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3. VECTORS AND MATRICES IN 3 DIMENSIONS §3.1. Extending the Theory of 2-dimensional Vectors x A point in 3-dimensional space can be represented by a column vector of the form y . z z-axis A E A y-axis z y x-axis x Most of the theory of 2-dimensional vectors can be extended in a straightforward way to 3dimensional vectors. In most cases the only change that needs to be made is to change “2” into “3” x2 x1 y and to put in an extra component. For example the distance between two points 1 and y2 is z2 z1 2 2 2 (x1 − x2) + (y1 − y2) + (z1 − z2) . A A E E A A E A A Some of the less obvious differences are listed here. The matrix corresponding to a rotation through an arbitrary angle about an arbitrary axis is too complicated to discuss here. However where the cos θ −sinθ 0 axis is the z-axis, and the angle of rotation is θ the matrix of the rotations is sin θ cos θ 0 . 0 0 1 We defined the determinant of a 2 × 2 matrix as the signed area of a certain parallelogram. We shall define the determinant of a 3 × 3 matrix as the signed volume of a certain parallelepiped. Of course these definitions can only work if the components of the matrices are real numbers. In a later chapter we’ll give an algebraic definition that will apply no matter what field the components come from. A E A So what is a parallelepiped? It’s a 3-dimensional analogue of a parallelogram. It’s like a box that has been squashed a bit. It has six faces, each of which is a parallelogram, and they are parallel in pairs. Now determinants can be positive or negative so we need to explain whether we take the actual volume or minus the volume. Three non-coplanar vectors a, b, c, in that order, form a positive triple if, when viewed from c, it takes a positive rotation (less than 180°) to rotate from a to b and a negative triple otherwise. A handy way to remember this is to imagine that you’re inside a room, looking up to a corner. The origin is that corner of a room and the vectors a, b, c are 37 the three edges coming out from that corner. If moving around the vectors in order you are moving in a positive (anti-clockwise) direction, the triple is a positive one. Otherwise it’s a negative one. b a a c c b (a, b, c) is a positive triple (a, b, c) is a negative triple Another way is the right hand rule. Hold your right hand, with thumb and forefinger outstretched, and try to point your thumb in the direction of the first vector, your forefinger in the direction of the second and your third finger in the direction of the third. If you can do it easily the triple is a positive triple. If the experiment would be easier using your left hand then it is a negative triple. X Y Z Note that if the three vectors lie in a plane they’re neither a positive nor a negative triple. So what sign do we give to the determinant? Fortunately if the parallelepiped lies in a plane the volume is zero, and −0 is the same as +0. The determinant of the 3 ×3 matrix (a, b, c), that is whose columns are the vectors a, b and c, is defined to be the signed volume of the parallelepiped whose vertices are 0, a, b, c, a + b, a + c, b + c and a + b + c. b+c a+b+c c a+c b a+b 0 a The signed volume is positive if the vectors form what is called a “positive triple” and negative if they form a “negative triple”. Adapting the argument for 2 × 2 matrices to the 3-dimensional case we can see that when a closed region, having a certain volume, is transformed by the linear transformation v → Av, where A is a 3 × 3 matrix, the volume of the resulting region will be |A| times the original one, if we ignore 38 the sign . This makes the fact that |AB| = |A|.|B| plausible. Multiplying volumes by |A| then multiplying by |B| is equivalent to multiplying by |A|.|B|. The only difficulty is in reconciling the signs. When we come to general n × n matrices we’ll give a proof that’s independent of geometry. §3.2. Vector Products We’ve defined a 3 × 3 determinant in terms of the volume of a parallelepiped. That is the area of the base times the perpendicular height. To find the perpendicular height we must be able to find a vector that is perpendicular to any two given vectors. Suppose n is orthogonal to both a and b. Then a.n = 0 and b.n = 0. That is, if the components of a are a1, a2, a3 and similarly for b and n: a1n1 + a2n2 + a3n3 = 0 and b1n1 + b2n2 + b3n3 = 0. We eliminate n3 from these equations by multiplying the first by b3 and the second by a3 and subtracting. Hence (a1b3 − a3b1)n1 + (a2b3 − a3b2)n2 = 0. Writing these coefficients as 2 × 2 determinants we get a1 b1 a2 b2 n 1+ a3 b3 a3 b3 n3 = 0. One of the many solutions to this equation is a2 b2 a1 b1 n1 = a b , n2 = −a b . 3 3 3 3 Eliminating n2 in a similar way we obtain a1 b1 a2 b3 n 1− a2 b2 a3 b3 n3 = 0 and taking the value of n1 chosen above, this gives a1 b1 n3 = a b . 2 2 a b a b a b 2 2 1 1 1 1 T Hence the column vector a b , −a b , a b is perpendicular to both a and b. 3 3 3 3 2 2 A E A A A E A E E A A A E E A E A A A E E A A A A We define the vector product of a and b to be a2 b2 a1 b1 a1 b1 T a × b = a b , −a b , a b . 3 3 3 3 2 2 A E E A Note that the 2 × 2 determinants that occur in this definition can be obtained by forming the 3 × 2 matrix (a, b), that is by writing the two vectors as the two columns, and then deleting respectively the first, the second and the third rows. a1 b1 a b 2 2 a3 b3 A 3 Example 1: If a = 5 −2 5 Solution: a × b = −2 A A E A E A a1 b1 a b 2 2 a3 b3 A E A A a1 b1 a b 2 2 a3 b3 E 1 and b = 0 , find a × b. 7 0 3 1 3 1 T , − , = (35, −23, −5)T. 7 −2 7 5 0 E A E A E A 39 §3.3. Geometrical Interpretation of Vector Products a1 a1 a Throughout this section we assume that the vectors a and b are a = 2 and b = a2 . a3 a3 2 2 2 Theorem 1: |a × b| = |a| |b| − (a.b) . a2 b22 a1 b12 a1 b12 Proof: |a × b|2 = a b + a b + a b 2 2 3 3 3 3 = (a2b3 − a3b2)2 + (a1b3 − a3b1)2 + (a1b2 − a2b1)2 = (a1b2)2 + (a1b3)2 + (a2b1)2 + (a2b32 + (a3b1)2 + (a3b2)2 − 2(a2a3b2b3 + a1a3b1b3 + a1a2b1b2) = (a12 + a22 + a32)(b12 + b22 + b32) − a12b12 − a22b22 − a3b32 − 2(a2a3b2b3 + a1a3b1b3 + a1a2b1b2) = |a|2|b|2 − (a1b1 + a2b2 + a3b3)2 = |a|2|b|2 − (a.b)2. A A E A E A A E A E A A E A A Theorem 2: |a × b| = |a|.|b| sin θ. Proof: |a × b|2 = |a|2|b|2 − (a.b)2 = |a|2|b|2 − (|a|.|b| cos θ)2 = |a|2|b|2 (1 − cos2θ) = (|a|.|b|sin θ)2. We take the square roots of both sides. Since |a × b| ≥ 0 and sin θ ≥ 0 (as 0 ≤ θ <π), the result follows from taking the non-negative square roots. Theorem 3: |a × b| is the area of the parallelogram with vertices 0, a, b and a + b. Proof: Taking the length of the base of this parallelogram to be |a|, the perpendicular height is |b| sin θ. The area of a parallelogram is the area of the base times the perpendicular height and so the result follows. b a+b |b| |b| sin θ θ a |a| The vector product a × b of two non-parallel vectors is perpendicular to the plane determined by a and b and its length is equal to the area of the parallelogram formed from a and b. There are two such vectors, one on either side of the plane. Which of the two is a × b? 0 a×b? b 0 a Without loss of generality rotate the system so that a lies along the positive half of a ×we b may ? the x-axis and b lies in the x-y plane, and hence we may take r cos θ s a = 0 and b = r sin θ 0 0 where θ is the angle from a to b (viewed from the positive half of the z-axis) and r, s are the distances of the points from the origin. A E A A 40 E b r θ a 0 s 0 Then a × b = 0 . If θ < π, as in the diagram, a × b comes out of the page and so a, b, a × b rs sin θ form a positive triple. If θ > π, a × b goes into the page, and once again a, b, a × b form a positive triple. θ s a 0 A E A r b We’ve thus proved the following theorem. Theorem 4: If a, b are not parallel vectors, the vector a × b is the unique vector c such that: (1) c is perpendicular to both a and b; (2) a, b, c is a positive triple; (3) |c| is the area of the parallelogram formed from a and b. Theorem 5: a × b = 0 if and only if a, b are parallel. Proof: This follows from (3) of Theorem 4. The area of a parallelogram is zero if and only if the adjacent sides are parallel. Theorem 6: a × b = −(b × a). Proof: |a × b| = |b × a| since they are both equal to the area of the same parallelogram. Both a × b and b × a are perpendicular to the plane determined by a and b. However since a, b, a × b is a positive triple, b, a, a × b is a negative triple and so the direction of b × a is opposite that of a × b. §3.4. Vector Products and Determinants We are now in a position to find the determinant of a 3 × 3 matrix. Theorem 7: If A = (a, b, c) is a 3 × 3 matrix then |A| = (a × b).c. Proof: Case I: a, b, c is a zero triple. Then |A| = 0 (for the parallelepiped has zero volume). The vectors a, b, c are coplanar. Since a × b is orthogonal to this plane it is orthogonal to c. Thus (a × b).c = 0. Case II: a, b, c is a positive triple. The perpendicular height is the length of the projection of c onto a × b. a×b c |c| cos θ θ 0 |c| 41 If θ is the angle between a × b and c then this length is |c| cos θ = |c|. (a × b).c (a × b).c = . |a × b|.|c| |a × b| E E A A A E A E Then |A| = volume of the parallelepiped formed from a, b, c = area of base × perpendicular height a×b . c = (a × b) . c. = |a × b| |a × b| Case III: a, b, c is a negative triple. This is as for case II except that the projection of c onto a × b (a × b).c and |A| is minus the volume of the is now in the opposite direction to a × b and so is − |a × b| parallelepiped. E A A A E A E a1 b1 c1 a1 b1 a1 b1 a2 b2 Theorem 8: a2 b2 c2 = c1 a b − c2 a b + c3 a b . 3 3 3 3 2 2 a3 b3 c3 a2 b2 a3 b3 c1 a1 b1 Proof: By Theorem 7 the determinant is equal to −a b . c2 . 3 3 c3 a 1 b1 a2 b2 A E A E A A A E A A A E A E E A A E A §3.5. The Adjoint of a Matrix The adjoint of a 3 × 3 matrix A = (a, b, c) is defined to be adjA = (b × c, c × a, a × b)T. Theorem 9: (adj A)A = |A|.I for all 3 × 3 matrices A. Proof: (adjA)A = (b × c, c × a, a × b)T (a, b, c) (b × c).a (b × c).b (b × c).c = (c × a).a (c × a).b (c × a).c (a × b).a (a × b).b (a × b).c |A| 0 0 = 0 |A| 0 since (b × c).a, (c × a).b and (a × b).c are all equal to the signed 0 0 |A| volume of the same parallelepiped, the one formed from a, b, c. (They all have the same orientation.) 1 Corollary: A 3 × 3 matrix is invertible if and only if |A| ≠ 0. For such matrices A−1 = |A| adj A. E E A A E A A Theorem 10: If |A| = 0 there exists a non-zero vector v such that Av = 0. Proof: Suppose A = (a, b, c) and |A| = 0. Then a, b, c are coplanar. λ If a, b are not parallel then c = λa + µb for some λ, µ. Hence (a, b, c) µ = 0. −1 A a c b 0 λa a 42 E A E A λ If a ≠ 0 and b = λa then (a, b, c)−1 = 0. 0 1 If a = 0 then (a, b, c)0 = 0. 0 E A A E A A Theorem 11: a × (b + c) = a × b + a × c. Proof: This can be proved geometrically, but it’s easily proved algebraically. A convenient way to compute vector products is to express them in terms of the three “standard basis vectors” i = (1, 0, 0), j = (0, 1, 0) and k = (0, 0, 1). A general vector product can be reduced, by the distributive law, to the vector products between i, j and k. For reference we note these below: i × j = k, j × k = i, k × i = j i × i = j × j = k × k = 0. Example 2: If a = (3, 5, 0) and b = (−2, 0, 1) find a × b. Solution: a × b = (3i + 5j) × (−2i + k) = 0 − 3j +10k + 5i = (5, −3, 10). Another useful way of calculating vector products is to set up a 3 × 3 matrix with the vectors i, j, k as the first row, with the components of the two vectors occupying the second and third rows. This will become more useful when we discuss the evaluation of determinants in general, in a later chapter. Example 3: Find (2, 6, −1) × (1, 3, 5). i j k 26 2 −1 6 −1 j + k i− Solution: The vector product is 2 6 −1 = 13 1 5 3 5 1 3 5 = 33i − 11j + 0k = (33, −11, 0). A E A E A A A E A A E A Since the cross product is linear in each variable the map v → a × v is a linear transformation and corresponds to multiplication by a 3 × 3 matrix X(a). Theorem 12: X(a × b) = X(a)X(b) − X(b)X(a). a2v3 − a3v2 0 −a3 a2 v1 v1 a1 Proof: Let a = a2 and v = v2 . Then a × v = a3v1 − a1v3 = a3 0 −a1 v2 . v3 a3 a1v2 − a2v1 −a2 a1 0 v3 0 −a3 a2 0 −b3 b2 So X(a) = a3 0 −a1 and X(b) = b3 0 −b1 . The conclusion of the theorem is now easily −a2 a1 0 −b2 b1 0 checked. Corollary: For all vectors a, b, c a × (b × c) + b × (c × a) + c × (a × b) = 0. Proof: a × (b × c) + b × (c × a) + c × (a × b) = X(a)X(b)c − X(b)X(a)c − X(a × b)c = 0. A A E A E A A E A E A A E A A E A E A A §3.6. Application of Vector Products to Electromagnetic Theory The vector product is a valuable tool in physics, especially in connection with electricity and magnetism. Typical of its use in this area is the formula for the voltage generated in a piece of wire that is moving in a uniform magnetic field: 43 E = ⌠(v ⌡ × B).dL A E where E is the voltage generated, in volts, in a certain direction (if E is negative this indicates that the flow is in the opposite direction), v is the velocity of the piece of wire (the magnitude is the speed in metres per second and the direction is the direction of movement of the wire), B is a vector representing the magnetic field (in webers per square metre) and L is a vector representing the length (in metres) and orientation of the piece of wire. B B L v One conclusion from this formula is that if the wire lies in a certain plane, and moves only in that plane, and if the magnetic field is parallel to that plane, the voltage generated is zero. This is because (v × B).dL is the signed volume of the parallelogram formed from v, B and dL, which is zero if these vectors are coplanar. Example 4: Find, as a function of time, the voltage generated in a circular loop of wire of radius r that is rotating with a constant angular velocity of ω radians per second about the z-axis in a uniform magnetic field of M webers per square metre in the direction of (a) the positive half of the z-axis; (b) the negative half of the x-axis. R Solution: z z dL v r sin ψ sin ωt y ψ r sin ψ cos ωt r 0 r sin ψ ωt y ωt x x Let r be the vector representing the point under consideration. sin ψ cos ωt Then R = r sin ψ sin ωt . cos ψ Since v is orthogonal to both R and i (i being the unit vector along the positive x-axis) and i, R, v form a positive triple, v has the same direction as i × R. Adjusting magnitudes we see that |v| v= (i × R) |i × R| −r sin ψ sin ωt ωr sin ψ r sin ψ cos ωt = |i × R| 0 −sin ψ sin ωt ωr sin ψ = r sin ψ cos ωt r sin ψ 0 A A E A E A E A A E A A E A E 44 −sin ωt = ωr sin ψ cos ωt . 0 Since dL is orthogonal to both R and v and R, v, dL form a positive triple |dL| dL = (R × v) |R × v| 2 −ωr cos ψ sin ω cos ωt r −ωr2 cos ψ sin ψ sin ωt dψ = |R × v| ωr2 sin2ψ cos ψ cos ωt r =− 2 ωr2 sin ψ cos ψ sin ωt dψ ωr sin ψ − sin ψ cos ψ cos ωt = − r cos ψ sin ωt dψ. −sin ψ cos ωt 0 (a) B = 0 . Then v × B = Mrω sin ψ sin ωt whence M 0 π π 2 2 E=−2⌠ ⌡sin 2ψ dψ = 0. ⌡Mr ω cos ψ sin ψ dψ = −Mr ω ⌠ 0 0 A A E A E A E E A A E A A A A A E E A E A A A E A E A A A A E A 0 −M (b) B = 0 . Then v × B = Mrω sin ψ 0 whence cos ωt 0 π π 2 2 2 ⌠ ⌠ E = 2 ⌡Mr ω sin ψ cos ωt dψ = 2Mr ω cos ωt ⌡sin2ψ dψ = πr2Mω cos ωt. 0 0 NOTE: Case (b) is a greatly simplified account of how alternating current is generated. A E A A A E A E A E A A EXERCISES FOR CHAPTER 3 Exercise 1: If a = (2 −1, 1). b = (−1, 2, 1) and c = (1, 2, −1) find (a) a × b; (b) a × (b + c); (c) a.(b × c); (d) a × (b × c). Exercise 2: If a = (−1, 1, 3), b = (2, 1, 2) and c = (0, 1, −1) find (a) a × b; (b) a × (b + c); (c) a.(b × c); (d) a × (b × c). 5 −2 Exercise 3: Evaluate (i) −2 3 0 2 A 0 2 ; 5 E A 45 3 2 7 (ii) 1 −1 −1 . −2 1 0 E A A Exercise 4: Find 3 values of λ such that |λI − A| = 0 where A is the matrix of Exerciseb3(i). For each value of λ find a corresponding non-zero vector v such that (λI − A)v = 0. Exercise 5: (a) Find a non-zero vector that is perpendicular to (3, 2, 7) and (1, −1, −1). (b) Find a non-zero vector that is perpendicular to 5i − 2k and −2i + j. 3 1 2 Exercise 6: If a = (0, 4, − 2) and b = (−3, 2, 1) find a × b and 0 4 −2 . −3 2 1 A E A Exercise 7: If a = (1, −1, 2), b = (3, 1, −1) and c = (0, 2, 3) find (i) a × b; (ii) b × c; (iii) c × a. If A = (a, b, c) find A−1. X x Exercise 8: Find a 3 × 3 real matrix A such that the linear transformation Y = Ay Z z 2 2 2 X Y Z transforms the unit sphere x2 + y2 + z2 = 1 into the ellipsoid a2 + b2 + c2 = 1. Find |A| and hence find the volume of the above ellipsoid. A A E A A E A A E E A A E A Exercise 9: The six faces of a parallelepiped lie on the following planes: 2x − y = 0 2x − y = 1 x+y=0 . x+y=1 2x + y − z = 0 2x + y − z = 1 Find a 3 × 3 matrix A which transforms this parallelepiped into a cube. Hence find the volume of the parallelepiped. A E A Exercise 10: Write the following system of equations 3x − 2y + 5z = 4 x + 7y = −9 −4x + 8y − 5z = −21 x in the form Av = b where v = y and A is a 3 × 3 matrix. z A A E A Exercise 11: Find the point of intersection of the planes 2x + 5y − z = −17 −3x + 7y + 4z = −13 . 12x + 5z = 10 A E 46 A E Exercise 12: (a) Find the area of the parallelogram whose vertices are (0, 0, 0), (2, 3, −3), (4, 1, 0) and (6, 4, −3). (b) Find the area of the parallelogram whose vertices are (1, −1, 2), (3, 2, −1), (5, 0, 2) and (7, 3, −1). (c) Find the area of the triangle whose vertices are (1, −1, 2), (3, 2, −1) and (5, 0, 2). Exercise 13: Find the area of the parallelogram formed from (1, 0, 2), (5, 3, 4) and (2, 1, 1). Exercise 14: Find the angle between the vectors (2, −2, 1) and (1, 0, 1). Exercise 15: Find the angles of the triangle whose vertices are A(1 − 2, 1), B(3, 0, 2) and C(4, 4, 1). Exercise 16: Prove that the area of the triangle with vertices a, b. c is ½ |a×b + b×c + c×a|. Exercise 17: Find the equation of the plane through 0 that is perpendicular to the vector (1, 2, 3). Exercise 18: Prove that the equation of the plane that passes through the points a, b and c is (a×b + b×c + c×a).v = a.(b×c). Exercise 19: Prove that if a, b are 3-dimensional column vectors and if A is the 3 × 2 matrix (a, b) then |ATA| = |a × b|2. Exercise 20: The 3-dimensional column vectors a, b are non-collinear and d is the foot of the perpendicular from c to the plane through 0, a and b. A is the 3 × 2 matrix (a, b). Prove that: (i) a.(c − d) = b.(c − d) = 0; (ii) ATc = ATd; λ (iii) d = A for some real λ, µ; µ (iv) d = A(ATA)−1ATc; (v) Explain why one can’t use the identity (MN)−1 = N−1M−1 to conclude that d = c. (vi) By considering the geometric interpretation of the linear transformation v → A(ATA)−1ATv, or otherwise, show that 0 and 1 are eigenvalues of A(ATA)−1AT and that a, b and a×b are its eigenvectors. A E A Exercise 21: A 3-dimensional vector function is a function that assigns to every value of a real variable t a 3-dimensional real vector r(t). If the components of r(t) are differentiable the derivative of r(t) with respect to t is defined by differentiating each component. Prove that if r(t) and s(t) are differentiable 3-dimensional vector functions then: d dr(t) ds(t) dt (r(t).s(t)) = dt .s(t) + r(t). dt . Prove that if the length of r(t) is constant as t varies then dr(t) dt is perpendicular to r(t) for every value of t. A E A A E A E A E A E Exercise 22: Prove that if a(t) and b(t) are differentiable 3-dimensional vector functions of t d da(t) db(t) then dt (a(t) × b(t)) = × b(t) + a(t) × dt . d t A E A A E A E Exercise 23: If a, b, c are 3-dimensional unit vectors that are not coplanar and if b and c are perpendicular prove that the angle between a and the plane through 0, b and c is θ where sin θ = |a.(b × c)|. 47 Exercise 24: Examine the following theorem and proof and comment on its validity. “Theorem: If a, b, c, d are non-zero vectors with (a × b) × (c × d) = 0 then 0, a, b, c, d lie in a plane. Proof: Suppose (a × b) × (c × d) = 0. Then a × b is parallel to c × d. Since these are perpendicular to the planes through 0, a, b and 0, c, d respectively, these planes must coincide and hence all five points must lie in this plane.” 48 SOLUTIONS FOR CHAPTER 3 Exercise 1: (a) (−3, −3, 3); (b) (−4, 0, 8); (c) −12; (d) (4, 4, −4). Exercise 2: (a) (−1, 8, 3); (b) (−5, 7, −4); (c) 11; (d) (−4, −7, 1). Exercise 3: (i) 35; (ii) 0. 0 λ − 5 2 Exercise 4: 2 λ − 3 −2 = λ3 − 13λ2 + 47λ − 35 = (λ − 1)(λ − 5)(λ − 7) so 0 −2 λ − 5 λ = 1, 5, 7 are the three values. 1 −1 −1 1 −1 −1 −4 2 0 −2 1 0 1 −1 −1 When λ = 1, |λI − A| = 2 −2 −2 → 1 −1 −1 → −2 1 0 → 0 −1 −2 → 0 1 2 0 0 0 0 −1 −2 0 −1 −2 0 −2 −4 0 −1 −2 −1 so −2 is a suitable vector. 1 0 2 0 1 1 −1 1 1 −1 0 1 0 2 2 −2 0 1 0 When λ = 5, |λI − A| = → → 0 1 0 0 0 0 0 −2 0 1 so 0 is a suitable vector. 1 2 2 0 1 1 0 1 1 0 1 1 0 0 1 −1 2 4 −2 1 2 −1 0 1 −1 When λ = 7, |λI − A| = → → → 0 0 0 0 −2 2 0 −1 1 0 −1 1 −1 so 1 is a suitable vector. 1 E A A A A E E E E A A E A A E A A E A A E A A A A E A A A A A E E A A E A E A A A E A A Exercise 5: (a) (3, 2, 7) × (1, −1, −1) = (5, 10, −5) is a suitable vector. (b) (5i − 2k) × (−2i + j) = −10(i × i) +5(i × j) + 4(k × i) − 2(k × j) = 0 + 5k + 4j + 2i = 2i + 4j + 5k. Exercise 6: a × b = (8, 6, 12). If c = (3, 1, 2) the determinant is (a × b).c = 54. Exercise 7: (i) (−1, 7, 4); (ii) (5, −9, 6); (iii) (7, 3, −2). adjA = (b × c, c × a, a × b)T. 5 −9 6 5 −9 6 1 −1 = 7 3 −2 |A| = (a × b).c = 26 so A = 26 7 3 −2 . −1 7 4 −1 7 4 x ax Exercise 8: The linear transformation takes y to by . z cz a 0 0 The required matrix is A = 0 b 0 and |A| = abc. 0 0 c A E A A A A E E A E A A A 49 E E A A A A E 4π 4π The volume of the unit sphere is 3 so the volume of the ellipsoid is 3 abc. A E A A E A Exercise 9: Let X = 2x − y, Y = x + y, Z = 2x + y − z. Then the six faces are X = 0, X = 1, Y = 0, Y = 1, Z = 0, Z = 1 which are the faces of a unit cube. 2 −1 0 X x The matrix of the linear transformation that takes y to Y is A = 1 1 0 . Z z 2 1 −1 Since |A| = −3, volumes are multiplied by 3 by the linear transformation. (The minus sign merely relates to the orientation of the vertices.) Hence the volume of the parallelepiped is 3. E A A E A A A E A 3 −2 5 x 4 Exercise 10: The system of equations can be written as 1 7 0 y = −9 . −4 8 −5 z −21 A E A E A A E A 2 5 −1 x −17 Exercise 11: We must solve the equation −3 7 4 y = −13 . 12 0 5 z 10 −1 2 5 7 Let a = −3 , b = , c = 4 . 0 5 12 35 −25 27 T Then the adjoint of the matrix is (b × c, c × a, a × b) = 63 22 −5 . −84 60 29 35 −25 27 1 63 22 −5 . The determinant is 469 so the inverse is 469 −84 60 29 35 −25 27 −17 0 1 63 22 −5 −13 = −3 The solution is 469 −84 60 29 10 2 so there is a unique solution x = 0, y = −3, z = 2. A A E A A E A A E E E A A A E A A A A A E A E A E E A A E E A A A E Exercise 12: (a) (2, 3, −13) × (4, 1, 0) = (13, −52, −10) whose length is 132 + 522 + 102 = 2973 . (b) We must translate (1, −1, 2) to the origin. The points become (0, 0, 0), (2, 3, −3), (4, 1, 0) and (6, 4, −3). This is the same parallelogram as before, so the area is 2973 . (c) The area of this triangle is half that of the previous parallelogram, that is it is ½ 2973 . A A E E A A E A A Exercise 13: Translating (1, 0, 2) to the origin we get (0, 0, 0), (4, 3, 2) and (1, 1, −1). (4, 3, 2) × (1, 1, −1) = (−5, 6, 1) whose length is 52 + 62 + 12 = 62 . A Exercise 14: cos θ = A E 3 1 = . 9 2 2 E A A E A 50 A A A E A E A Exercise 15: Translating (1, −2, 1) to the origin we get the vectors (2, 2, 1) and (3, 6, 0). 18 2 If α is the angle at A then cos α = = so α ≈ 41.8° 9. 45 5 Translating (3, 0, 2) to the origin we get the vectors (−2, −2, −1) and (1, 4, −1). 18 2 If β is the angle at A then cos α = = so α ≈ 41.8°. 9. 45 5 The remaining angle is 96.4°. A A A E A E A A A E A E Exercise 16: The area is ½ |(b − a) × (c − a)| = ½|(b × c) − (b × a) − (a × c) + (a × a)| = ½|(b × c) + (a × b) + (c × a)|. Exercise 17: The plane is x + 2y + 3z = 0. Exercise 18: Let v lie on the plane. Translate a to the origin. The resulting vectors are 0, b − a, c − a and v − a. A vector perpendicular to this plane is (b − a) × (c − a) = a×b + b×c + c×a. Since v lies on this plane (a×b + b×c + c×a).(v − a) = 0 so (a×b + b×c + c×a).v = (a×b + b×c + c×a).a = (b×c).a since the other terms are zero. 2 a.a a.b |a| a.b = Exercise 19: Suppose A = (a, b). Then ATA = a.b b.b a.b |b|2 so |ATA| = |a|2.|b|2 − (a.b)2 = |a|2.|b|2 − (|a|.|b|.cos θ)2 where θ is the angle between a, b = |a|2.|b|2(1 − cos2 θ) = |a|2.|b|2.sin2 θ = |a × b|2. A E A A E Exercise 20: (i) The line from c to d is perpendicular to both a, b so a.(c − d) = b.(c − d) = 0 T a a.c a.d (ii) ATc = T c = = = ATd. b b.c b.d (iii) Since d lies in the plane determined by 0, a, b we must have d = λa + µb for some real λ, µ. a1 If a = a2 etc then d1 = λa1 + µb1, d2 = λa2 + µb2 and d3 = λa3 + µb3. a3 λa1 + µb1 a1 b1 λ λ Hence d = λa2 + µb2 = a2 b2 = A . µ µ λa3 + µb3 a3 b3 λ λ λ (iv) ATc = ATd = ATA so = (ATA)−1ATc and hence d = A = A(ATA)−1ATc. µ µ µ T (v) A and A are 3 × 2 matrices and hence don’t have inverses. (vi) The linear transformation v → A(ATA)−1ATv projects a vector v onto the plane that passes through 0, a, b. Vectors lying on this plane are fixed and so a, b are eigenvectors corresponding to the eigenvalue 1. Vectors perpendicular to this plane, such as a × b, are mapped to 0 and so are eigenvectors corresponding to the eigenvalue 0. A A E E A A E A A E A A A A E A E A A A E E A E A A E A A A 51 E A r1(t) Exercise 21: Let r(t) = r2(t) etc. r3(t) Then r(t).s(t) = r1(t)s1(t) + r2(t)s2(t) + r2(t)s2(t). d d Hence dt (r(t).s(t)) = dt [r1(t).s1(t) + r2(t).s2(t) + r3(t).s3(t)] d d d = dt r1(t) s1(t) + r2(t) s2(t) + dt r3(t) s3(t)) + dt d d d r1(t) dt s1(t) s1(t) + r2(t) dt s2(t) s2(t) + r3(t) dt s3(t) s3(t)) dr1(t) dr2(t) dr3(t) = dt , dt , dt . (s1(t), s2(t), s3(t)) + ds1(t) ds2(t) ds3(t) (r1(t), r2(t), r3(t)) . dt , dt , dt dr(t) ds(t) = dt .s(t) + r(t). dt . E A A E A E A E A A A E A A E A A E A A E A E A A E A A E A A E A E E E A A E r1(t) Exercise 22: Let r(t) = r2(t) etc and let ′ denote derivatives with respect to t. r3(t) a2(t)b3(t) − a3(t)b2(t) Then a(t) × b(t) = a3(t)b1(t) − a1(t)b3(t) . a1(t)b2(t) − a2(t)b1(t) a2′(t)b3(t) − a3′(t)b2(t) + a2(t)b3′(t) − a3(t)b2′(t) d Hence dt (a(t) × b(t)) = a3′(t)b1(t) − a1′(t)b3(t) + a3(t)b1′(t) − a1(t)b3′(t) a1′(t)b2(t) − a2′(t)b1(t) + a1(t)b2′(t) − a2(t)b1′(t) a2′(t)b3(t) − a3′(t)b2(t) a2(t)b3′(t) − a3(t)b2′(t) = a3′(t)b1(t) − a1′(t)b3(t) + a3(t)b1′(t) − a1(t)b3′(t) a1′(t)b2(t) − a2′(t)b1(t) a1(t)b2′(t) − a2(t)b1′(t) db(t) da(t) = dt × b(t) + a(t) × dt . E A A A E A E E A E A E A E E A E A A A E E E A E Exercise 23: The volume of the parallelepiped formed from 0, a, b, c is |a.(b × c)|. However the area of the base is 1 and the perpendicular height is sin θ so the volume is sin θ Exercise 24: The given identity holds if a × b = 0, in which case a, b are parallel. However the line through these vectors need not lie in the plane 0, c, d. Counter-example (i × i) × (j × k) = 0, yet the points 0, i, j, k don’t lie in a plane. 52