Assignment 10 QA

#$&*

course Mth 164

4/7 6:00

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.

010. Dot product, vector algebra

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Question: `q001. The dot product of two vectors is equal to the product of the magnitudes of the vectors, multiplied by the cosine of the angle between the vectors. What is the dot product of vector v having magnitude 10 and angle 30 degrees, with vector w having magnitude 8 and angle 90 degrees?

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Your solution:

The angle between the two vectors is 60 degrees = 90 deg - 30 deg.

cos(theta) = (v * w) / (||v|| * ||w||) so

v * w = cos(theta) * (||v|| * ||w||)

v * w = cos(60 deg) * (10*8)

v * w = ½ * 80 = 40

confidence rating #$&*: 3

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Given Solution:

`aThe magnitudes are 10 and 8, and the angle between the vectors is the change in angle from 30 degrees to 90 degrees, or 60 degrees. The dot product is therefore

dot product = product of magnitudes * cos(angle) = 10 * 8 * cos(60 deg) = 40.

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Self-critique (if necessary):

OK

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Self-critique Rating: OK

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Question: `q002. What are the x and y components of the vector v having magnitude 10 and angle 30 degrees, and vector w having magnitude 8 and angle 90 degrees?

What do you get if you add the product of the two x components to the product of the two y components?

How is this result related to the answer to the preceding exercise?

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Your solution:

x = r cos(theta) & y = r sin(theta) so

x component of v = 10cos(30 deg) = 10 * `sqrt(3)/2 or approx. 8.66 and

y component of v = 10sin(30 deg) = 10 * ½ = 5

x component of w = 8cos(90 deg) = 8 * 0 and

y component of w = 8sin(90 deg) = 8 * 1 = 8

The result of adding the product of the two x components to the product of the two y components is as follows:

(10 * `sqrt(3)/2) * 0 + 5 * 8 = 40 = dot product = v * w = cos(theta) * (||v|| * ||w||) as stated in the previous problem.

confidence rating #$&*: 3

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Given Solution:

`aThe vector v has x component vx = 10 cos(30 deg) = 8.7, approx., and vy = 10 sin(30 deg) = 5. The vector w has x component wx = 8 cos(90 deg) = 0 and y component wy = 8 sin(90 deg) = 8.

The product of the two x components is vx * wx = 8.7 * 0 = 0, and the product of the y components is vy * wy = 5 * 8 = 40.

The sum of these products is 0 + 40 = 40, which is identical to the result of the preceding exercise in which we found the dot product of v and w.

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Self-critique (if necessary):

OK

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Self-critique Rating: OK

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Question: `q003. If vector v is represented by < v1, v2 > and vector w by < w1, w2 > then if the result of the preceding exercise is valid, how do we write in symbols the dot product of the two vectors?

In symbols how do we write the magnitudes of the two vectors?

How then do we write the statement that the dot product of the two vectors is equal to the product of the magnitudes of the vectors, multiplied by the cosine of the angle between the vectors?

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Your solution:

In symbols the dot product of the two vectors is as follows:

dot product = v * w = v1(w1) + v2(w2).

In symbols we write the magnitudes of the two vectors as follows:

||v|| = `sqrt(v1^2 + v2^2)

||w|| = `sqrt(w1^2 + w2^2).

We write the statement that the dot product of the two vectors is equal to the product of the magnitudes of the vectors, multiplied by the cosine of the angle between the vectors as:

v1(w1) + v2(w2) = cos(theta) * `sqrt(v1^2 + v2^2) * `sqrt(w1^2 + w2^2).

confidence rating #$&*: 3

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Given Solution:

`aIf the preceding exercise generalizes then the dot product is the sum of the product of the x components and the product of the y components. In this case we would therefore say that the dot product is v1 * w1 + v2 * w2.

The magnitudes of the two vectors are | v | = sqrt(v1^2 + v2^2) and | w | = sqrt(w1^2 + w2^2).

The statement therefore says that

v1 * w1 + v2 * w2 = sqrt(v1^2 + v2^2) * sqrt(w1^2 + w2^2) * cos(theta).

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Self-critique (if necessary):

OK

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Self-critique Rating: OK

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Question: `q004. Use the result of the preceding exercise to find the cosine of the angle between the vectors < 2, 3 > and < -7, 4 >.

What therefore is the angle between these vectors?

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Your solution:

From the preceding exercise we have:

v1(w1) + v2(w2) = cos(theta) * `sqrt(v1^2 + v2^2) * `sqrt(w1^2 + w2^2) so

cos(theta) = (v1(w1) + v2(w2)) / (`sqrt(v1^2 + v2^2) * `sqrt(w1^2 + w2^2)).

Using the vectors given we have:

cos(theta) = (2(-7) + 3(4)) / (`sqrt(2^2 + 3^2) * `sqrt((-7)^2 + 4^2))

cos(theta) = -2 / (`sqrt(13) * `sqrt(65)) or approx. -.069 and the angle(theta) between these vectors is found as follows:

theta = cos^-1 (-.069) or approx.1.64 or 93.9 deg

confidence rating #$&*: 3

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Given Solution:

`aThe cosine of the angle theta between the vectors is found using the fact that dot product is product of magnitudes multiplies by the cosine of the angle, expressed in detail as

v1 * w1 + v2 * w2 = sqrt(v1^2 + v2^2) * sqrt(w1^2 + w2^2) * cos(theta).

We easily rearrange the equation to get

cos(theta) = [ v1 * w1 + v2 * w2 ] / [ sqrt(v1^2 + v2^2) * sqrt(w1^2 + w2^2)].

In this case we have

cos(theta) = [ 2 * -7 + 3 * 4 ] / [ sqrt(2^3 + 3^2) * sqrt( (-7)^2 + 4^2) ] = -2 / ( sqrt(13) * sqrt(53) ] = -2 / 26.3 = .08 approx..

The angle is therefore the solution theta to the equation cos(theta) = .08, which gives us theta = 83 degrees, approx..

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Self-critique (if necessary):

The given solution is incorrect. The y component of the vector w was not squared. The magnitude of w should be `sqrt(65) = sqrt(49 + 16).

@& Very good.*@

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Self-critique Rating: 3

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Question: `q005. The vectors v = < 2, 4, 5 > and w = < -3, 7, 2 > exist in 3-dimensional space. What do you think are the magnitudes of these two vectors?

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Your solution:

The magnitude of v is figured as follows:

||v|| = `sqrt(2^2 + 4^2 + 5^2) = sqrt(45) = 3sqrt(5) or approx 6.71.

The magnitude of w is figured as follows:

||w|| = `sqrt((-3)^2 + 7^2 + 2^2) = sqrt(62) = 7.87.

confidence rating #$&*: 3

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Given Solution:

`aThe magnitude of a vector is found by taking the square root of the sum of the squares of its components, just as for a vector in 2-dimensional space. In this case we get magnitudes sqrt(2^2 + 4^2 + 5^2) = sqrt(45) = 6.7, approx., and sqrt((-3)^2 + 7^2 + 2^2) = sqrt(76) = 8.7 approx..

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Self-critique (if necessary):

The given solution is incorrect for ||w||.

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Self-critique Rating: 3

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Question: `q006. What is the dot product of the vectors v = < 2, 4, 5 > and w = < -3, 7, 2 > ? What therefore is the angle between these vectors?

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Your solution:

Dot product = v*w = v1(w1) + v2(w2) + v3(w3) so using the given vectors we have:

dot product = v*w = 2(-3) + 4(7) + 5(2) = 32.

The angle(theta) between these vectors is figured as follows:

cos(theta) = 32/(sqrt(45) * sqrt(62)) so

theta = cos^-1 (32/(sqrt(45) * sqrt(62))) or approx. .92 or approx.52.7 deg

confidence rating #$&*: 3

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Given Solution:

`aSince dot product = product of magnitudes * cos(theta) we have cos(theta) = dot product / product of magnitudes. In the preceding problem we found the magnitudes of the vectors to be sqrt(45) and sqrt(76). So if we can find the dot product we can find the cosine of the angle and therefore the angle.

The dot product is again the sum of the products of the invidual components, in this case 2 * (-3) + 4 * 7 + 5 * 2 = 32.

Thus we have

cos(theta) = 32 / ( sqrt(45) * sqrt(76) ) and

theta = arccos[ 32 / (sqrt(45) * sqrt(76) ) = 56.8 deg, approx..

Note that these vectors can actually be constructed in 3-dimensional space, and if the construction is accurate the angle will be as indicated.

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Self-critique (if necessary):

Again, the given solution is incorrect due to ||w|| being incorrect.

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Self-critique Rating: 3

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Question: `q007. Calculate the angle between the two vectors < 1, 7, -3, 4 > and < -3, -5, 2, 7 > .

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Your solution:

To calculate the angle(theta) between the two vectors < 1, 7, -3, 4 > and < -3, -5, 2, 7 > , we need the magnitudes of the vectors and the dot product.

We let the first vector listed be v and the second w.

We find the needed values as follows:

||v|| = sqrt(1^2 + 7^2 + (-3)^2 + 4^2) = sqrt(75)

||w|| = sqrt((-3)^2 + (-5)^2 + 2^2 + 7^2) = sqrt(87)

v * w = 1(-3) + 7(-5) + (-3)2 + 4(7) = -16

Using the values found we have:

cos(theta) = -16/( sqrt(75)* sqrt(87)) so

theta = cos^-1(-16/( sqrt(75)* sqrt(87)) or approx. 1.77 or 101.4 deg

confidence rating #$&*: 3

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Given Solution:

`aAs before we find that theta = arccos ( dot product / product of magnitudes ). In this case

theta = arccos [ (1 * -3 + 7 * -5 + -3 * 2 + 4 * 7) / ( sqrt(1^2 + 7^2 + (-3)^2 + 4^2 ) * sqrt( (-3)^2 + (-5)^2 + 2^2 + 7^2 ) ] =

arccos[ -16 / 80.8 ] =

101 degrees, approx..

Note that though these vectors occur in 4 dimensions and hence cannot be constructed in 3-dimensional space the calculation of the angle between them, extending in the most obvious manner the procedures of the preceding problems, give us a result with just a little additional calculation.

"

Self-critique (if necessary):

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Self-critique rating:

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Question: `q007. Calculate the angle between the two vectors < 1, 7, -3, 4 > and < -3, -5, 2, 7 > .

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Your solution:

To calculate the angle(theta) between the two vectors < 1, 7, -3, 4 > and < -3, -5, 2, 7 > , we need the magnitudes of the vectors and the dot product.

We let the first vector listed be v and the second w.

We find the needed values as follows:

||v|| = sqrt(1^2 + 7^2 + (-3)^2 + 4^2) = sqrt(75)

||w|| = sqrt((-3)^2 + (-5)^2 + 2^2 + 7^2) = sqrt(87)

v * w = 1(-3) + 7(-5) + (-3)2 + 4(7) = -16

Using the values found we have:

cos(theta) = -16/( sqrt(75)* sqrt(87)) so

theta = cos^-1(-16/( sqrt(75)* sqrt(87)) or approx. 1.77 or 101.4 deg

confidence rating #$&*: 3

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Given Solution:

`aAs before we find that theta = arccos ( dot product / product of magnitudes ). In this case

theta = arccos [ (1 * -3 + 7 * -5 + -3 * 2 + 4 * 7) / ( sqrt(1^2 + 7^2 + (-3)^2 + 4^2 ) * sqrt( (-3)^2 + (-5)^2 + 2^2 + 7^2 ) ] =

arccos[ -16 / 80.8 ] =

101 degrees, approx..

Note that though these vectors occur in 4 dimensions and hence cannot be constructed in 3-dimensional space the calculation of the angle between them, extending in the most obvious manner the procedures of the preceding problems, give us a result with just a little additional calculation.

"

Self-critique (if necessary):

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Self-critique rating:

#*&!

&#Good work. See my notes and let me know if you have questions. &#