Math Solution 1 - Homework due 26th January Solutions to...

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Homework due 26th January Solutions to compulsory questions Problem ( § 1.3, Exercise 24). Find an equation for the plane that passes through the point (2 , - 1 , 3) and is perpendicular to the line v = (1 , - 2 , 2) + t (3 , - 2 , 4). Solution. The plane has a normal vector 3 i - 2 j + 4 k , and so has equation given by ( ( x i + y j + z k ) - (2 i - j + 3 k ) ) · (3 i - 2 j + 4 k ) = 0 . Equivalently, 3 x - 2 y + 4 z = 20 . Problem ( § 1.3 Exercise 32). Given vectors a and b , do the equations x × a = b and x · a = k a k determine a unique vector x ? Argue geometrically and analytically. There are quite a few ways to approach this problem. Here are a few. Solution 1. Firstly, if a = 0 then any vector x satisfies the equations, and so the solution is not unique in this case. Suppose now that a 6 = 0 . If there are two solutions, x and y say, then x · a = k a k y · a = k a k and x × a = b y × a = b . Subtracting gives ( x - y ) · a = 0 and ( x - y ) × a = 0 , and so x - y is both perpendicular and parallel to a . Therefore x - y = 0 , and so x = y . Therefore the solution, if it exists, is unique provided a 6 = 0 . (The question did not actually ask you to prove the existence of a solution). Solution 2. A more geometric argument proceeds by determining the length and direction of x . If
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Math Solution 1 - Homework due 26th January Solutions to...

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