exam1sol - 18.02 Exam 1 Solutions Problem 1 a P =(1 0 0 Q...

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18.02 Exam 1 Solutions Problem 1. a) P = (1 , 0 , 0), Q = (0 , 2 , 0) and R = (0 , 0 , 3). Therefore −−→ = ˆ ı QR = j + 3 k ˆ . QP j and −−→ −−→ QR · −−→ QP 1 , 2 , 0 � · � 0 , 2 , 3 = 4 1 2 + 2 2 2 2 + 3 2 b) cos θ = 65 = QP −−→ QP −−→ Problem 2. a) −−→ PR = �− 1 , 0 , 3 . PQ = �− 1 , 2 , 0 , −→ −−→ PQ × −→ PR = ˆ ı ˆ j k ˆ 1 2 0 1 0 3 = ı + j + 2 k ˆ . 1 = 1 6 2 + 3 2 + 2 2 = 1 49 = 7 −−→ PQ × −→ PR Then area (Δ) = . 2 2 2 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. In conclusion the equation of the · · · plane is 6 x + 3 y + 2 z = 11 . c) The line is parallel to 2 1 , 2 2 , 0 3 = 1 , 0 , 3 .
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