#$&* course Mth 164 1:39 pm 3/13/15
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Given Solution: `aThe magnitudes are 10 and 8, and the angle between the vectors is the change in angle from 30 degrees to 90 degrees, or 60 degrees. The dot product is therefore dot product = product of magnitudes * cos(angle) = 10 * 8 * cos(60 deg) = 40. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I see, the angle is created by the two vectors and to find the angle we need to subtract one placement (degree) by the other to find the actual angle between the 2 vectors. ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q002. What are the x and y components of the vector v having magnitude 10 and angle 30 degrees, and vector w having magnitude 8 and angle 90 degrees? What do you get if you add the product of the two x components to the product of the two y components? How is this result related to the answer to the preceding exercise? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: V * w = 10 * 8 cos 60 degrees = 40 I think I’m supposed to find the x and y values of the angles given and multiply those together and then add them. So for a circle with magnitude 8 a 90 degree angle would have the x value of 0 and a y value of 8. For the 30 degree angle with a magnitude of 10 we obtain an x value of 5sqrt3 and a y value of 5 So this would be: 0 * 5 sqrt 3 + 8 * 5 = 0 + 40 = 40 This is the same result as when we used this formula: V1 * v2 = m1*m2 cos (theta2 - theta1) confidence rating #$&*: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& 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. ------------------------------------------------ Self-critique Rating: 3 ********************************************* 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY 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). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& 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. ------------------------------------------------ Self-critique Rating: OK
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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.. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& 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. ------------------------------------------------ Self-critique 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.. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& 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. ------------------------------------------------ Self-critique 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& 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. ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q007. Calculate the angle between the two vectors < 1, 7, -3, 4 > and < -3, -5, 2, 7 > . YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY 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): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q007. Calculate the angle between the two vectors < 1, 7, -3, 4 > and < -3, -5, 2, 7 > . YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY 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 #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aAs before we find that theta = arccos ( dot product / product of magnitudes ). In this case theta = arccos [ (1 * -3 + 7 * -5 + -3 * 2 + 4 * 7) / ( sqrt(1^2 + 7^2 + (-3)^2 + 4^2 ) * sqrt( (-3)^2 + (-5)^2 + 2^2 + 7^2 ) ] = arccos[ -16 / 80.8 ] = 101 degrees, approx.. Note that though these vectors occur in 4 dimensions and hence cannot be constructed in 3-dimensional space the calculation of the angle between them, extending in the most obvious manner the procedures of the preceding problems, give us a result with just a little additional calculation. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&! ********************************************* Question: `q007. Calculate the angle between the two vectors < 1, 7, -3, 4 > and < -3, -5, 2, 7 > . YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: So, we have the formula: vx * wx + vy * wy + vz * wz + va*wa= sqrt(vx^2 + vy^2 + vz^2 + va^2) * sqrt(wx^2 + wy^2 + wz^2 + wa^2) * cos(theta). (1*-3) + (7 * -5) + (3 *2) + (4 * 7) (-3) + (-35) + (6) + (28) = -4 -4 = sqrt(1^2 + 7^2 + 3^2 + 4^2) * sqrt(-3^2 + -5^2 + 2^2 + 7^2) cos(theta) -4 / [sqrt(1^2 + 7^2 + 3^2 + 4^2) * sqrt(-3^2 + -5^2 + 2^2 + 7^2)] = cos(theta) -4 / [sqrt(1 + 49 + 9 + 16) * sqrt( 9 + 25+ 4 + 49) = cos (theta) Arcos (Cos(theta) = -4 / [sqrt(75) * sqrt(87)] ) Theta = 92.8 degrees confidence rating #$&*: 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): ------------------------------------------------ Self-critique rating: #*&!#*&!