course mth164 i didnt use the correct fomula on some problems but am going to use given answers to study by copying out of my saved file. ????????????y?assignment #010
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21:38:04 `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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RESPONSE --> the dot product of v is (10+8)*cos60= cos60=1/2 18*1/2=9 confidence assessment: 2
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21:38:46 The 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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RESPONSE --> ok misunderstood that the 10 and 8 were multiplied not added self critique assessment: 3
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21:46:30 `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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RESPONSE --> the x and y are the sub products of the magnitudes you get the magnitudes these are related due to the x and y are the lengths of the sides confidence assessment: 2
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21:53:36 The 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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RESPONSE --> i see that the x and y by the sin and cos and their given deg. self critique assessment: 1
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22:04:47 `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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RESPONSE --> (v1*v2)+(w1*w2)=dot formula confidence assessment: 1
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22:08:24 If 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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RESPONSE --> ok self critique assessment: 2
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22:27:36 `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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RESPONSE --> The cos will change the result to the cos of the 2 and 3 and also the -7 and 4.The angle will be the differance between the two numbers for each set of vectors confidence assessment: 2
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22:30:11 The 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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RESPONSE --> ok i now see that the 2 and the -7 and the 3 and the 4 should have been arranged like so cosx=2*-7+3*4/sqrt2^2+3^2*-7^2+4^2 self critique assessment: 2
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22:31:22 `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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RESPONSE --> the magnitudes of these vectors are 11 and 6 confidence assessment: 1
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22:32:38 The 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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RESPONSE --> ok i now see that the magnitudes are derived by squaring the vectors and the taking the square root self critique assessment: 2
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22:33:34 `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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RESPONSE --> the dot product is o because no angle is given and it necessary to have confidence assessment: 1
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22:34:50 Since 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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RESPONSE --> i didnt know that the preceeding problem influenced this one so i didnt have angles to calculate self critique assessment: 2
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22:38:50 `q007. Calculate the angle between the two vectors < 1, 7, -3, 4 > and < -3, -5, 2, 7 > .
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RESPONSE --> the angle between the vectors is 7.54 and 4.35 confidence assessment: 1
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22:39:10 As 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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RESPONSE --> didnt use correct formula self critique assessment: 1
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?????Vw?????? assignment #010 010. Dot product, vector algebra Precalculus II 12-02-2008