#$&* course MTH 277 9/13 9:34 Question: Find v X w when v = sin(theta)i + cos(theta)j and w = -cos(theta)i + sin(theta)j (theta is any angle).YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
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Given Solution: The result is just the vector k: v X w = ( sin(theta) i + cos(theta) j ) X (-cos(theta) i + sin(theta) j ) = -sin(theta) cos(theta) i X i + sin(theta) sin(theta) i X j - cos(theta) cos(theta) j X i + cos(theta) sin(theta) j X j. i X i amd j X j are both zero, since sin(theta) = 0 for both of these computations. i X j = k by the right-hand rule, and likewise j X i . = -k, so the product is sin(theta) sin(theta) k - cos(theta) cos(theta) (-k) = sin^2(theta) i + cos^2(theta) k = (sin^2(theta) + cos^2(theta) ) * k = k &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: OK ********************************************* Question: Find sin(theta) where theta is the angle between v = -i + j and w = -i + j + 2k. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: || v X w || = || v || || w || sin(theta) v X w = [(1)(2) - (0)(1)]i - [(-1)(2) - (0)(-1)]j + [(-1)(1) - (1)(-1)]k v X w = (2 - 0)i - (-2 - 0)j + (-1 + 1)k v X w = 2i + 2j -k || v X w || = sqrt [ 2^2 + 2^2 + (-1)^2] || v X w || = sqrt [4 + 4 + 1] || v X w || = sqrt (6) || v || = sqrt [(-1)^2 + 1^2] || v || = sqrt [1 + 1] || v || = sqrt (2) || w || = sqrt [(-1)^2 + 1^2 + 2^2] || w || = sqrt [1 + 1 + 4] || w || = sqrt (6) sqrt 6 = (sqrt 2)(sqrt 6) (sin theta) 1/ (sqrt 2) = sin theta theta = 45 degrees confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): OK ------------------------------------------------ Self-critique rating: OK ********************************************* Question: Find a unit vector which is orthogonal to both v = 2i - j and w = 2j - k. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: v X w = [(-1)(-1) - (0)(2)]i - [(2)(-1) - (0)(0)]j + [(2)(2) - (-1)(0)]k v X w = [1 - 0]i - [-2 - 0]j + [4 + 1]k v X w = i + 2j + 5k || v X w || = sqrt [1^2 + 2^2 + 5^2] || v X w || = sqrt [1 + 4 + 25] || v X w || = sqrt (29) u = (1 / (sqrt 29)) (i + 2j + 5k) u = (i / (sqrt 29)) + (2j / (sqrt 29)) + (5k / (sqrt 29)) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: v X w is orthogonal to both v and w. v X w = i + 2 j +4 k A unit vector in this direction is (i + 2 j + 4 k ) / sqrt(1^2 + 2^2 + 4^2) = (i + 2 j + 4 k ) sqrt(21) / 21 . If we take the dot product of this vector with either of our original vectors we will get zero. You can verify that this vector is indeed a unit vector. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Although I did not get the correct solution, I worked through the correct sequence of steps. I messed up in one of the first parts of the problem: v X w = [1 - 0]i - [-2 - 0]j + [4 + 1]k. This equation should read v X w = [1 - 0]i - [-2 - 0]j + [4 + 0]k because (-1)(0) = 0, not -1. ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: Find the area of the triangle with vertices P(2,0,0), Q(1,1,-1), R(3,1,2). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: PQ = (1 - 2)i + (1 - 0)j + (-1 - 0)k PQ = -i + j - k PR = (3 - 2)i + (1 - 0)j + (2 - 0)k PR = i + j +2k PQ x PR = [(1)(2) - (-1)(1)]i - [(-1)(2) - (-1)(1)] j + [(-1)(1) - (1)(1)]k PQ x PR = (2 + 1)i - (-2 + 1)j + (-1 + 1)k PQ x PR = 3i + j A = (.5)(|| PQ X PR ||) A = (.5)(sqrt (3^2 + 1^2)) A = (sqrt 10) / 2 confidence rating #$&*:232;2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: 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 X r). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: u X (v X w) = vector use cab-bac formula: a x (b x c) = (c dot a)b - (b dot a)c u dot (v dot w) = neither (u X v) dot (w X r) = neither *** Unless the equation matches a formula given for a vector in the book, I do not know how to determine whether a product is a vector, a scalar, or neither. Confidence rating: 1
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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 - j) x ( i - (1/2)j + (1/2)k) = [(-1)(1/2) - (0)(-1/2)]i - [(-1)(1/2) - (0)(1)] j + [(-1)(-1/2) - (-1)(1)] k (-1/2)i - (-1/2)j + (3/2)k ((-1/2)i - (-1/2)j + (3/2)k) x (-2i -2j - 2tk) = [(1/2)(-2t) - (3/2)(-2)]i - [(-1/2)(-2t) - (3/2)(2)] j + [(-1/2)(-2) - (1/2)(2)]k (-t + 3)i - (t - 3)j + (1-1)k (3 - t)i + (-t + 3)j 0 = sqrt [ (3-t)^2 + (3-t)^2] 0 = sqrt [2 * (3 - t)^2] 0 = (3 - t) (sqrt 2) 0 = 3 - t t = 3 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating:" Self-critique (if necessary): ------------------------------------------------ Self-critique rating: Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&! Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!#*&!