#$&* course Mth 277 9/17 2 If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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Given Solution: The result is just the vector k: v X w = ( sin(theta) i + cos(theta) j ) X (-cos(theta) i + sin(theta) j ) = -sin(theta) cos(theta) i X i + sin(theta) sin(theta) i X j - cos(theta) cos(theta) j X i + cos(theta) sin(theta) j X j. i X i amd j X j are both zero, since sin(theta) = 0 for both of these computations. i X j = k by the right-hand rule, and likewise j X i . = -k, so the product is sin(theta) sin(theta) k - cos(theta) cos(theta) (-k) = sin^2(theta) i + cos^2(theta) k = (sin^2(theta) + cos^2(theta) ) * k = k &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: Find sin(theta) where theta is the angle between v = -i + j and w = -i + j + 2k. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: after looking at the initial equation, i now know enough to solve the problem. || v X w || = || v || || w || sin(theta) so v X w you get 2i + 2j + 0k. magnitude is sqrt(8) magnitude of v is sqrt(2) magnitude of w is sqrt(6) sqrt(8) = sqrt(2) * sqrt(6) sin(theta) arcsin(sqrt(8) / sqrt(2) * sqrt(6) = about 54.74 degrees confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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:3 ********************************************* Question: Find a unit vector which is orthogonal to both v = 2i - j and w = 2j - k. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: orthogonal meaning perpendicular meaning dot product is 0. v X w is orthogonal to both v and w. take cross product of v and w and get ( i + 2 j +4 k ) which is orthogonal 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 . confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: v X w is orthogonal to both v and w. v X w = i + 2 j +4 k A unit vector in this direction is (i + 2 j + 4 k ) / sqrt(1^2 + 2^2 + 4^2) = (i + 2 j + 4 k ) sqrt(21) / 21 . If we take the dot product of this vector with either of our original vectors we will get zero. You can verify that this vector is indeed a unit vector. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating:3 ********************************************* Question: Find the area of the triangle with vertices P(2,0,0), Q(1,1,-1), R(3,1,2). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: PQ is ( -1, 1, -1 ) with mag of sqrt(3) declare base QR is ( 2, 0, 3 ) with mag of sqrt(13) declare hypotenuse PR is ( 1, 1, 2 ) with mag of sqrt(6) declare adjacent stuck After getting the equation, i can further my steps, i did not know that equation and did not know that was true for finding area of triangle. An altitude from point R to the base then has magnitude || PR || sin(theta). Thus the area is || PQ X PR || . mag is sqrt(23) = 3sqrt(23)i/23 + sqrt(23)j/23 - 2sqrt(23)k/23 = area confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 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): ok I had to look at the equations that declare what area is for the triangle given the information, but its in my notes now and understand it and am easily able to calculate the area ------------------------------------------------ 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: (v X w) is a vector perpendicular to both v and w. so v X (vXw) is a vector perpendicular to (vXw) u dot (v dot w), a scalar because they can define where the dot product is 0 (u X v) dot (w X r) is a vecor perp to u and v, and the other is a vector perp to w and r. these two dot products will give a vecor parallel confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 see that the last one should have been perp instead of parallel. I will study up on these. it looks like they could get hairy ------------------------------------------------ Self-critique rating:3
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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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):Tricky, I looked for a long time in linear algebra book and the vector calc book looking for a logical explenation for the dot product of two cross products.. I remember parts of this from class but it wasn't fresh on my mind. ------------------------------------------------ Self-critique rating:3 " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!