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exam_1-version_a-solutions

# exam_1-version_a-solutions - Ma 231 Exam I Version A...

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Unformatted text preview: Ma 231 Exam I Version A Solutions (l) (20 points) (a) Let a=(—1,—2,2) and b=(—6,3,2). (i) —-3a + 2b = Answer: —-3a + 2b = -3(—1, —2, 2) + 2(—6, 3, 2) = (—9,12,-2) (ii) What is the unit vector in the same direction as a? Answer- l=.<:l’_‘_2’_22= <4 _2 g> ' la! J1+4+4 3’ 3’3 (iii) What is the scalar projection of b on a, comp,(b)? Answer: comp_ (b) = — a-b =6-6+4=5 lal 3 3 (iv) What is the vector projection of b on a, proj,(b)? Answer: proj_ (b) = (comp_ (b))1 = <—§, "3,3 (1)) Draw the vector projection of w on v : Answer: W i v l i (2) (20 points) (a) For the line through the points (2, —1, 3) and (—2, 2, 5), ﬁnd the vector, parametric and symmetric equations. Solution: Direction: D = (-2,2,5)— (2,—1,3)=(—4,3,2) Point: (2,—1,3) Vector parametric: r(t) = (2, -1,3) + t {—4, 3, 2) = (2 - 4t,—l + 3t, 3 + 2t) Parametric: x=2—4t, y =-1+3t, z = 3+2! (b) Check that the lines intersect, ﬁnd the point of intersection and the cosine of the angle between the lines: x—5_y+1_z—7 3 2 —1 . x—2_y+8_z+6 L" —1 — 1 5 Solution: Parametric Equations: Ll: x=3t+5, y=2t—1,z=—t+7; L2: x=—s+2,y=s—8, z=5s—6 3t+5=—s+2 Solve: 2t—1=s—8 —t+7=5s-6 Adding ﬁrst two, 5t + 4 = —6, t = —2, s = 3; Check third: - (—2) + 7 = 9, 5(3) - 6 = 9 Point of intersection: (—1, —5,9) Directions: Dl = (3, 2, —l), D2 = {—1,1, 5), Cosine of angle of intersection: cos(0)=D"D2=__3_:_2_"_51= 6 =_?_ IDIHDzI Jib—7 J1_4J2_7 42 (3) (20 points) Let R :(—1,1, 2), P2 :(1,—2,1), P3 :(2,2,4) bethreepoints in space. (:1) Find the equation of the plane through PUP2 and P3. Solution: RPZ=(2,-3,—l) and If}; =(3,1,2) i j k N=EEXR13=2 -3 —1 3 l 2 ,-3—1.2—1 2—3 ,. = —] +k =—51-7]+llk l 2 3 2 3 l Check: (2, —3, -1) . (—5, —7, 1 1)=o, (3,1,2) . (-5,—7,1 1) = 0 Equation of plane: —5(x+l)—7(y-l)+ll(z—2)=O or—Sx—7y+llz=20 (b) What is the area of the parallelogram with sides RE and EH? Solution: Area =|EP§><EE|=|—5i—7j+11k|=J25+49+12 =J195 (4) (15 points) For the quadratic surface x = y2 + 422 : (:1) Sketch the y = O and z = 0 slices: 7-0 z (b) Sketch the x = constant slices: (c) Sketch the surface : (5) (10 points) (a) Give a vector parametric representation for the curve formed by the intersection of the surfacex=y2 —1 and theplane 3x—2y+z = 4. Solution: y=t, x=t2 —1, x=4—3x+2y=7--3t2 +2! r(t) = (t2 —l,t,7-3t2 +2t> (b) Write the arclength of the curve x = cos(21), y = —3 sin(2t), 0 S t S It as an integral. (Do M evaluate the integral). Solution: r(t) = (005(21), —3sin(21)), r'(t) = (—Zsin(2t), — 6cos(2t)) f f Arclength = ﬂr'a) |dt = I‘/4sin2(21)+ 36cos2 (2t) dr 0 0 (6) (15 points) (a) An object in space has a velocity of v(t) = —4cos(t)i + 4sin(t) j + 31’]; and the initial position r(0) = —2i + j + 5k, ﬁnd the position, r(t), at time 1. Solution: r(t) = j v(t) dt = (—4sin(t) + C1 )i + (—4 cos(t) + 02 ) j + (t3 + C3)k —2i +j+ 5k = r(0) = Cli +(—4+C2)j+C3k :> C1: —2,C2 = 5,C3 = 5 r(t) = (—4sin(t) -— 2)i + (—4 cos(t) + 5) j + (t3 + 5)k (b) Find the tangential and normal components of the acceleration of the object in (a). Solution: v(t) = ~4cos(t)i + 4sin(t)j + 3t2k , v(t) = |v(t)| = J16 + 9:4 11(1) = 4sin(t)i + 4cos(t) j + 6tk 18t3 J16+9r4 2 2 2 324:6 256+ 720:4 “Hi/‘4 “r: (’6 36”‘16+9,+= W l 4 ""2 a a =v't =— 16+9t 36f = r () 2( ) ( ) + ...
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exam_1-version_a-solutions - Ma 231 Exam I Version A...

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