Assignment 1

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course Mth 277

6/12/2012 around 4:00 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

.

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.

qa 09_1

The magnitude of a vector `A = a_x `i + a_y `j in two-dimensional space is found using the Pythagorean Theorem to be sqrt(a_x^2 + a_y)^2.

The magnitude of a vector `A = a_1 `i + a_2 `j + a_3 `k, in 3-dimensional space, is found by two applications of the Pythagorean Theorem to be sqrt( a_1 ^ 2 + a_2 ^ 2 + a_3 ^2).

A unit vector is a vector whose magnitude is 1. If you divide any vector by its magnitude you get a unit vector, and its direction is the same as that of the original vector.

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Question: `q001: Let P = (3, 5, 9) and Q = (-4, 11, 3), with `A the vector whose initial point is P and whose terminal point is Q.

What are the `i , `j and `k components of ?

(NOTE: Temporarily, the ` mark is used to denote a vector, so that `A means the vector A, while `i, `j and `k denote the vectors i, j and k.)

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Your solution:

P= (3,5,9) Q= (-4,11,3)

v_1 = x_2- x_1 V_2 = y_2 - y_1 v_3 = z_2 - z_1

= (-4)-3 = 11-5 = 3 - 9

= -7 = 6 = -6

i = -7 j = 6 k = -6

PQ = -7i + 6j - 6k

confidence rating #$&*:

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Given Solution:

To move from P to Q we must move in the x direction from x = 3 to x = -4, a displacement of -7 units. So the `i component of the vector PQ is -7.

Reasoning similarly the j and k components are 6 and -6, respectively, so PQ = -7 i + 6 j - 6 k.

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Self-critique (if necessary):

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Self-critique rating:

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Question: `q002: What is the magnitude of the vector defined in the preceding problem?

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Your solution:

v = < -7,6,-6 >

l v l = sqrt( (v_1)^2 + (v_2)^2 + (v_3)^2 )

= sqrt ( (-7)^2 + (6)^2 + (-6)^2 )

= sqrt ( 49 + 36 + 36 )

= sqrt ( 121 )

= 11

confidence rating #$&*:

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Given Solution: The magnitude of the vector is found by the Pythagorean Theorem to be

|| PQ || = || -7 i + 6 j - 6 k || = sqrt( 7^2 + 6^2 + 6^2) = sqrt(121) = 11.

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Self-critique (if necessary):

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Self-critique rating:

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Question: `q003: What are the `i, `j and `k components of a unit vector in the direction of the vector PQ from the preceding two questions?

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Your solution:

unit vector = v / | v |

v = < -7,6,-6 >

so,

unit vector = -7i + 6j - 6k / 11

= -7i/11 + 6j/11 - 6k/11

decimal approx. = -.64 i + .55 j - .55 k

confidence rating #$&*:

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Given Solution:

When you divide a vector by a positive number it becomes shorter by a factor equal to that number. For example if you divide a vector by 2, you end up with a vector of half the magnitude.

A unit vector in the direction of a given vector is therefore obtained by dividing that vector by its magnitude.

A unit vector in the direction of PQ is therefore

u = PQ / || PQ || = (-7 i + 6 j - 6 k) / 11 = -7/11 i + 6/11 j - 6/11 k.

A decimal approximation to this vector is roughly -.64 i + .55 j - .55 k.

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Self-critique (if necessary):

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Self-critique rating:

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Question: `q004: What are the `i, `j and `k components of a vector parallel to PQ, having magnitude 20?

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Your solution:

unit vector = -7i/11 + 6j/11 -6k/11

the magnitude is 20 so you have to multiply the unit vector by the magnitude 20

= 20 (-7i/11 + 6j/11 - 6k/11)

= -140i/11 + 120j/11 - 120k/11

decimal approx. = -12.73i + 10.91j - 10.91k

confidence rating #$&*:

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Given Solution: We have found a unit vector in the direction of PQ.

To get a vector of magnitude 20 in the same direction we just multiply this unit vector by 20.

The vector

20 ( -7/11 i + 6/11 j - 6/11 k) = -140 / 11 i + 120 / 11 j - 120 / 11 k.

A decimal approximation is -12.7 i + 10.9 j - 10.9 k.

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Self-critique (if necessary):

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Self-critique rating:

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Question:

Let P = (3, 5) and Q = (-1, 10), with `A the vector whose initial point is P and whose terminal point is Q.

`q005. Let `v be the vector from the origin to point P. Sketch the points P and Q and the vectors `v and `A. Then sketch the points at the tip of each of the following vectors, provided the initial point of each is the origin:

`v + .5 `A, `v + 1.5 `A, `v + 2.5 `A.

Based on your sketch mark your estimated locations of the terminal points of each of the following, assuming the initial point for each to be the origin:

`v + 2 `A

`v + 3 `A

`v - 1.5 `A.

Estimate the coordinates of the terminal points of these vectors, based on your sketch.

Calculate the coordinates of the points.

How well can you fit a straight line to these points?

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Your solution:

Im not sure how to find the solution. If your giving the point P and Q, I thought you were able to find V1 and V2 and than use those two points on the vector to find the solution.

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To advise you properly I need to know a little more about what you're thinking and seeing.

Can you first describe your sketch?

How would you use your sketch to construct the vector `v + .5 `A?

What are the components of your vector `A?

What are the components of your vector `v?

*@

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

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Question: `q006. Is each of the following true or false, and why?

4 `i - 3 `j = 4 `i - 2 `j

3 x `i - 5 `j = 6 `i - 5 `j if, and only if, x = 2

|| 4 `i - 3 `j || = || 3 `i + 4 `j ||

c * (4 `i - 3 `j) = (12 `i + 9 `j) if c = 3.

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Your solution:

4 'i - 3 'j = 4 'i - 2 'j False: The problem states that both 'i and j' are identical on both sides but in theory both j' are not identical on both sides. on the left j' = -3 and on the right j' = -2

3 x 'i - 5 'j = 6 'i - 5 'j if, and only if, x = 2 True: 3x = 6 and -5 = -5. This is true only if x = 2

x = 2

|4 'i - 3 'j | = | 3 'i + 4 'j | True: sqrt ( 4^2 + -3^2 ) = sqrt ( 3^2 + 4^2 )

sqrt ( 25 ) = sqrt ( 25 )

5 = 5

c * (4 'i - 3 'j) = (12 'i + 9 'j) if c = 3 False: if c = 3 the left side is not identical to the right side which makes it false. The 'j component is = -9 on the left side & the j' component is =9 on the right side

3 * (4 'i - 3 'j)

= (12 'i - 9 'j)

Confidence rating: 3

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Given Solution:

The equation 4 `i - 3 `j = 4 `i - 2 `j states that the `i and `j components are identical on both sides. This is not so since the `j components on the left is -3 and the `j component on the right is -2.

The equation 3 x `i - 5 `j = 6 `i - 5 `j if, and only if, x = 2 is true if and only if 3x = 6 and -5 = -5. The latter is clearly true, and the former is true if and only if x = 2.

The equation || 4 `i - 3 `j || = || 3 `i + 4 `j || states that sqrt( 4^2 + (-3)^2) = sqrt( 3^2 + 4^2). Bot expression are equal to sqrt(25) = 5, so the equation is true.

If c = 3 the equation c * (4 `i - 3 `j) = (12 `i + 9 `j) becomes 3 * (4 `i - 3 `j) = (12 `i + 9 `j) . The left-hand side would have `j component -9 and the right-hand side would have `j component +9. This is clearly not so, so the equation would be false for c = 3. In fact there is no value of c which makes the equation true, since this would require that 4 c = 12 and -3 c = 9. These two equations yield different results for c.

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Self-critique (if necessary):

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Self-critique rating:

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Question: `q007. Find the value(s) of c that make each of the following true. If no such value exists, explain why this is the case:

c * ( 4 `i - 3 `j + 6 `k) = -48 `i + 36 `j - 72 `k

c * || 4 `i + 3 `j - 6 `k || = || 6 `i - 4 `j + 3 `k ||

|| 4 `i + 3 `j - 5 `k || = || 5 `i + 7 `j - c `k ||

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Your solution:

c * ( 4 `i - 3 `j + 6 `k) = -48 `i + 36 `j - 72 `k

solution:

c * (4i-3j+6k) = -48i+36j-72k

c = -12

c * || 4 `i + 3 `j - 6 `k || = || 6 `i - 4 `j + 3 `k ||

Solution:

c * sqrt( 4^2+3^2-6^2) = sqrt(6^2-4^2+3^2)

c * sqrt(61) = sqrt (61)

c = sqrt(61)/ sqrt(61)

c = 1

|| 4 `i + 3 `j - 5 `k || = || 5 `i + 7 `j - c `k ||

solution:

sqrt (4^2+3^2-5^2) = sqrt (5^2+7^2-c^2)

sqrt (50) = sqrt ( 74 + c^2 )

50 = 74 + c^2

c^2 = -24

confidence rating #$&*:

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Given Solution:

c * ( 4 `i - 3 `j + 6 `k) = -48 `i + 36 `j - 72 `k implies that 4 c = -48, -3 c = 36 and 6 c = -72. Each of these three equations yields solution c = -12, so this value of c solves the equation.

c * || 4 `i + 3 `j - 6 `k || = || 6 `i - 4 `j + 3 `k || becomes c * sqrt( 4^3 + 3^2 + 6^2) = sqrt(6^2 + 4^2 + 3^2), i.e., c * sqrt(61) = sqrt(61), with solution c = sqrt(61) / sqrt(61) = 1.

|| 4 `i + 3 `j - 5 `k || = || 5 `i + 7 `j - c `k || becomes sqrt(4^2 + 3^2 + 5^2) = sqrt(5^2 + 7^2 + c^2), i.e., sqrt(61) = sqrt(64 + c^2). This is true if and only if 61 = 64 + c^2, yielding c^2 = -3. Since the square of any real number is positive, there is no real-number solution to this equation.

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Self-critique (if necessary):

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Self-critique rating:

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Question: `q008. If theta = pi / 6, then what is the magnitude of the vector sin(theta) * `i + cos(theta) * `j?

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Your solution:

theta = pi/6

sin (theta) * 'i + cos (theta) * 'j

|sin (theta)i + cos (theta)j|

= sqrt (sin^2(pi/6) + cos^2(pi/6) )

= 1

confidence rating #$&*:

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Given Solution: sin(pi/6) `i + cos(pi/6) `j has magnitude

|| sin(pi/6) `i + cos(pi/6) `j || = sqrt( (sin^2(pi/6) + cos^2(pi/6) ).

Since sin^2(theta) + cos^2(theta) = 1 for any value of theta, we get

|| sin(pi/6) `i + cos(pi/6) `j || = sqrt( (sin^2(pi/6) + cos^2(pi/6) ) = 1.

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#*&!

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