surface given by y = cos x.

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Your solution:

First, it is necessary to know what the graph of y=cosx looks like. It is easy to describe this graph as being a sound wave. The period or wavelength is 2pi and crosses the y-axis at y=1. The crest of the wave is at y=1 while the trough is at y=-1.

confidence rating #$&*:

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Given Solution:

In the x-y plane y = cos(x) consists of a sinusoidal function oscillating between the lines y = -1 and y = 1, with period 2 pi radians, and containing the point (0, 1).

The surface in 3 dimensions repeats this same curve for every value of z, so that the graph represents a wavy curtain hanging vertically downward, intersecting the xy plane along the sinusoidal curve.

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Self-critique (if necessary):OK

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Self-critique rating:OK

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Question: Determine if u = 2i + 3j + -4k is parallel to v = <1,-3/2,2>.

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Your solution:

Yes, u and v are not parallel to eachother. This can easily be determined without much work. Since <2,3,-4> is not a scalar multiple of <1,-1.5,2> these are not parallel to eachother.

confidence rating #$&*:

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Given Solution:

Two vectors are parallel if the angle between them is 0 or pi radians (180 degrees), meaning that the cosine of the angle between them is 1 or -1.

u dot v = || u || || v || cos(theta)

so that

cos(theta) = u dot v / (|| u || || v || )

= (2 * 1 + 3 * (-3/2) + (-4 * 2) ) / ( sqrt(2^2 + 3^2 + 4^2) * sqrt( 1^2 + (3/2)^2 + 2^2) )

= (-21/2) / (sqrt( 29) sqrt(29/4).

This is not 1 or -1, so the cosine is neither 0 nor pi rad (i.e., 180 deg).

The vectors are therefore not parallel.

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Self-critique (if necessary):

Can you not figure this out using scalar multiples?

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Self-critique rating:3

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Scalar multiples will work, but you do need to be aware of the dot product test , which I chose to illustrate in this solution.

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Question: Find the lengths of the sides of the triangle and determine if the triangle with vertices A(3,0,0), B(7,1,4) and C(5,4,4) is a right triangle, isosceles triangle, both, or neither.

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Your solution:

It is easy to find the lengths of the traingles by taking the magnitudes of the three vectors. Then, you can compare the lengths of the triangles to see if it might be equilateral, isosceles or neither.

vector AB <4,1,4> magn.=sqrt(33)

vector BC <-2,3,0> magn.=sqrt(13)

vector CA <-2,-4,-4> magn.=sqrt(36)=6

It is obvious that none of the sides are equal so the triangle can not be equilateral or isosceles. However, I am not sure how to tell if the triangle is right triangle.

confidence rating #$&*:

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Given Solution:

The sides can be represented by the vectors

AB = < 4, 1, 4 >,

BC = < -2, 3, 0 > and

AC = < 2, 4, 4 >.

The magnitudes of these vectors are respectively

sqrt(33)

sqrt(13)

sqrt(36).

None of the sides are the same length so the triangle is not isosceles.

The sum of the squares of the shorter two side is 33 + 13 = 46, which is not equal to the sum of the longest, so the triangle is not a right triangle.

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Self-critique (if necessary):

After seeing the sloution, it makes so much sense how to find if it is a right triangle and should have been able to solve it without help.

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Self-critique rating:3

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Good work.

There were a couple of errors in the given solution.

The correct solution on that one problem, incidentally, is

v / || v || = < -1, sqrt(3), 4 > / sqrt(20) =

<-sqrt(20) / 20, sqrt(60) / 20, 4 sqrt(20) / 20)> =

< -2 sqrt(5) / 20, 2 sqrt(15) / 20, 8 sqrt(5) / 20> =

<-sqrt(5) / 10, sqrt(15) / 10, 2 sqrt(5) / 5 >.

Check my notes.

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