#$&* course MTH 277 9/13 6:41 Question: Find v dot w when v = 4i + j and w =3i + 2k.YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
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Given Solution: v dot w = 4 * 3 + 1 * 2 = 12 + 2 = 14. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ??? Why can you multiply the coefficient before the j in the first equation by the coefficient before the k in the second equation??? ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: Determine whether v = 5i - 5j + 5k and w = 8i - 10j -2k are orthogonal. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: v dot w = (5)(8) + (-5)(-10) + (5)(-2) v dot w = 40 + 50 - 10 v dot w = 90 - 10 v dot w = 80 NOT ORTHOGONAL confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Two vectors are orthogonal if the angle between them is 90 deg, i.e., if and onlye if their dot product is zero. The dot product of these vectors is 5 * 8 - 5 * (-8) + 5 * (-2) = 40 + 40 - 10 = 70. They are not orthogonal. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I agree with the given solution that the vectors are not orthogonal. However, when the dot product is solved, I got a solution of 80, not 70. I believe that my solution is correct because according to the given solution, the coefficient of j in the second equation would have been -8. However, in the problem, this coefficient is -10. ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: Find the angle between v = 2i +3 k and w = -j + 4k. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: cos theta = (1/ (||v|| * ||w||) (v * w) ||v|| = sqrt [ 2^2 + 3^2 ] ||v|| = sqrt [4 + 9] ||v|| = sqrt (13) ||w|| = sqrt [ (-1)^2 + 4^2 ] ||w|| = sqrt [1 + 16] ||w|| = sqrt (17) v dot w = (2)(0) + (0)(-1) + (3)(4) v dot w = 0 + 0 + 12 v dot w = 12 cos theta = [1/ ((sqrt 13)(sqrt 17))] * (12) cos theta = 12/ (sqrt 221) theta= 36.176 degrees confidence rating #$&*:232;2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Since v dot w = || v || || w || cos(theta) we have theta = cos^-1 ( v dot w ) || v || || w || = cos^-1 ( 10 / (sqrt(13) * sqrt( 17) ) = cos^-1 (.67) = 48 degrees, approx., or roughly.8 radians. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ??? Why is the product of the k coefficients in the second step of the given solution 10 instead of 12??? ------------------------------------------------ Self-critique rating: ********************************************* Question: Find two distinct unit vectors orthogonal to both v = i + 2j -2k and w = i + j - 2k. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: x dot v = 0 x dot w = 0 x dot v = x dot w (x1)(1) + (x2)(2) + (x3)(-2) = (x1)(1) + (x2)(1) + (x3)(-2) x1 + 2*x2 - 2*x3 = x1 + x2 - 2*x3 x2 = 0 x1 - 2*x3 = 0 x1 = 2* x3 s = 2s s*i + 2s*k = x || x || = sqrt [s^2 + (2s)^2] || x || = sqrt (5 * s^2) || x || = abs (s) * sqrt (5) x = (s*i + 2s*k ) / (abs (s) * sqrt (5)) x = (i / sqrt (5)) + (2k / sqrt (5)) OR x = (-i / sqrt (5)) - (2k / sqrt (5)) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Suppose a i + b j + c k is orthogonal to both. Then the dot product of this vector with each of the given vectors is zero, and we have a + 2 b - 2 c = 0 a + b - 2 c = 0 Subtracting the second equation from the first we get b = 0. With this value of b both our first and our second equation become a - 2 c = 0 so that a = 2 c. Any vector of the form 2c i + c k is therefore orthogonal to our two vectors. Any such vector has magnitude sqrt( (2 c)^2 + c^2) = sqrt( 5 c^2) = sqrt(5) | c |. If c is positive then | c | = c and our vector is (2 c i + c k ) / (sqrt(5) c) = 2 sqrt(5) / 5 i + sqrt(5) / 5 k. If c is negative then | c | = - c and our vector will be (2 c i + c k ) / (- sqrt(5) c) = - 2 sqrt(5) / 5 i - sqrt(5) / 5 k. Our two solution vectors are equal and opposite. Each is a unit vector perpendicular to the two given vectors. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): When solving this problem, I got stuck after I set x1 - 2*x3 equal to 0. After briefly looking at the given response, I understood how to continue working the problem. I do not know if I could work another problem of this sort on my own. However, I do believe that with more practice, it will become a logical series of operations. ------------------------------------------------ Self-critique rating: 3
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Given Solution: cos(theta) = v dot w / ( || v || || w ||) = 4 / (sqrt(18) sqrt(14) ) = 4 / (12 sqrt(7) ). The condition v orthogonal to s v - w is v dot (s v - w ) = 0 (i - j + 4 k ) dot ( (s - 1) i + (-s + 3) j + (4 s + 2) k ) = 0 which becomes s - 1 + s - 3 + 16 s + 8 = 0 so that 18 s = 4 and s = 4 / 18. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Although my answer does not match the given solution, I believe that my answer is correct. In the given solution, it says “(i - j + 4 k ) dot ( (s - 1) i + (-s + 3) j + (4 s + 2) k ) = 0.” However, I believe that the correct equation is (i - j + 4 k ) dot ( (s - (-1)) i + (-s + 3) j + (4 s + 2) k ) = 0. With this alteration, my answer is correct. ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: Find the work performed when a force F = (6/11)i - (2/11)j + (6/11)k is applied to an object moving along the line from P(3,5,-4) to Q(-4,-9,-11). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: W = F dot PQ w = [(6/11)i - (2/11)j + (6/11)k] dot [(-4 - 3)i + (-9 - 5)j + (-11 -(-4))k] w = [(6/11)i - (2/11)j + (6/11)k] dot (-7i - 14j - 7k) w = (6/11)(-7) + (-2/11)(-14) + (6/11)(-7) w = (-42/11) + (28/11) - (42/11) w = -56/11 confidence rating #$&*:232;2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The work is F dot `ds = ( (6/11)i - (2 / 11) j + (6 / 11) k ) dot (-7 i - 14 j - 7 k ) = -42/11 + 28 / 11 - 42 /11 = 28 / 11. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ??? In the final step of the given solution, why does -42 + 28 - 42 = 28??? Self-critique rating: 3" Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: Find the work performed when a force F = (6/11)i - (2/11)j + (6/11)k is applied to an object moving along the line from P(3,5,-4) to Q(-4,-9,-11). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: W = F dot PQ w = [(6/11)i - (2/11)j + (6/11)k] dot [(-4 - 3)i + (-9 - 5)j + (-11 -(-4))k] w = [(6/11)i - (2/11)j + (6/11)k] dot (-7i - 14j - 7k) w = (6/11)(-7) + (-2/11)(-14) + (6/11)(-7) w = (-42/11) + (28/11) - (42/11) w = -56/11 confidence rating #$&*:232;2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The work is F dot `ds = ( (6/11)i - (2 / 11) j + (6 / 11) k ) dot (-7 i - 14 j - 7 k ) = -42/11 + 28 / 11 - 42 /11 = 28 / 11. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ??? In the final step of the given solution, why does -42 + 28 - 42 = 28??? Self-critique rating: 3" Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!