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exam1sol

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

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± ² ³ ´ µ 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 ± , −→ × = 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 × Then area (Δ) = . 2 2 2 2 b) A normal to the plane is given by N = × 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 ± . = 6 6 = Since 0, the line is N · 1 , 0 , 3 ± parallel to the plane. Problem 3. a) AB OA = 10 t, 0 ± and = cos t, sin t ± , hence B OB = AB OA + = 10 t + cos t, sin t ± . O A The rear bumper is reached at time t = π and the position of B is (10 π 1 , 0). b) V = 10 sin t, cos t ± , thus 2 2 | V | 2 = (10 sin t ) 2 + cos t = 100 20sin t + sin 2 t + cos t = 101 t.

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

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