#$&* course Mth 277 9/20/2011 @ 9:47 p.m. If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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Given Solution: The standard form A x + B y + C z + D = 0 is easily found by applying the distributive law: We get 3 x - 6 - 2 y + 2 - 3 z + 15 = 0 which we simplify to get 3 x - 2 y - 3 z + 11 = 0. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: Find the equation of the plane containing the point P(-1,3,2) and having normal vector N = 3j - 1k. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: [(x+1)i + (y-3)j + (z - 2)k] dot (0i + 3j - 1k) = 0…if the dot product of two vectors equals 0 they are perpendicular which means that the starting point of vector P is contained within the plan that has the above normal vector. This dot product equals 3(y - 3)j - (z - 2)k = 0 and if we plug in our values for y and z we clearly see that this is true. confidence rating #$&*:y ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ------------------------------------------------ Self-critique rating: ********************************************* Question: Find two unit vectors perpendicular to the plane x + 3y - 4z = 2. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We first us this plane and put it in the form of a vector giving us 1i + 3j - 4k…I’ll call it vector v. ||v|| = sqrt.(1^2 + 3^2 + 4^2) = sqrt.(26) dividing the vector by the mag. Gives us the unit vector in the same direction: sqrt.26/26i + 3sqrt.26/26 j - 4sqrt.26/26k. I’m not totally sure if this thinking is accurate, but if we extended this unit vector through the plane it would also be perpendicular. (which seems like it would have sign changes) confidence rating #$&*:rly ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: A vector perpendicular to the plane is `i + 3 `j - 4 `k. A unit vector in this direction is `i sqrt(26) / 26 + 3 j sqrt(26) / 26 - 4 k sqrt(26) / 26 = `i sqrt(26) / 26 + 3 j sqrt(26) / 26 - 2 k sqrt(26) / 13. Another vector perpendicular to the plane is the negative of the preceding. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I realized the perpendicular thing, but I wasn’t 100% sure if all signs would change or just in the k or something else entirely. But, this does make sense. ------------------------------------------------ Self-critique rating: ********************************************* Question: Find the distance between the point (-1,2,1) and the plane which contains the point (3,3,-2) and is normal to the vector N = -2i + j + 3k. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ------------------------------------------------ Self-critique rating:
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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): our numbers vary quite a bit, but that’s it. ------------------------------------------------ Self-critique rating: ********************************************* 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: I think if we dot the vector -2i + 7j + k with the vector -I -2j + 7k (found via the distance from the vector I + 4j -2k and a random point on the plane (2, 2, 5) we will get the distance from the center to the plane, a.k.a. the radius. This gives us a radius of 5 which means that the equation is (x+2)^2 + (y - 7)^2 + (z - 1)^2 = 25. confidence rating #$&*: too confident. ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): My general thinking seems sort of in line with the solution, but I don’t know that I dotted the vector from the center with the correct vector.