# emt drill solution - D1.1 Given points M-1,2,1,N(3-3,0 and...

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D1.1 Given points M(-1,2,1),N(3,-3,0) and P(-2,-3,-4),find (a) R MN (b) R MN + R MP (c) |r M | (d) a MP (e) |2 rP -3 rN |. SOLUTION: M M M P 2 6 MP 2
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r P = -2ax 3ay-4az 2r P = -4ax-6ay-az r N = 3ax -3ay+0az 3r N = 9ax-9ay+0az 2 rP -3 rN = (-4ax-6ay-az) (9ax-9ay+0az) = -13ax+3ay-8az ______________ |2 rP -3 rN |= (-13) 2 + (3) 2 +(8) 2 = 242 |2 rP -3 rN | = 15.56 ANSWER D1.2. A vector field S is expressed in rectangular coordinates as S = {125/[(x 1)2+(y 2)2+(z+1)2]}{(x 1)ax +(y 2)ay +(z+1)az}. (a) EvaluateS at P(2, 4, 3). (b) Determine a unit vector that gives the direction of S at P. SOLUTION: S 2 AC; R R
(c) an angle θBAC at vertex A; (d) the vector projection of AB on AC. SOLUTION: A A
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(c) (d) ANSWER D1.4. The three vertices of a triangle are located at A (6 , 1 , 2), B ( 2 , 3 , 4),and C ( 3 , 1 , 5). Find: ( a ) R AB × R AC ; ( b ) the area of the triangle; ( c ) a unit vector perpendicular to the plane in which the triangle is located. A A A C 9 2 3 x y z   a a a cos AB AC AB AC BAC R R R R cos AB AC BAC AB AC R R R R 2 2 2 2 2 2 ( 8 4 6 ) ( 9 2 3 ) ( 8) (4) ( 6) ( 9) (2) (3) x y z x y z     a a a a a a 62 116 94 1 cos (0.594) BAC 53.56 0.594 on AB AC AB AC AC R R R a a 2 2 2 2 2 2 ( 9 2 3 ) ( 9 2 3 ) ( 8 4 6 ) ( 9) (2) (3) ( 9) (2) (3) x y z x y z x y z a a a a a a a a a ( 9 2 3 ) 62 94 94 x y z a a a 5.963 1.319 1.979 x y z   a a a
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Professor Aatif Saeed Preston University Islamabad Page 4 R AC =r C - r A R AC = (-3,1,5)-(6,-1,2) A = (-3-6),(1+1),(5-2) R Ac r A = -9,2,3 = -9ax+2ay+3az C r C B R AB R AC = ax ay az -8 4 -6 -9 2 3 = ax(4x3 -2(-6))-ay(-8x3 (-6)(-9))+az(2(-3)-4(-9)) =ax(12+12)-ay(-24-54)+az(-16+36) = 24ax -78ay+20az A (b) Area =1/2│ R AB × R AC│ = 1/2│24ax - 78ay+20az│ =1/2 (24) 2 + (-78) 2 + (20) 2 B C =1/2 576+6084+400 =1/2 7060 =1/2(84.0) = 42.0 (c)