#$&* course Mth 277 query_09_6*********************************************
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Given Solution: The standard form A x + B y + C z + D = 0 is easily found by applying the distributive law: We get 3 x - 6 - 2 y + 2 - 3 z + 15 = 0 which we simplify to get 3 x - 2 y - 3 z + 11 = 0. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK ********************************************* Question: Find the equation of the plane containing the point P(-1,3,2) and having normal vector N = 3j - 1k. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: N=<0,3,-1> the vector containing the point is v=(x+1)i+(y-3)j+(z-2)k To find the equation of the plane we would do the dot product. N dot v=0(x+1)+3(y-3)-1(z-2)=0 N dot v=3y-9-z+2=3y-z-7=0 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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:OK ********************************************* Question: Find two unit vectors perpendicular to the plane x + 3y - 4z = 2. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Since we know the direction of the vectorr <1,3,-4> We can easily find the unit vector by dividing by its magnitude. <1/sqrt(26),3/sqrt(26),-4/sqrt(26)> The other vector is found by negating the vector found. <-1/sqrt(26),-3/sqrt(26),4/sqrt(26)> confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: A vector perpendicular to the plane is `i + 3 `j - 4 `k. A unit vector in this direction is `i sqrt(26) / 26 + 3 j sqrt(26) / 26 - 4 k sqrt(26) / 26 = `i sqrt(26) / 26 + 3 j sqrt(26) / 26 - 2 k sqrt(26) / 13. Another vector perpendicular to the plane is the negative of the preceding. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK ********************************************* Question: Find the distance between the point (-1,2,1) and the plane which contains the point (3,3,-2) and is normal to the vector N = -2i + j + 3k. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To find the distance, we need to find the projection of u onto N and take the magnitude of the vector. u=(3-(-1))i+(3-2)j+(-2-1)k=4i+j-3k u dot N/(magn(N))= (-8+1-9)/(sqrt(14))=-16/sqrt(14) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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):OK ------------------------------------------------ Self-critique rating:OK ********************************************* 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: The equation can easily be found with the center point given. (x+2)^2+(y-7)^2+(z-1)^2=r^2 I am unsure if this is right, but I am going to use the points of the plane and plug in to find r. (9)+(9)+(1)=r^2 sqrt(19)=r
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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):OK ------------------------------------------------ Self-critique rating:OK