Solutions8

# Solutions8 - Math 601 Solutions to Homework 8 1 Find the...

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Math 601 Solutions to Homework 8 1. Find the distance between the following two lines: x y z = 0 1 2 + t 1 1 0 x y z = 1 - 1 3 + t 1 1 0 Answer: First, note that the two lines are parallel, since both lines are in the direction given by the vector a = 1 1 0 . A point on the first line is P 0 = (0 , 1 , 2) and a point on the second line P 1 = (1 , - 1 , 3). Thus, the vector --→ P 0 P 1 = 1 - 2 1 goes from one line to the other line. We can project this vector onto one of the lines, as in the following picture: T T We find the vector projection of --→ P 0 P 1 onto the vector a : ˆ --→ P 0 P 1 · a k a k ! a k a k = - 1 / 2 - 1 / 2 0 Then, if we subtract this result from --→ P 0 P 1 , we will get a vector perpen- dicular to the two lines, which goes between the two lines. The length of this vector will be distance between the two lines. 1 - 2 1 - - 1 / 2 - 1 / 2 0 = 3 / 2 - 3 / 2 1 The length of this vector is r 9 4 + 9 4 + 1 = 22 2 . Thus, the distance between the two lines is 22 2

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2. Consider the four points P 1 = (5 , 2 , 0), P 2 = (1 , 5 , - 1), P 3 = (1 , 7 , 3), and P 4 = (4 , 6 , 3) in R 3 . (a) Find the equation for the plane consisting of all points equidistant from P 1 and P 2 . (b) Do the same for P 1 and P 3 and for P 1 and P 4 . (c) Suppose that all four points lie on the surface of a sphere S . De- termine the center point of S . Answer: (a) The plane equidistant to the two points is the plane that lies di- rectly in between the two points. This plane will be perpendicular to the line segment between the two points, and it will contain the midpoint of that line segment. Thus, the vector --→ P 1 P 2 = - 4 3 - 1 will be perpendicular to the plane. So, the equation for the plane will be of the form: - 4 x + 3 y - z = D We just need to solve for D . The midpoint of the line segment from P 1 to
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