#$&* course MTH 277 6/13 between 2-2:30 Question: Find v dot w when v = 4i + j and w =3i + 2k.YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
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Given Solution: v dot w = (4i + j) dot (3i + 2k ) = (4i + j + 0 k) dot (3i + 0 j + 2k ) = 4 * 3 + 1 * 0 + 0 * 2 = 12. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique rating: ok ********************************************* Question: Determine whether v = 5i - 5j + 5k and w = 8i - 10j -2k are orthogonal. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Two vectors are orthogonal if their dot product is zero. The dot product of these vectors is 5 * 8 - 5 * (-10) + 5 * (-2) = 40 + 50 - 10 = 80. They are not orthogonal. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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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 * (-10) + 5 * (-2) = 40 + 50 - 10 = 80. They are not orthogonal. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: ok ********************************************* Question: Find the angle between v = 2i +3 k and w = -j + 4k. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Since v dot w = || v || || w || cos(theta) we have theta = cos^-1 ( v dot w ) || v || || w || = cos^-1 ( 12 / (sqrt(13) * sqrt( 17) ) = cos^-1 (.79) = 38 degrees. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Since v dot w = || v || || w || cos(theta) we have theta = cos^-1 ( v dot w ) || v || || w || = cos^-1 ( 12 / (sqrt(13) * sqrt( 17) ) = cos^-1 (.79) = 38 degrees, approx., or roughly.6 radians. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: ok ********************************************* Question: Find two distinct unit vectors orthogonal to both v = i + 2j -2k and w = i + j - 2k. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Suppose a i + b j + c k is orthogonal to both. a + 2 b - 2 c = 0 a + b - 2 c = 0 Subtracting the second equation from the first we get b = 0. With this value of b both our first and our second equation become a - 2 c = 0 so that a = 2 c. Any vector of the form 2c i + c k is therefore orthogonal to our two vectors. Any such vector has magnitude sqrt( (2 c)^2 + c^2) = sqrt( 5 c^2) = sqrt(5) | c |. If c is positive then | c | = c and our vector is (2 c i + c k ) / (sqrt(5) c) = 2 sqrt(5) / 5 i + sqrt(5) / 5 k. If c is negative then | c | = - c and our vector will be (2 c i + c k ) / (- sqrt(5) c) = - 2 sqrt(5) / 5 i - sqrt(5) / 5 k. Our two solution vectors are equal and opposite. Each is a unit vector perpendicular to the two given vectors. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ok ------------------------------------------------ Self-critique rating: ok ********************************************* Question: Let v = i - j + 4k and w = -i + 3j + 2k. Find cos(theta). Find s such that v is orthogonal to sv - w. Also find t such that v - tw is orthogonal to w YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your 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 = 2/9. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: cos(theta) = v dot w / ( || v || || w ||) = 4 / (sqrt(18) sqrt(14) ) = 4 / (12 sqrt(7) ). The condition v orthogonal to s v - w is v dot (s v - w ) = 0 (i - j + 4 k ) dot ( (s - 1) i + (-s + 3) j + (4 s + 2) k ) = 0 which becomes s - 1 + s - 3 + 16 s + 8 = 0 so that 18 s = 4 and s = 4 / 18 = 2/9. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique rating:ok ********************************************* Question: Find the work performed when a force F = (6/11)i - (2/11)j + (6/11)k is applied to an object moving along the line from P(3,5,-4) to Q(-4,-9,-11). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The work is F dot `ds = ( (6/11)i - (2 / 11) j + (6 / 11) k ) dot (-7 i - 14 j - 7 k ) = -42/11 + 28 / 11 - 42 /11 = -56 / 11. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 = -56 / 11. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: ok " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ok " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!