#$&* course MTH 277 09/21/2011 4:11 am query_09_4*********************************************
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Given Solution: The result is just the vector k: v X w = ( sin(theta) i + cos(theta) j ) X (-cos(theta) i + sin(theta) j ) = -sin(theta) cos(theta) i X i + sin(theta) sin(theta) i X j - cos(theta) cos(theta) j X i + cos(theta) sin(theta) j X j. i X i amd j X j are both zero, since sin(theta) = 0 for both of these computations. i X j = k by the right-hand rule, and likewise j X i . = -k, so the product is sin(theta) sin(theta) k - cos(theta) cos(theta) (-k) = sin^2(theta) i + cos^2(theta) k = (sin^2(theta) + cos^2(theta) ) * k = k &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): It looks like I need to go over the rules in 3.4 a little more in-depth. ------------------------------------------------ Self-critique rating: 2
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Given Solution: || v X w || = || v || || w || sin(theta) so sin(theta) = || v X w || / (|| v || || w || ) = || 2 j + 2 i || / (sqrt(2) sqrt(6) ) = 2 sqrt(2) / ( sqrt(2) sqrt(6) ) = 2 / sqrt(6) = 2 sqrt(6) / 6 - sqrt(6) / 3. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I stopped one step short. I thought all I had to do to correctly answer the question was find the cross product. I didn’t connect the two ideas that I had to use the dot product as well to find the magnitude and then the sin(theta) ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: Find a unit vector which is orthogonal to both v = 2i - j and w = 2j - k. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I = 2, 0 J = -1, 2 K = 0, -1 (0 + 1)i - (0 - 2)j + (0 + 4)k = i - (-2)j + 4k i + 2j + 4k confidence rating #$&*:232; 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): . . . The answer provided below seems to show a different method of solving cross-products than what was given in the book. If that is indeed the case, then I need to figure out exactly what the method is. Upon further examination, I have been adding the second cross products, rather than subtracting them. (a2*b3 + a3*b2) is what has been happening. ------------------------------------------------ Self-critique rating: 3
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Given Solution: Consider the vector PQ = < -1, 1, -1 > to be the base of the triangle, which therefore has magnitude || PQ ||. The vector PR = < 1, 1, 2 > then forms a side adjacent to the base. An altitude from point R to the base then has magnitude || PR || sin(theta). Since PQ X PR has magnitude || PQ || || PR || sin(theta), which is just the product of the triangle's base and altitude. Thus the area is || PQ X PR || . Having calculated this quantity you will have the area of the triangle. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Right method, but I used the values given, forgetting to calculate the vectors themselves. ------------------------------------------------ Self-critique rating: 3 ********************************************* 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 X r). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 1st: vector. This uses the cab-bac formula and turns the entire equation into a few dot products. Dot products always return a vector, thus the answer is a vector. 2nd: vector. Solve the dot product in parenthesis, end up with a vector, then solve the second dot product, end up with a vector. 3rd: scalar. Assume (w X r = height of parallelogram) then you end up with (u X v) dot (w) which is the equation for a parallelogram, AKA scalar confidence rating #$&*:232; 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: (v X w) is a vector perpendicular to both v and w , so u X (v X w) is a vector perpendicular to both u and v X w . (v dot w) is a scalar (i.e., just a number), so u dot (v dot w) is a dot product of a vector with a scalar. Dot products are just defined between vectors, so this expression is not well-defined. That is, this is a meaningless expression. Both of the cross products (u X v) and (w X r) are vectors, so (u X v) dot (w X r) is a vector perpendicular to both of these vectors. All these answers assume that none of the vectors is zero, and that none of the cross products are of parallel vectors. In those cases each meaningful calculation would be zero. All t &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I blanked on the dot product, which made it a mistake. I also completely forgot the definition of scalar, which made defining it difficult. ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: Find a number t such that the vectors -i - j, i - (1/2) j + (1/2)k and -2i -2j - 2tk all lie in the same plane. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I don’t really know how to solve this problem. There is nothing in the book about solving a problem of this sort. I think I have a basic idea as to what the solution may be though. If I find the cross product of the two given vectors, I will end up with a vector that is orthogonal to them. The angle at which the orthogonal vector lies should ensure that the two vectors lie on the same plane. However, I am unsure how to get the number t after finding that. confidence rating #$&*:232; 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Any two of these vectors define the orientation of a plane. The direction perpendicular to that plane is perpendicular to all vectors in the plane. If the third vector is also in the plane, it will also be perpendicular to that direction. Assuming that none of the vectors are zero and that none are parallel to any of the others, we can pick any two of the vectors and find their cross product, which will be perpendicular to the plane. Then the third vector will be in the same plane, provided it is perpendicular to that cross product. If the vectors are u, v and w, then, our test would be any of the following: u dot (v X w ) = 0 v dot (u X w ) = 0 w dot (u X v ) = 0. If any of the vectors is zero, or if any two of the vectors are parallel, then the condition must hold, and you should justify. f &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Looks like I had the theory mostly correct . . . (smacks forehead) SCALAR triple product duh. I missed it because they used it to solve the area of a solid, but that makes sense. ------------------------------------------------ Self-critique rating: 3 " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!