#$&* course Mth 277 6/18 4 pm 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): ------------------------------------------------ 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: If the vector is normal to the point's plane, then the two vectors' dot products should equal 0. The first vector:
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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: A vector perpendicular to that plane would be i + 3j - 4k. The unit vector of that would be equal to sqrt(26)/26 i + 3*sqrt(26)/26 j - 4*sqrt(26)/26 k. Any unit vector parallel to this is also perpendicular to the original plane, so another example would be sqrt(26)/13 i + 3*sqrt(26)/13 j - 4*sqrt(26)/13 k [the above vector multiplied by 2]
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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): ------------------------------------------------ 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: The distance to the plane is equal to the length of the vector projection from the point to the plane onto the normal vector. The vector from the point to the plane PS: (3-(-1)) i + (3-2) j + (-2-1) k 4 i + j - 3 k The scalar projection of PS onto N: (PS dot N)/|N| = (-8 + 1 - 9)/sqrt((-2^2) + 1^2 + 3^2) = -16/sqrt(14) 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: ********************************************* 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: The lines can be represented with vectors: -2 i - 2 j - k and 5 i + 2 j + 3k The cross product of these vectors will give us a vector that is perpendicular to both. (Handy in this case.) <-2, -2, -1> X <5, 2, 3> | -2 -1 |i + | -2 -1 |j + | -2 -2 |k | 2 3 | | 5 3 | | 5 2 | (-6-(-2))i + (-6-(-5))j + (-4-(-10))k -4i - j + 6k We can use the two equations to give us two points on each line: (-1,-2,-1) and (4,-1,1) are points on each line. The vector between these two points is: <4-(-1), -1-(-2), 1-(-1)> <5, 1, 2> Now the distance between the two lines will be the scalar projection of that vector onto the cross product. = <5,1,2> dot <-4,-1,6> / |<-4,-1,6>| = (-20 - 1 + 12)/sqrt(16 + 1 + 36) = (-9)/sqrt(53) confidence 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): After looking over this problem, it seemed as though I did the cross product incorrectly: <-2, -2, -1> X <5, 2, 3> | -2 -1 |i + | -2 -1 |j + | -2 -2 |k | 2 3 | | 5 3 | | 5 2 | (-6-(-2))i + (-6-(-5))j + (-4-(-10))k -4i - j + 6k I've looked it over several times, but I can't see what I'm doing wrong. The j coefficient *should* be 1, but I keep getting -1. Can you point me in the right direction?
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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:" Self-critique (if necessary): ------------------------------------------------ Self-critique rating: Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!