#$&* course Mth 277 9/13/2011 @ 11:22 p.m. If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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Given Solution: The result is just the vector k: v X w = ( sin(theta) i + cos(theta) j ) X (-cos(theta) i + sin(theta) j ) = -sin(theta) cos(theta) i X i + sin(theta) sin(theta) i X j - cos(theta) cos(theta) j X i + cos(theta) sin(theta) j X j. i X i amd j X j are both zero, since sin(theta) = 0 for both of these computations. i X j = k by the right-hand rule, and likewise j X i . = -k, so the product is sin(theta) sin(theta) k - cos(theta) cos(theta) (-k) = sin^2(theta) i + cos^2(theta) k = (sin^2(theta) + cos^2(theta) ) * k = k &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I don’t quite see how iXi AND jXj are both 0 since sin(theta) = 0…especially the jXj. Also, I completely overlooked the iXj = k…but I do know this. ------------------------------------------------ Self-critique rating: Much better. ********************************************* Question: Find sin(theta) where theta is the angle between v = -i + j and w = -i + j + 2k. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: ||w X v|| = ||w||*||v|| sin(theta). Sin(theta) = (2i + 2j) / sqrt.6 * sqrt.2. sin(theta) = 0. Theta = 0. confidence rating #$&*:y ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: || v X w || = || v || || w || sin(theta) so sin(theta) = || v X w || / (|| v || || w || ) = || 2 j + 2 i || / (sqrt(2) sqrt(6) ) = 2 sqrt(2) / ( sqrt(2) sqrt(6) ) = 2 / sqrt(6) = 2 sqrt(6) / 6 - sqrt(6) / 3. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I think our answers are the same... ------------------------------------------------ Self-critique rating: ********************************************* Question: Find a unit vector which is orthogonal to both v = 2i - j and w = 2j - k. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: After some reviewing on the internet, since I didn’t know the answer to this in the last Query, I now understand the cross products a bit more. u X w = i + 2j + 4k, this vector is perpendicular to our original 2. To make it into a unit vector, we divide by the magnitude. (1/sqrt.21)i + (2/sqrt.21)j + (4/sqrt.21)k. confidence rating #$&*:y ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: v X w is orthogonal to both v and w. v X w = i + 2 j +4 k A unit vector in this direction is (i + 2 j + 4 k ) / sqrt(1^2 + 2^2 + 4^2) = (i + 2 j + 4 k ) sqrt(21) / 21 . If we take the dot product of this vector with either of our original vectors we will get zero. You can verify that this vector is indeed a unit vector. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: Find the area of the triangle with vertices P(2,0,0), Q(1,1,-1), R(3,1,2). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: PQ = -i + j -k, ||PQ|| = 1.73. QR = 2i + 3k, ||QR|| = 3.61. PR = i + j + 2k, ||PR|| = 2.45. These magnitudes are the lengths of the sides. This triangle is scalene, so I will use Heron’s formula to calculate area (if I can remember how to apply it correctly) s = (a + b + c) / 2 where a b and c are sides. S = 3.9. A = sqrt.(s(s-a)(s-b)(s-c)) = sqrt.(3.9 * 1.73 * 3.61 * 2.45) = 1.89 units ^2. confidence rating #$&*:erately. ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: 8) Determine if each of the following products is a vector, scalar, or not defined at all. Explain why. u X (v X w) , u dot (v dot w), (u X v) dot (w Xr). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: u X (v X w) is a vector that is perpendicular to a vector perpendicular to (v X w). u dot (v dot w) is a scalar used to find the angle between two vectors. (u X v) dot (w X r) would be two vectors being multiplied together, which is just a cross product. confidence rating #$&*:m pretty sure about the first two answers, but not at all sure about the third one. ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: Find a number t such that the vectors -i - j, i - (1/2) j + (1/2)k and -2i -2j - 2tk all lie in the same plane. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I thought on this one for about 20 minutes searching for how to do it. I feel like I need a vector parallel to the resultant vector of the first two vectors, or something close to this idea. confidence rating #$&*: at all. ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: