CHEM 151B Midterm: Chem 151B UCSC Exam 2 2016
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You have 60 minutes to answer all four questions. Let x = (4; 4; 0), y = (1; 1; 1), and z = (1; 1; 0). Solutions: two directions in the plane are (0; 0; 1) and (3; 3; 0). So a normal to the plane is (1; (cid:0)1; 0). Hence an equation of the plane is: (1; (cid:0)1; 0) (cid:1) (u1 (cid:0) 1; u2 (cid:0) 1; u3) = 0 or u1 (cid:0) u2 = 0: you need a line through x with direction (1; (cid:0)1; 0). Explain why the same question for xtbx does not make sense. Solutions: a is symmetric, hence it is diagonalizable. B has only one distinct eigenvalue, so it is not diagonalizable (see below). 1 (cid:0)2(cid:19) = 0 that can be rewritten as (4 (cid:0) (cid:21))((cid:0)2 (cid:0: a(cid:146)s characteristic equation is det(cid:18)4 (cid:21)) (cid:0) 1 = 0; the solutions of this equation are 1 (cid:6) (1; (cid:0)3 + p10) and (3 (cid:0)
