#$&* course mth 277 12/7 11 If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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Given Solution: The result is just the vector k: v X w = ( sin(theta) i + cos(theta) j ) X (-cos(theta) i + sin(theta) j ) = -sin(theta) cos(theta) i X i + sin(theta) sin(theta) i X j - cos(theta) cos(theta) j X i + cos(theta) sin(theta) j X j. i X i amd j X j are both zero, since sin(theta) = 0 for both of these computations. i X j = k by the right-hand rule, and likewise j X i . = -k, so the product is sin(theta) sin(theta) k - cos(theta) cos(theta) (-k) = sin^2(theta) i + cos^2(theta) k = (sin^2(theta) + cos^2(theta) ) * k = k &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):Okay ------------------------------------------------ Self-critique rating:3 ********************************************* Question: Find sin(theta) where theta is the angle between v = -i + j and w = -i + j + 2k. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: ||vXw||=||v||*||w||*sin(theta) Sin(theta)= ||vXw||/(||v||*||w||) Sin(theta)=||2j+2i||/(sqrt(2)*sqrt(6)) Sin(theta)= 2*sqrt(2)/(sqrt(2)*sqrt(6)) (theta)=sin^-1(2sqrt(6)/6-sqrt(6)/3) Theta=sin^-1(0)=0 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): I think I went too far but I’m unsure. ------------------------------------------------ Self-critique rating: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):okay ------------------------------------------------ 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=<-1,1,-1> ,PR=<1,1,2> im going to use PQ as the base of the triangle (1/2) ||PQ X PR||= <-1,1,-1>X<1,1,2> =(1/2)( 3i+j) im unsure if I need to take the magnitude then multiply it by (1/2) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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):I could use a little help im unsure what to do after I crossed the vectors. ------------------------------------------------ Self-critique rating:3
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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):the second one was the only one I was unsure about your final answer. ------------------------------------------------ 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. f &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: