Solutions to Assignment #01. (a) Use tail-to-head addition and the Parallelogram Law to nd the resultant, (a + b) = (b + a) :

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1 Solutions to Assignment # Puhalskii/Kawai Section 8. (I) Demonstrate the vector sums/di erences on separate aes on the graph paper side of our engineering pad. Let a = h; i = i j and b = h; i = i + j: (a) Use tail-to-head addition and the Parallelogram Law to nd the resultant, (a + b) = (b + a) : We place a and b tail-to-tail at rst. To form (a + b) ; we translate b so that its tail coincides with a s head. Similarl, we form (b + a) b translating the tail of a onto the head of b: We connect the tail of the rst vector to the head of the last vector to form the resultant h; i + h; i = h; i = i + j: - - (b) If a and b are tail-to-tail, then show the di erence vector (a b) : Connect the heads. The vector (a b) starts at the head of b and runs to the head of a:

2 (c) Geometricall form the sum (a + b) ; and then algebraicall evaluate ka + bk : We need two copies of a and one cop of b: Use tail-to-head addition h; i + h; i = h; 6i + h; i = h; i = i j: - - kh; ik = + ( ) = p 6: (d) Find the unit vector which has the same direction as v = a + b: Divide v b its magnitude. u = p 6 h; i : (II) Eamine the two pairs of vectors below. ONE pair of vectors is parallel. Which pair? EXPLAIN wh one pair is parallel and the other is not. First pair: u = h6; ; i and v = h; ; i Second pair: u = h 9; ; i and v = h ; ; i : The vectors in the SECOND pair are PARALLEL. The vector u is a scalar multiple of v : h 9; ; i = h ; ; i ) u = v :X In the rst pair, we have 6 = ; but ( ) = : The scalar multiplier is not the same for all three components. (III) Let A be the initial point and B be the terminal point of vector v = AB: Find the vector which has the same direction as v; but has magnitude. The two points are A (; 7) and B (; ) : We have AB = h; i and the unit vector is u = p h; i : Now stretch it out to the correct length b multipling b the scalar. [I just assumed that it was a force vector.] F = p h; i = p h; i = p Dp p E h; i = ; :

3 (IV) Complete Problem #9 on p. 69. The weight is a downward force (negative -ais component). w = h; 7i : Now add the other two forces. The resultant is the net unbalanced force on the crate. h 6; i + h77; 77i + h; 7i = h; 7i pounds. The magnitude is p + 7 = p 8 : = : pounds. The unit direction vector is p 8 h; 7i : (V) Complete Problem #9 on p. 69. [Same warning given as in the previous problem] If we diagram this on the standard aes, then let us assume that the banks of the river run verticall (north and south) and that the left bank is represented b the -ais. The still water velocit will be a vector which points to the right, and so I have designated that as the still water vector, s = h; i ft/sec. If we assume the river current runs from north to south (downward), then we have c = h; i ft/sec. Their vector sum is the actual velocit: v = s + c = h; i So in m set-up, the resultant velocit vector points into Quadrant IV The component vectors form the right triangle as seen here. The angle with respect to the -ais (the bank) is tan (=) = : = 76 : -

4 Section 8. (VI) We have two points in R : P ( ; ; ) and Q ( ; ; ) : (a) Find the distance between the points P and Q: This is the etended Pthagorean Theorem. ( ( )) + ( ( )) + ( ) = p + + = : (b) Find the component form of the vector P Q and its engineering form. Subtract the coordinates. [Terminal minus initial...] P Q = h ( ) ; ( ) ; i = h; ; i : (c) Find the unit vector which points in the opposite direction of P Q: Part (a) tells us that the magnitude of P Q is, so the unit vector in the same direction as P Q is h; ; i = ; ; ; but we want the vector in the opposite direction. Multipl b ( ) : h; ; i = ; ; :X (VII) This is similar to Problem #8 in this section. Write the given vector as the PRODUCT of its magnitude and its associated unit direction vector. v = h ; ; 6i : We have the magnitude: kh ; ; 6ik = ( ) + + ( 6) = 7: The unit vector is and thus, we can rewrite v as h ; ; 6i = 7 7 ; 7 ; 6 7 v = 7 7 ; 7 ; 6 :X 7 (VIII) This is similar to Problem # in this section. Identif the geometric shape described b the euation: + + z + z = : THIS IS A SPHERE Rewrite this euation in standard form [Complete the suare] If the coe cient of the suared term is one, then we simpl take one-half of the coe cient of the rst degree term and then suare that. + = :

5 Thus, in the rst case, we add to both sides. + + The center is located at C ( ) + to both side, and then, in the second case, we add + z + z + ; = + + z + = = : ; and its radius is R = : + (IX) Complete Problem # on p. 7. If the points are colinear (on the same line), then the vectors P Q and QR must be PARALLEL (scalar multiples of each other). P Q = h ; ; i = h ; ; i QR = h ; ; i = h; ; i : Clearl, the scalar multiplier is di erent for each component, so the are not parallel, and thus, the three points do NOT lie on the same line in D space. (X) Complete Problem # on p. 7. The weight vector points downward onl in the negative z direction. The net force on the crate is the resultant. w = h; ; i : F = h; ; i + h ; 8; 6i + h; ; i = h ; ; i : This is the net unbalanced force on the crate. Its magnitude is The unit direction vector is ( ) + + ( ) = p : = 9 pounds. p h ; ; i = p h ; ; i ; so the force vector points in this direction. [The answer in the back of the book was incorrect.] So we can sa that the general direction is h ; ; i ; since it has convenient integers as components.