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# finalexamsol - 18.02 Final Exam Solutions Problem 1 a Line...

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± ± ± ± 18.02 Final Exam Solutions Problem 1. a) Line L has direction vector v = 1 , 2 , 3 ± which lies in P . To get a point P 0 on L take t = 0 P 0 = (1 , 1 , 2) . −−→ P 0 Q = 1 , 1 , 2 ± − 1 , 1 , 2 ± = 2 , 0 , 0 ± also lies in P . A normal to P is = i (0) j ( 6) + k (4) = 0 , 6 , 4 ± . i j k P 0 Q = 1 2 3 2 n = v × 0 0 So, the equation of P is 0( x 1) + 6( y 1) + 4( z 2) = 0 or 6 y = 4 z = 14 or 3 y + 2 z = 7 . v = 14 1 , 2 , 3 ± 1 6 2 , 1 , 1 ± , 1 n = Component of n ± on v ± is 1 3 n v = b) n Q = 2 , 1 , 1 ± ⇒ v = 1 , 2 , 3 ± ⇒ 6 14 (2 + 2 3) = −√ 84 · · Problem 2. a) Direction vector for L : v = 1 , 2 , 0 ± . P 0 = (0 , 0 , 1) equation for L : r = x,y,z ± = 0 , 0 , 1 ± + t 1 , 2 , 0 ± or x = t, y = 2 t, z = 1 . b) n = normal vector for P = 1 , 2 , 0 ± since L ⊥ P . P 0 = (0 , 0 , 1) 1( x 0) + 2( y 0) + 0( z 1) or x + 2 y = 0. c) P on L P = ( t, 2 t, 1) for some t = 0 (part (a)) ² P = ( t, 2 t, 1) since then dist( P 0 ,P ) = dist( P 0 ) = | t | 5. Problem 3. 1 0 3 = 1 0 2 1 1 2 + 3 = 3 3 = 0 . 1 1 2 2 1 1 2 1 1 1 2 a) 1 1 1 b) get a non-zero solution take the cross-product of any two rows of A 2 ; for example i j k 1 0 3 = 3 , 5 , 1 ± 2 1 1

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3 This implies all solutions to A x = 0 are x = t 5 . 1 c) 1 0 0 A 1 1 A 1 = I 3 = 0 1 0 0 0 1 ⎞⎛ ∗ ∗ 1 0 3 ∗ ∗ p 5 ⎠⎝ 1 = 3 2 1 3 2 p 5 ∗ ∗ ⇒ − 8 2 p = 0 p = 4 . ∗ ∗ 1 1 1 ∗ ∗ Problem 4. t a) r ( t ) = sin(e t )e t , cos(e t )e t , e ± ⇒ | r ( t ) | = e t 1 + 1 = e t 2 r ( t ) 1 T ( t ) = r ( t ) = 2 sin(e t ) , cos(e t ) , 1 ± . | | b) 1 e t T ( t ) = 2 cos(e t ) , sin(e t ) , 0 ± = −√ 2 cos(e t ) , sin(e t ) , 0 ± .
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finalexamsol - 18.02 Final Exam Solutions Problem 1 a Line...

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