#$&* course MTH 277 7/13/2012 335AM 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 standard form A x + B y + C z + D = 0 is easily found by applying the distributive law: We get 3 x - 6 - 2 y + 2 - 3 z + 15 = 0 which we simplify to get 3 x - 2 y - 3 z + 11 = 0. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): My textbook says A x + B y + C z + D = 0 is general form and a(x-x_1) + b(y-y_1) + c(z-z_1) = 0 is standard form. ------------------------------------------------ Self-critique rating: 3
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Given Solution: (x, y, z) lies on the plane if and only if the vector (x + 1) `i + (y - 3) `j + (z - 2) `k, from P to (x, y, z), is perpendicular to N. This condition is ((x + 1) `i + (y - 3) `j + (z - 2) `k ) dot (3 `j - `k) = 0 giving us 3 ( y - 3 ) - (z - 2) = 0, which simplifies to 3 y - z - 7 = 0. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: ********************************************* Question: Find two unit vectors perpendicular to the plane x + 3y - 4z = 2. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Normal Vector N= <1, 3, -4> || N || = sqrt(26) unit vector = 1/sqrt(26) < 1, 3, -4> =
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Given Solution: A vector perpendicular to the plane is `i + 3 `j - 4 `k. A unit vector in this direction is `i sqrt(26) / 26 + 3 j sqrt(26) / 26 - 4 k sqrt(26) / 26 = `i sqrt(26) / 26 + 3 j sqrt(26) / 26 - 2 k sqrt(26) / 13. Another vector perpendicular to the plane is the negative of the preceding. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: ********************************************* Question: Find the distance between the point (-1,2,1) and the plane which contains the point (3,3,-2) and is normal to the vector N = -2i + j + 3k. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: P(-1,2,1) N = <-2,1,3> point on plane (3,3,-2) plane: a(x-x_1) + b(y-y_1) + c(z-z_1) = 0 = -2(x-3) + (y-3) + 3(z+2) = 0 = -2x + 6 +y -3 + 3z +6 = 0 = -2x + y + 3z +9 find another point on plane by setting x and y to 0. z= -3 Q(0,0,-3) Distance = ||proj_nPQ|| = |PQ dot N| / ||N|| = <-1,2,4> dot <-2,1,3>/ sqrt(4+1+9) = 16/sqrt(14) = 8sqrt(14)/7 = approximately 4.28 units which is realistic confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: A vector from the first point to the second is `u = (3 - (-1) ) `i + (3 - 2) `j + (-2 - 1) `k = 4 `i + `j - 3 `k. The component of this vector perpendicular to the plane is found by projecting `u onto the normal vector. The magnitude of the projection is ( `u dot `N / || `N || ) = -16 / sqrt(14), which can easily be simplified and approximated. This is the distance between the first point and the plane. Note on vector projection: We don't need it here, but the vector projection of `u onto `N is ( `u dot `N / || `N || ) * `N / || `N || = (-16 / sqrt(2^2 + 1^2 + 3^2) ) * (-2i + j + 3k) / sqrt(2^2 + 1^2 + 3^2) = -16 / 14 * (-2i + j + 3k). The magnitude of this vector is the requested distance. Note that ( `u dot `N / || `N || ) is the magnitude of the projection of `u onto `N. This is multiplied by the unit vector `N / || `N || to get a vector of the appropriate magnitude in the direction of `N. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): -16 / 14 * (-2i + j + 3k) The magnitude of the vector is approximately 4.28 units ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: Find the distance between the lines (x+1)/(-2) = (y+2) / (-2) = (z+1)/(-1) and (x-4)/5 = (y+1)/2 = (z-1)/3 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (x+1)/(-2) = (y+2) / (-2) = (z+1)/(-1) v = <-2,-2,-1> and P(-1,-2,-1) (x-4)/5 = (y+1)/2 = (z-1)/3 w = <5,2,3> and Q(4,-1,1) find a vector perpendicular to these two lines N = v X w = -4'i + 'j -14'k
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Given Solution: The lines are in the directions of the respective vectors `u = -2 `i - 2 `j - `k and `v = 5 `i + 2 `j + 3 `k. The distance between the lines is measured perpendicular to both lines, in the direction of `u X `v = -4 `i + `j + 6 `k. Any vector from a point of one line to a point of the other will project onto this vector in such a way that the magnitude of the projection is equal to the distance between the lines. The point (-1, -2, -1) is on the first line, and the point (4, -1, 1) is on the second. A vector from the first to the second is therefore `w = 5 `i - `j - 2 `k The magnitude of the projection of this vector onto `u X `v is `w dot (`u X `v) / || `u X `v || = (5 * -4 + -1 * 1 + (-2) * 6 ) / sqrt(4^2 + 1^2 + 6^2) = -33 / sqrt(53), which can be put into standard form and approximted (approximate value is between 4 and 5, so the lines are between 4 and 5 units apart). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I keep getting 3.22, but thats becuase I am projecting N onto PQ. N is the line perpendicular to both lines and by finding the projection of N onto PQ(any line connecting the two lines) you find the distnance of N. The given solution says to use the formula `w dot (`u X `v) / || `u X `v ||, but isn't that the volume formula???????? ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: Find the equation of the sphere with center C(-2,7,1) and tangent the the plane x + 4y - 2z = 10. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: C(-2,7,1) x + 47 - 2z =10 N = <1,4,-2> x^2 +y^2 +z^2 = r^2 Find any point, B, on the sphere x and y = 0, therefore B(0,0, PQ = (-2, 7, 1) dot (0, 0, -5) = <-2, 7, 6> confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The sphere has equation (x - (-2)) ^ 2 + (y - 7) ^2 + (z - 1)^2 = r^2, where r is its presently unknown radius. The sphere is tangent to the plane, which by the geometry of circles and spheres implies that a vector from the center of the sphere to the point of tangency is perpendicular to the plane. It follows that the magnitude of that vector is equal to the distance from the point to the plane. So to find r we need only find the distance from (-2, 7, 1) to the plane x + 4 y - 2 z = 10. We do this by finding some point, any point, on the plane, and projecting the vector from (-2, 7, 1) to that point onto the vector `i + 4 `j - 2 `k which is normal to the plane. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating:OK " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!