Assignment 10 QA

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course Mth 164

1:39 pm 3/13/15

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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:

So vector 1 times vector 2 equals magnitude1 times magnitude 2 multiplied by cos (theta1) times cos(theta2)

V1 * v2 = m1*m2 cos theta1 * cos theta2

V * w = 10 * 8 cos 30 cos 90 = 0

confidence rating #$&*: 2

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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):

I see, the angle is created by the two vectors and to find the angle we need to subtract one placement (degree) by the other to find the actual angle between the 2 vectors.

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Self-critique Rating: 3

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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:

V * w = 10 * 8 cos 60 degrees = 40

I think I’m supposed to find the x and y values of the angles given and multiply those together and then add them. So for a circle with magnitude 8 a 90 degree angle would have the x value of 0 and a y value of 8. For the 30 degree angle with a magnitude of 10 we obtain an x value of 5sqrt3 and a y value of 5

So this would be:

0 * 5 sqrt 3 + 8 * 5 =

0 + 40 = 40

This is the same result as when we used this formula:

V1 * v2 = m1*m2 cos (theta2 - theta1)

confidence rating #$&*: 2

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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):

I used the exact value of the 30 degree angle although it did not interfere with my results. I was glad to see I grasped the idea at hand.

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Self-critique Rating: 3

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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:

I believe this would be:

V1 * w1 + v2 * w2 = m1 * m2 cos (theta2 - theta 1)

confidence rating #$&*: 2

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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):

Does this mean, to get the absolute value of the magnitudes we square them and then find the square root????

I see that we only need the angle between the vectors and it is unnecessary to subtract one from the other within the formula.

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Self-critique Rating: OK

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The magnitude is an absolute value.

To get the magnitude from the components of a vector you use the Pythagorean Theorem, adding the squares of the components and taking the square root.

Be sure you understand how this is related to the Pythagorean Theorem.

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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:

I’m not so sure I remember how to draw these but I believe they both originate at the origin and create angles from there. The angle appears to be 90 degrees but I assume using the formulas from the former problem:

vx * wx + vy * wy = sqrt(vx^2 + vy^2) * sqrt(wx^2 + wy^2) * cos(theta).

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

-14 + 12 = sqrt(4 + 9) * sqrt(49 + 16) * cos(theta)

-2 = sqrt(13) * sqrt (65) * cos(theta)

-2/29.06 = cos(theta) find arcos

Theta = 93.95 approximately

confidence rating #$&*: 2

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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):

??????When I find the value for

sqrt( (-7)^2 + 4^2) ]

I get sqrt(49 + 16) = sqrt(65) not 53. Is this a typo or did I complete a step incorrectly??????

Otherwise my steps seems to prove successful in finding the solution.

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Self-critique Rating: 3

@&

It looks like I added 4 to 49 rather than adding 16.

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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:

Well before we used a formula of |v| = sqrt(vx^2 + vy^2) for the magnitude of a two dimensional vector so I would assume that there is a possibility of find a 3 dimensional vector in a similar manner as follows:

|v| = sqrt(2^2 + 4^2 + 5^2)

= sqrt(4 + 16 + 25)

=sqrt(45)

= 6.71 approx.

And

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

= sqrt (9 + 49 + 4)

= sqrt (62)

= 7.87 approx.

confidence rating #$&*: 2

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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):

Once again I see a variance between one of my solutions and the one given although I followed similar steps.

????What am I doing wrong????

Other than this variance my solution and method was correct.

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Self-critique Rating: 3

@&

It looks like I was on a roll when I wrote this one.

I suspect that I typed 8.7 when I meant to type 7.8, but that's only one of a number of possible explanations.

*@

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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:

Following the previous pattern I would think for this we see:

Adding the products from the x values to the y values and what I will call the z values. So:

(vx * wx) + (vy * wy) + (vz * wz)

(2 * -3) + (4 * 7) + (5 * 2)

(-6) + (28) + (10)

= 32

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):

I did not find the angle theta. I only found what the dot product is and accidentally forgot to find the angle. This was negligence on my part.

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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:

So, we have the formula:

vx * wx + vy * wy + vz * wz + va*wa= sqrt(vx^2 + vy^2 + vz^2 + va^2) * sqrt(wx^2 + wy^2 + wz^2 + wa^2) * cos(theta).

(1*-3) + (7 * -5) + (3 *2) + (4 * 7)

(-3) + (-35) + (6) + (28)

= -4

-4 = sqrt(1^2 + 7^2 + 3^2 + 4^2) * sqrt(-3^2 + -5^2 + 2^2 + 7^2) cos(theta)

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

-4 / [sqrt(1 + 49 + 9 + 16) * sqrt( 9 + 25+ 4 + 49) = cos (theta)

Arcos (Cos(theta) = -4 / [sqrt(75) * sqrt(87)] )

Theta = 92.8 degrees

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.

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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:

So, we have the formula:

vx * wx + vy * wy + vz * wz + va*wa= sqrt(vx^2 + vy^2 + vz^2 + va^2) * sqrt(wx^2 + wy^2 + wz^2 + wa^2) * cos(theta).

(1*-3) + (7 * -5) + (3 *2) + (4 * 7)

(-3) + (-35) + (6) + (28)

= -4

-4 = sqrt(1^2 + 7^2 + 3^2 + 4^2) * sqrt(-3^2 + -5^2 + 2^2 + 7^2) cos(theta)

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

-4 / [sqrt(1 + 49 + 9 + 16) * sqrt( 9 + 25+ 4 + 49) = cos (theta)

Arcos (Cos(theta) = -4 / [sqrt(75) * sqrt(87)] )

Theta = 92.8 degrees

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:

So, we have the formula:

vx * wx + vy * wy + vz * wz + va*wa= sqrt(vx^2 + vy^2 + vz^2 + va^2) * sqrt(wx^2 + wy^2 + wz^2 + wa^2) * cos(theta).

(1*-3) + (7 * -5) + (3 *2) + (4 * 7)

(-3) + (-35) + (6) + (28)

= -4

-4 = sqrt(1^2 + 7^2 + 3^2 + 4^2) * sqrt(-3^2 + -5^2 + 2^2 + 7^2) cos(theta)

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

-4 / [sqrt(1 + 49 + 9 + 16) * sqrt( 9 + 25+ 4 + 49) = cos (theta)

Arcos (Cos(theta) = -4 / [sqrt(75) * sqrt(87)] )

Theta = 92.8 degrees

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:

#*&!#*&!

&#Your work looks good. See my notes. Let me know if you have any questions. &#