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6.1 Change of Basis So far, our main way of representing vectors involves the horizontal and vertical components of a vector. Indeed, we have introduced three different ways of writing a vector in terms of its components: " v a Figure 1: Any vector v can be written as a linear combination of i and j. vx vy # ( vx , vy ) vx i + vy j. Figure 1 shows the geometry of these components. The idea is that any vector v can be decomposed into a horizontal vector and a vertical vector, i.e. a multiple of i and a multiple of j, and the components tell you how much of each you need to make v. However, the vertical and horizontal directions are not always the most important directions to consider. For example, when analyzing the motion of an object on an inclined plane, it might make more sense to use the direction of the incline as one of the primary directions. Similarly, a geometry problem involving a diagonal line might be easier to solve if we use the direction of the line as one of our primary directions. In this case, it makes sense to use another pair of vectors a, b as our two basis vectors, instead of the horizontal and vertical vectors i and j. Any vector v can be written as a linear combination of the two basis vectors: v sa + tb. a Figure 2: Any vector v can be written as a linear combination of the basis vectors a and b. The coefficients s and t in this linear combination are called the components of v with respect to the basis a, b. For example, Figure 2 shows a vector v whose components are 4 and 2 with respect to a certain basis a, b. EXAMPLE 1 " # Find the components of the vector v 2 8 " # with respect to the basis a 1 ,b 1 " # −1 . 1 SOLUTION We wish to express v and a linear combination of a and b. This means that we want to find scalars s and t so that " # 2 8 " # s 1 1 " +t # −1 . 1 This vector equation can be made into a system of linear equations 2 s−t 8 s+t Solving yields s 5 and t 3, i.e. " # 2 8 a Figure 3: The vector v can be expressed as 5a + 3b. " # 5 1 1 " +3 # −1 . 1 Thus the components of v are 5 and 3 with respect to this basis. This decomposition of v is illustrated in Figure 3. Linear Independence It is not always possible to use a given pair of vectors a, b as a basis for R2 . For example, if the two vectors a and b point in the same direction, then they cannot be a basis, since any linear combination of a and b will point in that same direction. CHANGE OF BASIS 2 Two vectors a and b in the plane are said to be linearly independent if they can serve as a basis for R2 . There is a simple criterion for linear independence: Linear Independence for Two Vectors Two vectors a and b are linearly independent if and only if: 1. Neither a nor b is the zero vector, and 2. The vectors a and b are not scalar multiples of one another. Stated differently, two vectors a and b are linearly independent as long as they point in two completely different directions. Using Other Bases It is often helpful in geometry problems to use a basis other than the standard basis i, j. The following examples illustrate this technique. EXAMPLE 2 Figure 4 shows a right triangle in the plane. Find the coordinates of the point p. SOLUTION Note that the vector (3, 1) is parallel to the line, while (−1, 3) is perpendicular. We wish to express the vector (10, 10) as a linear combination of these basis vectors: " a Figure 4: The right triangle for 10 10 # " # 3 s 1 " # −1 +t . 3 This vector equation can be made into a system of linear equations Example 2. 10 3s − t 10 s + 3t Solving yields s 4 and t 2, i.e. " 10 10 # " # 4 3 1 " +2 # −1 . 3 This decomposition of (10, 10) is shown in Figure 5. It follows that a p 4 (3, 1) (12, 4) . Figure 5: The decomposition of (10, 10) . EXAMPLE 3 Figure 6 shows a parallelogram in the plane. Find the coordinates of the point p. Note that the vector (4, 1) is parallel to the line y x/4, while (1, 2) is parallel to the line y 2x. We wish to express the vector (11, 8) as a linear combination of these basis vectors: " # " # " # 11 4 1 s +t . 8 1 2 SOLUTION This vector equation can be made into a system of linear equations a Figure 6: The parallelogram for Example 3. 11 4s + t 8 s + 2t CHANGE OF BASIS 3 Solving yields s 2 and t 3, i.e. " 11 8 # " # 2 " # 4 1 +3 1 . 2 This decomposition of (11, 8) is shown in Figure 5. It follows that p 2 (4, 1) (8, 2) . a Figure 7: The decomposition of (11, 8). EXERCISES Find the components of the vector v with respect to the basis a, b. 1–2 " 1. v # 10 ; a −9 " # " # 2 1 " # 1 ; a 0 2. v 3 2 and " # 4 1 " # " # " # 5. # 2 , b −1 3 , b 2 " # 2 3 Determine whether the given pair of vectors are linearly independent. 3–6 3. " 1 3 " # and 2 8 4. " 0 0 6. 30 42 7. For what values of k are the vectors # 3 12 " and # 45 64 # " # " # 1 3 " and and 4 k linearly independent? 8. The following figure shows a right triangle in the plane. Find the coordinates of the point p. 9. The following figure shows a parallelogram in the plane. Find the coordinates of the point p. 6.2 Dot Product The dot product is an operation that takes two vectors as input and outputs a scalar. It is defined by the formula ( a, b ) · ( c, d ) ac + bd. That is, corresponding components of the two vectors are multiplied, and then the products are added together. For example, (2, 3) · (5, 4) (2)(5) + (3)(4) 10 + 12 22. Because the result is a scalar, the dot product is also known as the scalar product of vectors. It is also sometimes called the inner product of vectors, or more properly the Euclidean inner product. Note that the dot product of a vector with itself gives the square of the magnitude: ( a, b ) · ( a, b ) a 2 + b 2 | ( a, b ) | 2 . That is, v · v |v| 2 for any vector v. The dot product is often thought of as a kind of multiplication of vectors, though the result is a scalar as opposed to a vector. Algebraically, it satisfies several rules analogous to the rules for multiplication, as listed below. Algebraic Properties of the Dot Product The dot product has the following properties: 1. v · w w · v for any two vectors v and w. 2. ( kv) · w v · ( kw) k (v · w) for any scalar k and any vectors v and w. 3. u · (v + w) u · v + u · w for any three vectors u, v, and w. Because the dot product obeys the distributive law, we can distribute dot products of sums just as we do in normal algebra. For example, (a + b) · (c + d) a · c + a · d + b · c + b · d for any four vectors a, b, c, d. Geometric Interpretation The importance of the dot product stems from its geometric interpretation. Geometric Interpretation of Dot Product If v and w are vectors, then v · w |v| |w| cos θ, a Figure 1: The angle between v and w. where θ is the angle between v and w, as shown in Figure 1. DOT PRODUCT 2 This formula follows from the law of cosines from trigonometry. To see this, recall that the law of cosines states that c 2 a 2 + b 2 − 2ab cos θ a Figure 2: Triangle for the law of cosines. for any triangle with side lengths a, b, c, where θ is the angle opposite the side of length c, as shown in Figure 2. To derive the desired formula for v · w, we apply the law of cosines to the triangle of vectors shown in Figure 3. This gives the formula |w − v| 2 |v| 2 + |w| 2 − 2 |v| |w| cos θ. But expanding |w − v| 2 using the dot product gives |w − v| 2 (w − v) · (w − v) w·w − w·v − v·w + v·v a Figure 3: A triangle of vectors. |v| 2 + |w| 2 − 2v · w, and it follows that v · w |v| |w| cos θ. EXAMPLE 1 Two vectors v and w have magnitudes 5 and 6, and the angle between them is 60◦ . What is the value of v · w? We have v · w |v| |w| cos θ (5)(6) cos (60◦ ) (5)(6)(1/2) 15. SOLUTION One way the formula v · w |v| |w| cos θ is useful is that it allows us to find the angle between two vectors. To use it in this fashion, we solve for the angle θ: Because of the inverse cosine, this formula always yields an angle between 0◦ and 180◦ . θ cos−1 v·w |v| |w| EXAMPLE 2 Find the angle between the vectors (3, 2) and (1, 5) . SOLUTION Let v (3, 2) and w (1, 5) . Then v · w (3)(1) + (2)(5) 13, a Figure 4: The angle between the vectors (3, 2) and (1, 5) is 45◦ . so θ cos −1 v·w |v| |w| |v| −1 cos √ 13, 13 √ √ 13 26 |w| and ! −1 cos 1 √ 2 √ 26, ! 45◦ . Thus the angle between the vectors is 45◦ , as shown in Figure 4. Another application of the formula v · w |v| |w| cos θ is that it gives us a general rule for the sign of v · w. In particular, assuming |v| > 0 and |w| > 0, the dot product v · w must have the same sign as cos θ, as shown in Figure 5. Thus: • v · w > 0 if the angle between v and w is acute. • v · w 0 if the angle between v and w is a right angle. a Figure 5: The cosine is positive for acute angles and negative for obtuse angles. • v · w < 0 if the angle between v and w is obtuse. DOT PRODUCT 3 d Figure 6: The sign of the dot product depends on the angle between the two vectors. This trichotomy is illustrated in Figure 6. Of these three cases, the most important case is where the dot product is equal to zero. Two vectors v and w are said to be orthogonal if v · w 0. For example, the vectors (2, 5) and (−5, 2) are orthogonal, since Note that the zero vector 0 is orthogonal to any other vector, since 0 · v 0 for any vector v. (2, 5) · (−5, 2) (2)(−5) + (5)(2) 0. Geometrically, two nonzero vectors are orthogonal if and only if their directions are perpendicular. EXERCISES " # 1. Compute the dot product of the vectors 5 2 " and # 3 . −4 2. Suppose that two vectors v and w meet at a 30◦ angle. If v · v 4 and w · w 12, what is the value of v · w? 3. Compute the angle between the vectors 2i − j and −i + 3j. 4. Compute the angle between the vectors (2, 5) and (−1, 4) , correct to the nearest degree. 5. The following figure shows four vectors a, b, c, and d. (a) For which two of these vectors is the dot product positive? (b) Which two of these vectors are orthogonal? 6. Let a (−9, −3) , b (3, −8) , c (7, 2) , and d (−2, 6) . (a) Which two of these vectors are orthogonal? (b) Which two of these vectors meet at an acute angle? 7. Suppose |v| 4 and w · w 9, and that v · w 0. What is the value of |v + w|? 6.3 Three-Dimensional Space So far we have only considered vectors and vector geometry in two-dimensional space. In this section we expand our point of view to consider three-dimensional space. Three-Dimensional Coordinates a Figure 1: Each point in three-dimensional space has three coordinates. Just as a point on the plane can be described by two coordinates x and y, a point in three-dimensional space can be described by three coordinates x, y, and z. The geometric meaning of these coordinates is shown in Figure 1. Using these coordinates, we can identify the points in three-dimensional space with ordered triples ( x, y, z ) of real numbers. In linear algebra and advanced mathematics, the set of all such ordered triples is usually denoted R3 (pronounced “arr-three”). The space R3 has three coordinates axes, namely the x-axis, the y-axis, and z-axis. These three axes are all mutually perpendicular, and they meet at the origin (0, 0, 0) . We usually think of the x and y axes as horizontal, while the z-axis is vertical. There are also three coordinate planes, namely the x y-plane, the xz-plane, and the yz-plane, as shown in Figure 2. Each of these planes contains two of the coordinate axes, and is perpendicular to the third. These planes partition R3 into eight octants, similar to the four quadrants in R2 . Though there is not a standard numbering for these octants, everyone agrees that the first octant is the portion of R3 consisting of all points whose coordinates are positive, i.e. the visible portion in Figure 2. Vectors in R3 a Figure 2: The three coordinate planes. Every vector v in R3 has three components vx , vy , and vz . As in R2 , we think of vectors and points as the same thing, so vx v vy ( vx , vy , vz ) . vz Addition and scalar multiplication work the same way that they do in R2 , with a a0 a + a0 b + b 0 b + b 0 0 0 c c c + c and ka a k b kb . c kc EXAMPLE 1 Figure 3 shows a rectangle in R3 . Find the coordinates of the point p. SOLUTION Let v be the vector from (7, 2, 3) to (4, 6, 4) . Then v (4, 6, 4) − (7, 2, 3) (−3, 4, 1) . This same vector lies along the top edge of the rectangle from (5, 0, 5) to p, so p (5, 0, 5) + v (5, 0, 5) + (−3, 4, 1) (2, 4, 6) . a Figure 3: A rectangle in R . 3 Distance in R3 There is a nice formula for distances in R3 , which is based on a three-dimensional version of the Pythagorean theorem. THREE-DIMENSIONAL SPACE 2 Three-Dimensional Pythagorean Theorem For a rectangular box with length a, width b, and height c, as shown in Figure 4, the distance d between opposite corners of the box obeys the formula a2 + b2 + c2 d2 . a Figure 4: A rectangular box. This theorem can be justified using two applications of the usual Pythagorean theorem, as shown in Figure 5. The two right triangles in this figure give us the equations a 2 + b 2 L2 and L2 + c 2 d 2 , and combining these gives the equation a 2 + b 2 + c 2 d 2 . Based on this theorem, the magnitude of a vector v in R3 should be given by the formula |v| a Figure 5: Justification for the three-dimensional Pythagorean theorem. q vx2 + vy2 + vz2 In analogy with the two-dimensional dot product, we can also define a threedimensional dot product by the formula v · w vx w x + vy w y + vz w z . This has all the same properties as the two-dimensional dot product. In particular, v · v |v| 2 and v · w |v| |w| cos θ EXAMPLE 2 Find the angle between the vectors v (1, 1, 0) and w (2, 1, 1) . SOLUTION Observe that |v| p √ 2, 12 + 12 + 02 |w| p 22 + 12 + 12 and v · w (1)(2) + (1)(1) + (0)(1) 3. Plugging these into the formula v · w |v| |w| cos θ gives √ √ 3 2 6 cos θ. Simplifying and solving for cos θ yields cos θ √ 3 , 2 and therefore θ 30◦ . Bases for R3 The standard basis for R3 consists of the vectors Note that the vectors i (1, 0, 0) and j (0, 1, 0) in R3 are three-component vectors, meaning that they are different from the standard basis vectors i (1, 0) and j (0, 1) in R2 . 1 i 0 , 0 0 j 1 , 0 0 and k 0 1 √ 6 THREE-DIMENSIONAL SPACE 3 Any vector in R3 can be written as a linear combination of the standard basis vectors: v vx i + vy j + vz k. More generally, three vectors a, b, c form a basis for R3 if every vector v can expressed as a linear combination of a, b, and c: v sa + tb + uc. In this case, the scalars s, t, u are called the components of v with respect to this basis. EXAMPLE 3 Find the components of v (12, 9, 3) with respect to the basis a (1, 2, 2) , b (2, 1, −2) , c (2, −2, 1) . SOLUTION We wish to express v as a linear combination of a, b, and c: 12 1 2 2 9 s 2 + t 1 + u −2 . 3 2 −2 1 This vector equation is equivalent to three scalar equations: 12 s + 2t + 2u 9 2s + t − 2u 3 2s − 2t + u With three equations and three unknowns, we ought to be able to solve for s, t, and u. Solving for u in the third equation gives u 3 − 2s + 2t. We substitute this formula for u into each of the first two equations: 12 s + 2t + 2 (3 − 2s + 2t ) 9 2s + t − 2 (3 − 2s + 2t ) Rearranging and simplifying yields 6 −3s + 6t 15 6s − 3t Solving gives s 4 and t 3, and hence u 1. Thus 12 1 2 2 9 4 2 + 3 1 + −2 . 3 2 −2 1 EXERCISES 1. Find the distance between the points (6, 3, 2) and (2, 7, 9) . 1 0 2. Find the angle between the vectors 1 and 1 . −4 1 THREE-DIMENSIONAL SPACE 4 3. Find a value of t such that (2, 3, t ) is orthogonal to 5i − 2j + 2k. 4. The following figure shows a rectangular box in R3 . (a) Find the coordinates of the point p. (b) What is the volume of the box? 5. The following figure shows two spheres in R3 that are tangent at a point r. The larger sphere is centered at p (−5, 0, 5) , and has a radius of 9. The smaller sphere is centered at q (5, 5, −5) . (a) What is the radius of the smaller sphere? (b) Find the coordinates of the point r at which the two spheres are tangent. 6. The following figure shows a right triangle in R3 . (a) Determine the value of t. (b) What is the area of the triangle? 7. Find the components of the vector (7, 5, 8) with respect to the basis (1, 1, 0) , (1, 0, 1) , (0, 1, 1) . 8. Find the components of the vector (2, −10, 7) with respect to the basis (−8, 4, 1) , (4, 7, 4) , (1, 4, 8) .