Department of Physics, Korea University
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1 Name: Department: Notice +2 ( 1) points per correct (incorrect) answer. No penalty for an unanswered question. Fill the blank ( ) with (8) if the statement is correct (incorrect).!!!: corrections to an incorrect answer. Student ID #: Like the displacement vector, the vector AB is identified by the difference between the initial point A and the final point B. The starting point A is called the tail of a vector and the destination B is called the head of a vector. We put an arrow to the head B of the straight line that connects A and B. 5. ( ) : solution. Textbook: Walker, Halliday, Resnick, Principles of Physics, Tenth Edition, John Wiley & Sons (2014). 3-1 and Their Components 1. ( ) A vector is a physical quantity that has both magnitude and direction. The most fundamental vector is the displacement. For example, velocity, acceleration, force, linear momentum, torque, and angular momentum are all vectors. 2. ( ) A scalar is a physical quantity that has only the magnitude. For example, mass, length, time, temperature, energy are all scalar. 3. ( ) A vector is written in the following form: ~v or Two vectors a and b are equivalent, a = b, if one of them can be translated to be exactly overlapped onto the other. Here, the translation denotes moving a vector keeping both the magnitude and the direction. 6. ( ) v. We can either put an arrow on top of an italic letter or write the letter in bold italic. 4. ( ) We denote the vector space, the set of vectors, by V. Let a and b are two vectors that are not parallel or antiparallel. Suppose that the tails of the two vectors are at the same point. Let a0 and b0 that are translated from a and b so that their tails are placed at the heads of b and a, Page 1 of 9
2 respectively. Then a, b, a0, b0 are four sides of a parallelogram. In addition, a = a0, a0 a and (b and parallelogram. b0 ) b = b0. are opposite sides of a single If x = 0, then xa = 0 is the zero vector (null vector). If x > 0, then xa is parallel to a. If x < 0, then xa is antiparallel (opposite) to a. The magnitude of xa is 7. ( ) xa = x a, x R, a V. 13. ( ) Distributive law is effective for scalar multiplications: For all x, y R and a, b V (x + y)a = xa + ya, x(a + b) = xa + xb. 14. ( ) The null vector 0 is the additive identity. a + 0 = 0 + a = a, The sum, ~a + ~b or a + b, of two vectors ~a (a) and ~b (b) is defined as follows. 15. ( ) The vector a ( 1)a is the additive inverse of a. Translate b so that the tail of b meet the head of a. Connect the tail of a and the head of b. The tail of the resultant vector a + b is that of a. The head of the resultant vector a + b is that of b. a V. a + ( a) = ( a) + a = 0, a V. 3-2 Unit, Adding by Components 1. ( ) Any two vectors are coplanar. 2. ( ) 8. ( ) The three vectors a, b, and a + b is always coplanar, placed on a single plane. They always make a triangle. 9. ( ) The addition of two vectors is commutative: a + b = b + a. 10. ( ) The addition of three vectors is associative: (a + b) + c = a + (b + c). 11. ( ) The magnitude a of a vector a is the length of the vector. 12. ( ) A real number x ( R) can be multiplied to a vector a. The product xa is also a vector. The scalar product a b of two vectors a and b is defined by a b = a b cos (a, b), where (a, b) is the angle between a and b. Page 2 of 9
3 3. ( ) The unit vector a is defined by a = The magnitude of 1 a. a a is unity and a is parallel to a. 4. ( ) The Cartesian coordinate axes which is also called the rectangular coordinate system consists of three orthogonal coordinate axes x, y, and z. 5. ( ) If a is on the xy plane and the angle between a and i is θ, then x and y components of a vector a are defined, respectively, by ax = a i = a cos θ, ay = a j = a cos π2 θ = a sin θ. The Pythagoras theorem states that q a = a2x + a2y, cos2 θ + sin2 θ = 1, ay. tan θ = ax 8. ( ) The i, j, and k are the unit vectors along the x, y, and z axes, respectively. Because they are orthonormal (any two elements are orthogonal and any one is of length unity), we find that i i = 1, i j = 0, i k = 0, j i = 0, j j = 1, j k = 0, k i = 0, k j = 0, k k = ( ) The x, y, and z components of a vector a are defined, respectively, by ax = a i, ay = a j, az = a k. 7. ( ) By applying the Pythagoras theorem twice, we find that the magnitude of a vector a in three dimensions is q a = a2x + a2y + a2z, Page 3 of 9
4 where ax = a i, ay = a j, az = a k. 13. ( ) 9. ( ) If we make use of the multiplication table for the scalar product of Cartesian unit vectors, i i = 1, i j = 0, i k = 0, j i = 0, j j = 1, j k = 0, k i = 0, k j = 0, k k = 1, then we can find that the scalar product a b is a b = ax bx + ay by + az bz. 10. ( ) The scalar product of a and itself is also denoted by a a = a2 = a ( ) In three dimensions, a point x on a plane perpendicular to the unit vector n is (x a) n = 0, where a is a point on the plane. 12. ( ) Angles formed by drawing lines from the ends of the diameter of a circle to its circumference form a right angle. This theorem can be proved in a straightforward way if we make use of the scalar product. Let a, b, and c three vectors from the center to three points A, B, and C on a circle. a2 = b2 = c2 = r2, where r is the radius of the circle. Let AB be a diameter. Then b = a. The two chords can be expressed as the following vectors: AC = c a, BC = c b = c + a. Let a be a vector and n be an arbitrary constant unit vector. The a can be decomposed into two pieces a = ak + a, where ak is parallel to n and a is perpendicular to n. Then, we find that ak = (a n )n, a = a (a n )n. (1) (2) The scalar product of the two chord vectors is AC BC = (c a) (c + a) = c2 a2 = r2 r2 = 0. Thus the angle between the two chords is ( ) Page 4 of 9
5 on a circle of radius r on the xy plane whose center is at a = ax i + ay j. The vector x satisfies the following constraint equation, (x a)2 = r ( ) The unit vector both i and j is n that makes the angle 45 with 1 2 n = (i + j ). The unit vector, on the xy plane, perpendicular to n is 1 ± (i j ). 2 Let i 0 and j 0 be the vectors obtained by rotating i and j, respectively, by an angle θ counterclockwise on the xy plane. Then, i 0 = cos θi + sin θj, 0 j = sin θi + cos θj. 15. ( ) 17. ( ) Consider an arbitrary point, x = xi + y j, Page 5 of 9
6 Let a0 be the vector obtained by rotating a by an angle θ counterclockwise on the xy plane. Then, a0x = ax cos θ ay sin θ, a0y = ax sin θ + ay cos θ. 18. ( ) 1. ( ) The cross product (vector product) a b of two vectors a and b is defined by a b = a b n sin (a, b), where (a, b) is the angle between a and b. n is the unit vector normal to the plane spanned by a and b. There are two normal directions. The direction of n is chosen according to the right-handed-screw rule: (a) Sweep from a to b with the fingers of your right-hand. Your outstretched thumb indicates the direction of n. 2. ( ) The components of a constant vector a are given by ax = a i, ay = a j. We keep the vector a invariant and rotate the frame of reference with the new Cartesian basis vectors i 0 and j 0 that are obtained by rotating i and j, respectively, by an angle θ counterclockwise on the xy plane: i 0 = cos θi + sin θj, 0 j = sin θi + cos θj. Then, the components of the same vector in terms of the new coordinate system are given by a0x = a i 0 = ax cos θ + ay sin θ, a0y = a j 0 = ax sin θ + ay cos θ. 3-3 Multiplying The cross product is anticommutative: b a = a b. 3. ( ) The cross product vanishes if a and b are collinear. 4. ( ) The i, j, and k are the unit vectors along the x, y, and z axes, respectively. Because they are orthonormal (any two elements are orthogonal and any one is of length unity), we find that i i = 0, i j = k, i k = j, j i = k, j j = 0, j k = i, k i = j, k j = i, k k = ( ) Page 6 of 9
7 Let a and b be two sides of a triangle. Then the area of the triangle is 1 S = a b ( ) Let a, b, and c are three adjacent sides (edges) of a parallelepiped. Then the volume of the parallelepiped is V = a (b c). 9. ( ) By making use of the identity sin2 θ = 1 cos2 θ, we find that (a b)2 = a b 2 = a2 b2 (a b) Consider a triangle ABC. The following vectors are defined by a = BC, b = CA, c = AB. Let a and b be two adjacent sides of a parallelogram. Then the area of the parallelogram is S = a b. 7. ( ) If we make use of the multiplication table for the cross product of Cartesian unit vectors, i i = 0, i j = k, i k = j, j i = k, j j = 0, j k = i, k i = j, k j = i, k k = 0, then we can find that the cross product a b is a b = (ay bz az by )i +(az bx ax bz )j +(ax by ay bx )k. Verify the following statements. (a) ( ) a + b + c = 0. (b) ( ) a b = b c = c a. (c) ( ) The following three quantities are all equal. a b = ab sin C, b c = bc sin A, 8. ( ) c a = ca sin B, Page 7 of 9
8 where A = (b, c), B = (c, a), and C = (a, b). (d) ( ) The law of sine in Euclidean geometry can be proved immediately from the above identities as 12. Consider two unit vectors a and b that can be obtained by rotating i by angles α and β, respectively, counterclockwise. We assume that α > β. a b c = =. sin A sin B sin C 11. Consider two unit vectors a and b that can be obtained by rotating i by angles α and β, respectively, counterclockwise. (a) ( ) If we take into account the angle between a and b, we find that a b = k sin(α β). (b) ( ) If we compute a b by making use of Eq. (3), then we find that a b = k [cos α sin β sin α cos β]. (a) ( ) as a and b are expressed in terms of i and j a = cos αi + sin αj, b = cos β i + sin β j. (3a) (c) ( ) Thus we have proved the addition rule for the sine function by employing the scalar product: sin(α β) = cos α sin β sin α cos β. (3b) (b) ( ) If we take into account the angle between a and b, we find that 13. Consider two vectors a and b whose tails are at the same point and they make the right angle. a b = cos α β. (c) ( ) If we compute a b by making use of Eq. (3), then we find that a b = cos α cos β + sin α sin β. (d) ( ) Thus we have proved the addition rule for the cosine function by employing the scalar product: cos α β = cos α cos β + sin α sin β. Page 8 of 9
9 (a) ( ) a b = 0. (b) ( ) a, b, and c = a b make three sides of a right triangle. Thus we can prove the Pythagoras theorem by computing c 2 as c 2 = a 2 + b Consider a triangle ABC and three vectors a = BC, b = CA, and c = AB. Because a + b + c = 0, a + b = c, (4a) b + c = a, (4b) c + a = b. (4c) We define α = CAB, β = ABC, and γ = BCA. We introduce another way to prove the law of cosine in Euclidean geometry. (a) ( ) Squaring both sides of Eq. (4), we find that 2a b = c 2 a 2 b 2, (5a) 2b c = a 2 b 2 c 2, (5b) 2c a = b 2 c 2 a 2. (5c) (b) ( ) The scalar products in Eq. (5) can be expressed as a b = ab cos(π γ) = ab cos γ, (6a) b c = bc cos(π α) = bc cos α, (6b) c a = ca cos(π β) = ca cos β. (6c) (c) ( ) Thus the cosines of the angles α, β, and γ are given by cos α = b2 + c 2 a 2, 2bc cos β = c2 + a 2 b 2, 2ca cos γ = a2 + b 2 c 2. 2ab Page 9 of 9
Department of Physics, Korea University Page 1 of 8
Name: Department: Student ID #: Notice +2 ( 1) points per correct (incorrect) answer No penalty for an unanswered question Fill the blank ( ) with ( ) if the statement is correct (incorrect) : corrections
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