Auroux - prac1B solutions

Auroux - prac1B solutions - Math 53 Practice Midterm 1 B...

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Unformatted text preview: Math 53 Practice Midterm 1 B Solutions Problem 1. a) P = (1 , , 0), Q = (0 , 2 , 0) and R = (0 , , 3). Therefore QP = 2 and QR = 2 + 3 k. b) cos = QP QR vextendsingle vextendsingle vextendsingle QP vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle QP vextendsingle vextendsingle vextendsingle = ( 1 , 2 , ) ( , 2 , 3 ) 1 2 + 2 2 2 2 + 3 2 = 4 65 Problem 2. a) PQ = ( 1 , 2 , ) , PR = ( 1 , , 3 ) . So PQ PR = vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle k 1 2 1 3 vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle = 6 + 3 + 2 k . Then area () = 1 2 vextendsingle vextendsingle vextendsingle PQ PR vextendsingle vextendsingle vextendsingle = 1 2 radicalbig 6 2 + 3 2 + 2 2 = 1 2 49 = 7 2 . b) A normal to the plane is given by N = PQ PR = ( 6 , 3 , 2 ) . Hence the equation has the form 6 x + 3 y + 2 z = d . Since P is on the plane, d = 6 1 + 3 1 + 2 1 = 11. So the equation of the plane is 6...
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Auroux - prac1B solutions - Math 53 Practice Midterm 1 B...

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