Solutions to Assignment #02 MATH u v p 59. p 72. h 3; 1; 2i h4; 2; 5i p 14. p 45. = cos 1 2 p!

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1 Solutions to Assignment #0 MATH 41 Kawai/Arangno/Vecharynski Section 1. (I) Comlete Exercises #1cd on searation to TWO decimal laces. So do NOT leave the nal answer as cos 1 (something) : (c) The dot roduct is You will need some technology to nd the angle of u v = h1; ; 7i h8; ; i = (1) (8) + ( ) ( ) + (7) ( ) = 0: Thus, the angle of searation is (d) Same. cos () = u v kuk kvk = = 0 ) = cos 1 (0) = = 90 : cos () = h ; 1; i h4; ; 5i = = 0 70 = 70 1 : The inverse cosine value is = cos 1! 70 : = 14:8 : 1 (II) Comlete Exercise #19ab on If you can t take a guess (conjecture) for art (b), then don t worry about it. Just calculate the two angles of interest to the NEAREST DEGREE. (a) Assuming that this is a unit cube, we have d = h1; 1; 1i and u = h1; 1; 0i : cos () = h1; 1; 1i h1; 1; 0i = 6 = 6 ) = cos 1 := 6 5 : (b) The vector u is in the xy-lane and we have u = h1; 1; 0i : Vector v runs from (1; 1; 0) to (0; 1; 1) ; so we have u = h 1; 0; 1i : The answer of searation here is h1; 1; 0i h 1; 0; 1i cos () = = ( 1) 1 ) = cos 1 = 10 : Why might you guess that this is true? If we connect the tail of u to the head of v; the three sides form an equilateral triangle. Each angle of the triangle is 60 ; and by translation, the angle of searation is the sulement. 1

2 (III) Comlete Exercise #5a on ALSO, nd orth b v = v (#5a) These are D vectors. h; 1; i h1; ; i roj b v = h1; ; i h1; ; i orth b v = h; 1; i = roj b v: Show that orth b v and roj b v are orthogonal! h1; ; i = 6 9 h1; ; i = 4 ; 7 ; 5 : : Is roj b v orthogonal to orth b v? YES. 4 ; 7 ; 5 = = 0:X Section 1.4 (IV) Comlete Exercise #10 on. 8. We need TWO UNIT vectors. I m only giving the result of the cross roduct here. h 7; ; 1i h; 0; 4i = h1; 0; 6i Since the comonents are rather large, we should factor out 6 rst: 6 h; 5; 1i Now we can nd the unit vectors associated with h; 5; 1i: 1 q h; 5; 1i = 1 h; 5; 1i : ( 1) 0 (V) Comlete Exercise #1 on. 8. I gave the determinant formulation for this cross roduct on the answer sheet. h1; 1; i h0; ; 1i = h 7; 1; i The magnitude of this vector is the area of the arallelogram. q kh 7; 1; ik = ( 7) + ( 1) + = 59: (VI) Comlete Exercise #7 on. 8. This is the torque roblem. The scalar moment is the magnitude of torque: P! Q F. Careful! Look at the vector which runs from the center of the bolt to the end of the wrench where F is alied. Since these are all D vectors, just tack on a zero k-comonent! In the diagram, we assume that the head of the bolt is located at the origin and that all this action is occuring in the xy-lane. Thus, the force vector is alied at the oint Q (0:; 0:0) meters.! The wrench vector is P Q = h0:; 0:0; 0i : We tacked on the extra zero comonent for k:

3 The force vector has magnitude 00 and all we need is the standard osition angle. Since it is 18 o the vertical in Quadrant I, the angle must be = = 7 : Thus, the comonent form of the force vector is F = h00 cos (7 ) ; 00 sin (7 ) ; 0i : Now we can erform the cross roduct. i j k 0: 0: cos (7 ) 00 sin (7 ) 0 = k 0: 0:0 00 cos (7 ) 00 sin (7 ) : The i and j comonents are both zero since there are two zeros under the k: Both of those minors are zero. We recall that if both vectors are in the xy-lane, then the cross roduct must be normal to the xy-lane. It will only have a k comonent. Thus, the torque vector must be = h0; 0; 40 sin (7 ) 6 cos (7 )i : = h0; 0; 6:i newton-meters. Additional question: In WHICH DIRECTION does the torque vector oint? The ositive z-axis oints OUT of the aer, which means, for a standard threading on the bolt, the bolt is LOOSENED and the rotation is counterclockwise by the Right-Hand Rule. Section 1.5 (VII) Comlete Exercise #b on. 88. Assume that the base oint is (1; 1; 1) : We only want the line SEGMENT, so it turns out that we want 0 t 1 for our arameter. (L 1 ): The direction vector oints straight down. We have v = h0; 0; 1i : x = (0) t + 1 y = (0) t + 1 z = ( 1) t + 1 (L ): The direction vector oints back into the lane of the age. We have v = h 1; 0; 0i : x = ( 1) t + 1 y = (0) t + 1 z = (0) t + 1 (L ): The direction vector oints to the left. We have v = h0; 1; 0i : x = (0) t + 1 y = ( 1) t + 1 z = (0) t + 1

4 (L 4 ): The direction vector oints back to the origin. We have v = h 1; 1; 1i : (VIII) Comlete Exercises #14 & #17 on. 89. x = ( 1) t + 1 y = ( 1) t + 1 z = ( 1) t + 1 (#14) We want arametric equations, not just the usual rectangular equation to describe the tangent line. In D, vectors can be regarded as having sloe: ha; bi, m = change in vertical change in horizontal = b a : So by Calculus I, we have y = x ) y 0 = x: The sloe of the tangent line at ( ; 4) is m = ( ) = 4: So I can choose the vector h1; 4i which has sloe m = 4 as my direction vector for the D line: (#17) If the line is arallel to x = (1) t y = ( 4) t + 4: x = () t + 1 y = ( 1) t + 4 z = () t + 6 then we can use the same direction vector v = h; 1; i : We want our line to ass through ( ; 0; 5): (IX) Comlete Exercises #8 & #9 on. 89. x = () t y = ( 1) t + 0 z = () t + 5: (#8) Show that the lines L 1 and L are skew. [No intersection, and NOT arallel.] For L ; we will robably want to change the name of the arameter to something else, like s: Here are the arametric equations for L : x = (8) s + y = ( ) s + 5 z = (1) s + 6 So comare the direction vectors for the two lines: h8; 8; 10i vs. h8; ; 1i : Clearly, these are not scalar multiles of each other (their rst comonents are identical!), so the vectors cannot be arallel. 4

5 Now we will show that there is no intersection. Set the x s equal to each other, etc. 8t + = 8s + 8t + 6 = s t = s + 6 If add the rst two equations together, we eliminate t: 8 = 5s + 8 ) s = 0 (and t = 1=8) : We attemt to substitute these values into the third equation and we see that the last equation does not balance. 10t = 10 (1=8) = 5 4 s + 6 = (0) + 6 = 6: Thus, we say that the system of equations is inconsistent, and there is no intersection. The two lines are skew.x (#9) Are the two lines arallel? Comare their direction vectors... h ; 1; 1i vs. h 4; ; i : Yes! We see that they are scalar multile of each other. Section 1.6 Thus, the two lines are arallel. (X) Comlete Exercises #4 & #11 on. 87. h 4; ; i = h ; 1; 1i :X Let P ( ; 1; 1) ; Q (0; ; ) ; and R (1; 0; 1) be the names of the three oints.! Please use the vectors P Q and P! R for the cross roduct calculation! We have! P Q = h; 1; i and P! R = h; 1; i : The cross roduct vector will be orthogonal to both vectors, and will serve as the normal vector to the lane. i j k 1 = i j + k 1 1 = i ( ( )) j ( 4 6) + k ( ) = h0; 10; 5i : 5

6 We can use any of the three onits as the base oint for the lane. standard form for the lane is If we use P; then the 0 (x ( )) + 10 (y 1) 5 (z 1) = 0 and this can be reduced to (y 1) 1 (z 1) = 0: Using Q or R will eventually give us the same form: 0 (x 0) + 10 (y ) 5 (z ) = 0 ) (y ) 1 (z ) = 0 0 (x 1) + 10 (y 0) 5 (z ( 1)) = 0 ) (y 0) 1 (z ( 1)) = 0: 6