#$&* course MTH 277 09/21/2011 11:08 amThere were a couple of questions I had regarding this particular assignment, I will probably ask about them during or after class query_09_6
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Given Solution: The standard form A x + B y + C z + D = 0 is easily found by applying the distributive law: We get 3 x - 6 - 2 y + 2 - 3 z + 15 = 0 which we simplify to get 3 x - 2 y - 3 z + 11 = 0. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK ********************************************* Question: Find the equation of the plane containing the point P(-1,3,2) and having normal vector N = 3j - 1k. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: QP = (x + 1)i + (y - 3)j + (z - 2)k QP = (x + 1)(0) + (y - 3)(3) + (z - 2)(-1) QP = 0 + 3y - 9 - z + 2 Plane QP = 3y - z - 7 confidence rating #$&*:232; 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): Forgot the = 0 part of the equation ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: Find two unit vectors perpendicular to the plane x + 3y - 4z = 2. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Perpendicular vector = orthogonal vector x + 3y - 4z + 2 = 0 I’m not sure what to do with the equation after this point confidence rating #$&*:232; 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 am totally lost. Where did the sqrt(26)/26 come from? I see where you got the number in the equation, but where the heck did the answers come from? ------------------------------------------------ Self-critique rating: 1
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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): I had the equation subtract j instead of adding it. I’ll need to double-check and see exactly why it needed to be that way. Otherwise, the problem looks good. ------------------------------------------------ Self-critique rating: 2
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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 didn’t know how to pull the vector from the given line. It looks like all I had to do was use the denominators? I’ll probably ask about that during or after class. ------------------------------------------------ Self-critique rating: 2
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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): Apparently a lot simpler than I thought it was going to be. Everything makes sense now that it’s laid out. I also thought that the equation had to be equal to one, not the radius squared. ------------------------------------------------ Self-critique rating: 2 " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!