query_09_2

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course Mth 277

9/14 12am

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

.

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_2

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

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Your solution:

u+v=0

u-v=2i+4j-6k

(5/2)u=2.5i+5j-(15/2)k

2u+3v=-i-2j+3k

confidence rating #$&*:

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

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Your solution:

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

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):OK

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Self-critique rating:OK

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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:

(1,3,-6) r=3sqrt(7)

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:

PQ=5i-5j-7k

length=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:

=(-1/2sqrt(5))i+(sqrt(15)/10)j+2sqrt(5)/5k

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):OK

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Self-critique rating:OK

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

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Your solution:

A wavy wall

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:

NO. The components do not even have the same signs

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):Ok I understand how to find it using dot product.

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Self-critique rating:2

@& Another alternative would be the cross product. If the vectors are parallel the cross product is zero.*@

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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:

AB=sqrt(33)

BC=sqrt(13)

AC=6

Not right

Not Isosceles

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):OK

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Self-critique rating:OK

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@& Very good. Check my one note for still another way to tell if vectors are parallel.*@