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**14 Vectors in Three-dimensional Space Case Study**

14.1 Vectors in Three-dimensional Rectangular Coordinate System 14.2 Vector Product and Scalar Triple Product Chapter Summary

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Case Study Captain, please follow the following flight route. A plane is approaching Hong Kong International Airport. The flight- control operator of the control tower is trying to give instructions to the pilot to provide a safe route for landing. We are now ready for landing. Please indicate a flight route. To describe the position and the route of the plane, we can introduce the three-dimensional rectangular coordinate system as shown in the figure. Let the airport be the origin of the coordinate system. Then we use the triplet (x, y, z) to describe the horizontal position (x, y as in the two-dimensional case) and the height of the plane (z). The plane is located at the point A(2, 1, 4), B(−1, −2, 3), C(2, 0, 2), D(1, 1, 1) are points in space, such that the plane follows the route A ® B ® C ® D ® O.

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**14.1 Vector in Three-dimensional Rectangular Coordinate System**

A. Vectors in Three-dimensional Space For any two points A and B in space, the directed line segment from A to B is called the vector from A to B, and is denoted by The magnitude of is denoted by , which is the same as the vectors on the plane defined before. The addition, subtraction, scalar multiplication, negative, parallelism of vectors, and the rules of operations of vectors are also defined in the same way as in the case of plane vectors. For example, for the cube ABCDEFGH, we have 1. (equal vectors) 2. (negative vectors) 3. (parallel vectors) 4. (addition of vectors ) 5. (subtraction of vectors)

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**Example 14.1T 14.1 Vector in Three-dimensional**

Rectangular Coordinate System A. Vectors in Three-dimensional Space Example 14.1T The figure shows a cube ABCDEFGH. Let = a, = b and = c. Express the following in terms of a, b and c. (a) (b) (c) Solution: (a) (b) (c)

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**Example 14.2T 14.1 Vector in Three-dimensional**

Rectangular Coordinate System A. Vectors in Three-dimensional Space Example 14.2T The figure shows a cube ABCDEFGH. Prove that Solution:

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**14.1 Vector in Three-dimensional Rectangular Coordinate System**

B. Representation of Vectors in Three-dimensional Rectangular Coordinate System The three-dimensional Cartesian coordinate system R3 consists of three mutually perpendicular axes: x, y and z. The directions of these axes are aligned in such a way that they obey the right-hand rule. If the x- and y-axes are represented by the index finger and the middle finger respectively, then the thumb represents the z-axis.

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**14.1 Vector in Three-dimensional Rectangular Coordinate System**

B. Representation of Vectors in Three-dimensional Rectangular Coordinate System Every point in space can be represented in the three-dimensional coordinate system by the triplet (x, y, z), where x, y and z represent the directed distances from the yz-, zx- and xy-planes respectively. These three values are called x-, y- and z-coordinates of the point respectively. The point of intersection of the three axes is called the origin O and its coordinates are (0, 0, 0). In the figure, i, j and k are the unit vectors in the positive directions of x-, y- and z-axes respectively. They have a common starting point at the origin, and their terminal points are (1, 0, 0), (0, 1, 0) and (0, 0, 1) respectively. P(x, y, z) is a point in R3, so we can express the position vector as By Pythagoras’ theorem, we have

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**14.1 Vector in Three-dimensional Rectangular Coordinate System**

B. Representation of Vectors in Three-dimensional Rectangular Coordinate System Now if two points A(x1, y1, z1) and B(x2, y2, z2) in R3 are given, the vector from A to B can be found by subtracting the position vector from which is the same as we did in the case of R2, then we have

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**Example 14.3T 14.1 Vector in Three-dimensional**

Rectangular Coordinate System B. Representation of Vectors in Three-dimensional Rectangular Coordinate System Example 14.3T Given two points P(–3, –2, 8) and Q(0, –5, 4). Find the unit vector in the direction of Solution: \ Unit vector

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**14.1 Vector in Three-dimensional Rectangular Coordinate System**

B. Representation of Vectors in Three-dimensional Rectangular Coordinate System Consider two vectors pi + qj + rk and si + tj + uk in R3. As i, j and k are non parallel vectors, we have Property 14.1 pi + qj + rk = si + tj + uk if and only if p = s, q = t and r = u, and (b) pi + qj + rk = 0 if and only if p = q = r = 0.

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**Example 14.4T 14.1 Vector in Three-dimensional**

Rectangular Coordinate System B. Representation of Vectors in Three-dimensional Rectangular Coordinate System Example 14.4T Given three points A(–3, 1, 5), B(2, 5, –1) and C(–6, 4, 3). Find the coordinates of a point D if (a) ABCD forms a parallelogram, (b) ABDC forms a parallelogram. Solution: Let the coordinates of D be (x, y, z). (a) If ABCD forms a parallelogram, \ We have \ The coordinates of D are (11, 0, 9).

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**Example 14.4T 14.1 Vector in Three-dimensional**

Rectangular Coordinate System B. Representation of Vectors in Three-dimensional Rectangular Coordinate System Example 14.4T Given three points A(–3, 1, 5), B(2, 5, –1) and C(–6, 4, 3). Find the coordinates of a point D if (a) ABCD forms a parallelogram. (b) ABDC forms a parallelogram. Solution: (b) If ABDC forms a parallelogram, \ We have \ The coordinates of D are (1, 8, 3).

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**Example 14.5T 14.1 Vector in Three-dimensional**

Rectangular Coordinate System B. Representation of Vectors in Three-dimensional Rectangular Coordinate System Example 14.5T Consider the three vectors a = 3i – 4j + 2k, b = i – 3j – k and c = 5i + 2j + k. If m = –16i – 8j – 7k, express m in terms of a, b and c. Solution: Let m = a a + b b + g c. Consider the determinant of the coefficient matrix: By Cramer’s rule,

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**Example 14.6T 14.1 Vector in Three-dimensional**

Rectangular Coordinate System B. Representation of Vectors in Three-dimensional Rectangular Coordinate System Example 14.6T Given two points A and B with = 3i – 8j + 5k and = 6i + 2j – 7k. C is a point on the line segment AB. Find if (a) C is the mid-point of AB, (b) C divides AB in the ratio 2 : 1. Solution: (a) (b)

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**14.1 Vector in Three-dimensional Rectangular Coordinate System**

C. Scalar Product In the three-dimensional rectangular coordinate system R3, as the unit base vectors i, j and k are mutually perpendicular, we have i i = j j = k k = 1 i j = j i = 0 j k = k j = 0 i k = k i = 0 We also have the following properties of scalar product: a b = b a a (b + c) = a b + a c (ka) b = k(a b) = a (kb) If a = x1i + y1j + z1k and b = x2i + y2j + z2k are two non-zero vectors, then a b = x1x2 + y1y2 + z1z2, where is the angle between a and b.

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**Example 14.7T 14.1 Vector in Three-dimensional**

Rectangular Coordinate System C. Scalar Product Example 14.7T Two vectors r = 2i + 3j – k and s = i + 2k are given. (a) Find the value of r s. (b) Hence find the angle between r and s. Solution: (a) (b) \ The angle between r and s is 90.

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**Example 14.8T 14.1 Vector in Three-dimensional**

Rectangular Coordinate System C. Scalar Product Example 14.8T If the vectors ci + 5j – 3k and 2ci + cj + k are perpendicular to each other, find the value(s) of c. Solution: ∵ ci + 5j – 3k and 2ci + cj + k are perpendicular to each other.

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**Example 14.9T 14.1 Vector in Three-dimensional**

Rectangular Coordinate System C. Scalar Product Example 14.9T Given three points A(2, 1, 6), B(– 5, 3, 5) and C(0, –6, 5) are vertices of DABC. Solve DABC. (Give the answers in surd form or correct to the nearest degree.) Solution: (cor. to the nearest degree)

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**Example 14.9T 14.1 Vector in Three-dimensional**

Rectangular Coordinate System C. Scalar Product Example 14.9T Given three points A(2, 1, 6), B(– 5, 3, 5) and C(0, –6, 5) are vertices of DABC. Solve DABC. (Give the answers in surd form or correct to the nearest degree.) Solution: (cor. to the nearest degree) (cor. to the nearest degree)

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**Example 14.10T 14.1 Vector in Three-dimensional**

Rectangular Coordinate System C. Scalar Product Example 14.10T Given two vectors x = 6j – 5k and y = 3i – 4j. (a) Find the angle between x and y, correct to the nearest degree. (b) Find the length of the projection of y on x. Solution: (a) Let be the angle between x and y. (cor. to the nearest degree) \ The angle between x and y is 128. (b) The length of the projection of y on x

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**14.2 Vector Product and Scalar Triple Product**

A. Definition of Vector Product Suppose we have two non-zero vectors a and b in the three-dimensional space. The vector product of a and b, denoted by a b, is the vector which is perpendicular to both a and b, with the magnitude equal to |a b| = |a||b|sin, where is the angle between a and b (with 0° ≤ ≤ 180°). Particularly, the direction of a b is defined in such a way that a, b and a b always obey the right-hand rule. In conclusion, a b = |a||b|sin , where is the angle between a and b, and is a unit vector whose direction is defined by the right-hand rule. Note: 1. a b is read as ‘a cross b’. Therefore the vector product is also called the cross product. 2. The vector product is only defined in the three-dimensional space. 3. In contrast to the scalar product of two vectors, the vector product is a vector while the scalar product is a scalar.

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**14.2 Vector Product and Scalar Triple Product**

A. Definition of Vector Product For and two non-zero vectors a and b, a b = 0 if and only if a and b are parallel to each other. In particular, if b = a, we have a a = 0. For the unit vectors i, j and k: i i = j j = k k = 0 i j = k j i = k j k = i k j = i k i = j i k = j

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**14.2 Vector Product and Scalar Triple Product**

B. Properties of Vector Product Property 14.2 Properties of Vector Product (a) b a = (a b) (b) (a + b) c = a c + b c (c) a (b + c) = a b + a c (d) (ka) b = a (kb) = k(a b) (e) |a b|2 = |a|2|b|2 – (a b)2 Proof of (a): \ b a = (a b)

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**14.2 Vector Product and Scalar Triple Product**

B. Properties of Vector Product Proof of (d): If k = 0 or a = 0 or b = 0, then (ka) × b = a × (kb) = (k a × b) = 0. Assume that k ¹ 0 and a and b are non-zero. Let q be the angle between a and b, and be the unit vector in the direction of a × b. When k > 0, When k < 0, Similarly, it can be proved that a × (kb) = k(a × b). \ (ka) × b = a × (kb) = k(a × b)

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**14.2 Vector Product and Scalar Triple Product**

B. Properties of Vector Product Proof of (e): Since |a × b| = |a||b|sin q, |a × b|2 = (|a||b|sin q)2 = |a|2|b|2sin2q = |a|2|b|2 − |a|2|b|2cos2q sin2q = 1 – cos2q = |a|2|b|2 – (a b)2 a – b = |a||b|cos q

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**14.2 Vector Product and Scalar Triple Product**

C. Calculation of Vector Product We can use the determinant to represent the vector product: If a = x1i + y1j + z1k and b = x2i + y2j + z2k, then

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**Example 14.11T 14.2 Vector Product and Scalar Triple Product Solution:**

C. Calculation of Vector Product Example 14.11T For the following pairs of vectors m and n, find the vector products m × n. (a) m = 3i + 8j, n = 6k (b) m = –4i + 2j + 6k, Solution: (a) (b)

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**Example 14.12T 14.2 Vector Product and Scalar Triple Product Solution:**

C. Calculation of Vector Product Example 14.12T P, Q and R are three points with position vectors i + j + k, –2j and –i + 3j – k respectively. Find the unit vectors which are perpendicular to and Solution: Unit vectors which are perpendicular to and

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**14.2 Vector Product and Scalar Triple Product**

D. Applications of Vector Product Consider a parallelogram ABCD. Area of the parallelogram ABCD = Since the area of ABD is half that of parallelogram ABCD, we can obtain a formula for the area of triangle: Area of ABD = The above formula can be further rewritten as Area of ABD =

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**Example 14.13T 14.2 Vector Product and Scalar Triple Product Solution:**

D. Applications of Vector Product Example 14.13T Find the area of the triangle formed by vertices X(2, 1, 1), Y(0, –1, 0) and Z(–2, 1, –1). Solution: Area of XYZ

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**14.2 Vector Product and Scalar Triple Product**

E. Scalar Triple Product The volume of a parallelepiped (a prism with all faces are parallelograms) with sides a, b and c is given by |a (b × c)|. The expression a (b c) is called the scalar triple product of a, b and c. Since the scalar product of two vectors is a scalar, thus a (b c) is a scalar as the name suggests. In the three-dimensional rectangular coordinate system, suppose a = x1i + y1j + z1k, b = x2i + y2j + z2k and c = x3i + y3j + z3k, then Note: If a (b c) = 0, the volume of the parallelepiped with sides a, b and c equals zero. This only when a, b and c are coplanar.

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**14.2 Vector Product and Scalar Triple Product**

E. Scalar Triple Product Since a determinant is unchanged when interchanging the rows twice and for any non-zero vectors x and y, x y = y x, we have the following properties of the scalar triple product: Property 14.3 Properties of Scalar Triple Product (a) (a b) c = a (b c) (b) a (b c) = b (c a) = c (a b)

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**Example 14.14T 14.2 Vector Product and Scalar Triple Product Solution:**

E. Scalar Triple Product Example 14.14T If p = 2i + j + 3k, q = 3i – j – 2k and r = –i + 2j – k, find (a) r × p, and (b) q (r × p). Solution: (a) (b)

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**Example 14.15T 14.2 Vector Product and Scalar Triple Product Solution:**

E. Scalar Triple Product Example 14.15T Consider A(2, 1, 0), B(–3, 4, 5), C(0, –2, 4) and D(1, 2, 5). Find the volume of the parallelepiped with sides , and Solution: \ Volume of the parallelepiped

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**Chapter Summary 14.1 Vectors in Three-dimensional**

Rectangular Coordinate System 1. Every point in the space can be represented in the three-dimensional coordinate system by the triplet (x, y, z), where x, y and z represent the directed distances from the yz-, zx- and xy-planes respectively. 2. The distance between two points A(x1, y1, z1) and B(x2, y2, z2) is given by

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**Chapter Summary 14.1 Vectors in Three-dimensional**

Rectangular Coordinate System 1. The rules of operations and properties of vectors in the space are the same as vectors on a plane. 2. In R3, we define three mutually perpendicular unit vectors i, j and k, which point in the positive direction of x-, y- and z-axes respectively. For a point P(x, y, z) in R3, the position vector can be expressed as , where

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**Chapter Summary 14.1 Vectors in Three-dimensional**

Rectangular Coordinate System Scalar Product If a = x1i + y1j + z1k and b = x2i + y2j + z2k, are two non-zero vectors, then where q is the angle between a and b.

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**Chapter Summary 14.2 Vector Product and Scalar Triple Product**

1. If a = x1i + y1j + z1k and b = x2i + y2j + z2k, are non-zero vectors and q is the angle between them, then 2. Area of DABC

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**Chapter Summary 14.2 Vector Product and Scalar Triple Product**

1. If a = x1i + y1j + z1k and b = x2i + y2j + z2k and c = x3i + y3j + z3k are non-zero vectors, then 2. Volume of the parallelepiped with sides a, b and c = |a (b c)|.

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**Follow-up 14.1 14.1 Vector in Three-dimensional**

Rectangular Coordinate System A. Vectors in Three-dimensional Space Follow-up 14.1 The figure shows a cube ABCDEFGH. Let = a, = b and = c. Express the following in terms of a, b and c. (a) (b) (c) Solution: (a) (b) (c)

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**Follow-up 14.2 14.1 Vector in Three-dimensional**

Rectangular Coordinate System A. Vectors in Three-dimensional Space Follow-up 14.2 The figure shows a tetrahedron ABCD. M and N are the mid-points of BC and AD respectively. Prove that Solution:

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**Follow-up 14.3 14.1 Vector in Three-dimensional**

Rectangular Coordinate System B. Representation of Vectors in Three-dimensional Rectangular Coordinate System Follow-up 14.3 Given two points M(–1, 4, 2) and N(3, 1, –3). Find the unit vector in the direction of Solution: \ Unit vector

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**Follow-up 14.4 14.1 Vector in Three-dimensional**

Rectangular Coordinate System B. Representation of Vectors in Three-dimensional Rectangular Coordinate System Follow-up 14.4 Given three points A(1, 4, −2), B(6, 3, 5) and C(−3, 6, 3). Find the coordinates of a point D if ABCD forms a parallelogram. Solution: Let the coordinates of D be (x, y, z). If ABCD forms a parallelogram, \ We have The coordinates of D are (8, 7, 4).

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**Follow-up 14.5 14.1 Vector in Three-dimensional**

Rectangular Coordinate System B. Representation of Vectors in Three-dimensional Rectangular Coordinate System Follow-up 14.5 Given three vectors a = – 2i + 4j + k, b = i – 2j + 5k and c = 4i + j – 3k. If r = 14i + 8j, express r in terms of a, b and c. Solution: Let r = a a + b b + g c. Consider the determinant of the coefficient matrix: By Cramer’s rule,

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**Follow-up 14.6 14.1 Vector in Three-dimensional**

Rectangular Coordinate System B. Representation of Vectors in Three-dimensional Rectangular Coordinate System Follow-up 14.6 Given two points A and B with = –5i + 2j – 4k and = 3i + 8j + 8k. C is a point on the line segment AB. Find if (a) C is the mid-point of AB, (b) C divides AB in the ratio 3 : 2. Solution: (a) (b)

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**Follow-up 14.7 14.1 Vector in Three-dimensional**

Rectangular Coordinate System C. Scalar Product Follow-up 14.7 Given two vectors m = i – 2j + 4k and n = 3i + 4j + 3k. (a) Find the value of m n. (b) Hence find the angle between m and n, correct to the nearest degree. Solution: (a) (b) Let be the angle between m and n. (cor. to the nearest degree) \ The angle between m and n is 75.

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**Follow-up 14.8 14.1 Vector in Three-dimensional**

Rectangular Coordinate System C. Scalar Product Follow-up 14.8 If the vectors bi – 3j + b2k and i + 2bj – 3k are perpendicular to each other, find the value(s) of b. Solution: ∵ and are perpendicular to each other.

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**Follow-up 14.9 14.1 Vector in Three-dimensional**

Rectangular Coordinate System C. Scalar Product Follow-up 14.9 Three points A(−3, 3, 4), B(0, −5, 2) and C(4, 1, 7) are vertices of DABC. Solve DABC. (Give the answers in surd form or correct to the nearest degree.) Solution:

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**Follow-up 14.9 14.1 Vector in Three-dimensional**

Rectangular Coordinate System C. Scalar Product Follow-up 14.9 Three points A(−3, 3, 4), B(0, −5, 2) and C(4, 1, 7) are vertices of DABC. Solve DABC. (Give the answers in surd form or correct to the nearest degree.) Solution: (cor. to the nearest degree) (cor. to the nearest degree) (cor. to the nearest degree)

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**Follow-up 14.10 14.1 Vector in Three-dimensional**

Rectangular Coordinate System C. Scalar Product Follow-up 14.10 Given two vectors a = i + 3j + 5k and b = 2i + 4j + 6k. (a) Find the angle between a and b, correct to the nearest degree. (b) Find the length of the projection of a on b. Solution: (a) Let be the angle between a and b. (cor. to the nearest degree) (b) The length of the projection of a on b

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**Follow-up 14.11 14.2 Vector Product and Scalar Triple Product**

C. Calculation of Vector Product Follow-up 14.11 For the following pairs of vectors p and q, find the vector products p × q. (a) p = 3i + j + 2k, q = i + 4j – k (b) p = i – 4j – 6k, q = 2i + 5k (c) p = 2i – 5j + k, q = –6i + 15j – 3k Solution: (a) (b) (c)

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**Follow-up 14.12 14.2 Vector Product and Scalar Triple Product**

C. Calculation of Vector Product Follow-up 14.12 X, Y and Z are three points with position vectors i – k, 3i + j – 2k and – j + k respectively. Find the unit vectors which are perpendicular to and Solution: Unit vectors which are perpendicular to and

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**Follow-up 14.13 14.2 Vector Product and Scalar Triple Product**

D. Applications of Vector Product Follow-up 14.13 Find the area of the triangle with vertices P(0, 1, –1), Q(–2, –1, 1) and R(1, –2, 0). Solution: Area of PQR

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**Follow-up 14.14 14.2 Vector Product and Scalar Triple Product**

E. Scalar Triple Product Follow-up 14.14 If x = 3i + j + k, y = –i – j + 3k and z = 2i + 2j – 2k, find (a) x × y, and (b) (x × y) z. Solution: (a) (b)

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**Follow-up 14.15 14.2 Vector Product and Scalar Triple Product**

E. Scalar Triple Product Follow-up 14.15 The coordinates of P, Q, R and S are (0, 0, 2), (3, 0, 4), (1, 2, 3) and (0, 1, –2) respectively. Find the volume of the parallelepiped with sides , and Solution: \ Volume of the parallelepiped

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©Brooks/Cole, 2001 Chapter 12 Derived Types-- Enumerated, Structure and Union.

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