#$&* course MTH 277 09/14/20119:38 am If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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Given Solution: v dot w = 4 * 3 + 1 * 2 = 12 + 2 = 14. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I thought we had to stick to the same signs, so I didn’t add the others. ------------------------------------------------ Self-critique rating:2
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Given Solution: Two vectors are orthogonal if the angle between them is 90 deg, i.e., if and onlye if their dot product is zero. The dot product of these vectors is 5 * 8 - 5 * (-8) + 5 * (-2) = 40 + 40 - 10 = 70. They are not orthogonal. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK ********************************************* Question: Find the angle between v = 2i +3 k and w = -j + 4k. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Cos(theta) = (v dot w)/(||v|| ||w||) V dot w = -2 +12 = 10 ||v|| = sqrt(13) ||w|| = sqrt(17) ||v|| ||w|| = sqrt(30) 10/sqrt(30) roughly= 1.82574 cos(theta) roughly= 1.82574 confidence rating #$&*: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): Screwed up with the calculations of the denominator. Added the numbers together instead of multiplying the square roots. ------------------------------------------------ Self-critique rating: 2 ********************************************* Question: Find two distinct unit vectors orthogonal to both v = i + 2j -2k and w = i + j - 2k. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 1: origin (0,0,0) 2: <2, 0, 1> confidence rating #$&*:232; 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Suppose a i + b j + c k is orthogonal to both. Then the dot product of this vector with each of the given vectors is zero, and we have a + 2 b - 2 c = 0 a + b - 2 c = 0 Subtracting the second equation from the first we get b = 0. With this value of b both our first and our second equation become a - 2 c = 0 so that a = 2 c. Any vector of the form 2c i + c k is therefore orthogonal to our two vectors. Any such vector has magnitude sqrt( (2 c)^2 + c^2) = sqrt( 5 c^2) = sqrt(5) | c |. If c is positive then | c | = c and our vector is (2 c i + c k ) / (sqrt(5) c) = 2 sqrt(5) / 5 i + sqrt(5) / 5 k. If c is negative then | c | = - c and our vector will be (2 c i + c k ) / (- sqrt(5) c) = - 2 sqrt(5) / 5 i - sqrt(5) / 5 k. Our two solution vectors are equal and opposite. Each is a unit vector perpendicular to the two given vectors. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I clearly had no idea what I was doing here. ------------------------------------------------ Self-critique rating: 1 ********************************************* Question: Let v = i - j + 4k and w = -i + 3j + 2k. Find cos(theta). Find s such that v is orthogonal to sv - w. Also find t such that v - tw is orthogonal to w cos(theta) = v dot w / ( || v || || w ||) = 4 / (sqrt(18) sqrt(14) ) = 4 / (12 sqrt(7) ). The condition v orthogonal to s v - w is v dot (s v - w ) = 0 (i - j + 4 k ) dot ( (s - 1) i + (-s + 3) j + (4 s + 2) k ) = 0 which becomes s - 1 + s - 3 + 16 s + 8 = 0 so that 18 s = 4 and s = 4 / 18. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: First part: (v dot w)/(||v|| ||w||) [1 + (-3) + 8] = 6 ||v|| = sqrt(18) ||w|| = sqrt(14) 6/(sqrt(18) * sqrt(14)) cos(theta) = 0.377964 Second Part: W dot (v - tw) = 0 (-i + 3j + 2k) dot (i - j + 4k - t(-i + 3j + 2k)) = 0 (-i + 3j + 2k) dot ((1 - t)i + ( confidence rating #$&*:232; ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& 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: [(42/11) + (28/11) - (42/11)] confidence rating #$&*:232; 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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):OK ------------------------------------------------ Self-critique rating:OK " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!