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Chapter 7: Vectors and the Geometry of Space

Chapter 7: Vectors and the Geometry of Space. Section 7.2 Space Coordinates and Vectors in Space. Written by Karen Overman Instructor of Mathematics Tidewater Community College, Virginia Beach Campus Virginia Beach, VA With Assistance from a VCCS LearningWare Grant.

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Chapter 7: Vectors and the Geometry of Space

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  1. Chapter 7: Vectors and the Geometry of Space Section 7.2 Space Coordinates and Vectors in Space Written by Karen Overman Instructor of Mathematics Tidewater Community College, Virginia Beach Campus Virginia Beach, VA With Assistance from a VCCS LearningWare Grant

  2. In this lesson you will learn: • 3 Space - The three-dimensional coordinate system • Points in space, ordered triples • The distance between two points in space • The midpoint between two points in space • The standard form for the equation of a sphere • Vectors in 3 Space • Different forms of vectors • Vector operations • Parallel vectors • Applications of vectors

  3. Three-Dimensional Space Previously you studied vectors in the Cartesian plane or 2-dimensions, now we are going to expand our knowledge of vectors to 3-dimensions. Before we discuss vectors, let’s look at 3-dimensional space. To construct a 3-dimensional system, start with a yz plane flat on the paper (or screen). Next, the x-axis is perpendicular through the origin. (Think of the x-axis as coming out of the screen towards you.) For each axis drawn the arrow represents the positive end. z y x

  4. z This is considered a right-handed system. To recognize a right-handed system, imagine your right thumb pointing up the positive z-axis, your fingers curl from the positive x-axis to the positive y-axis. y x In a left-handed system, if your left thumb is pointing up the positive z-axis, your fingers will still curl from the positive x-axis to the positive y-axis. Below is an example of a left-handed system. z x Throughout this lesson, we will use right-handed systems. y

  5. The 3-dimensional coordinate system is divided into eight octants. Three planes shown below separate 3 space into the eight octants. The three planes are the yz plane which is perpendicular to the x-axis, the xy plane which is perpendicular to the z-axis and the xz plane which is perpendicular to the y-axis. Think about 4 octants sitting on top of the xy plane and the other 4 octants sitting below the xy plane. z y yz plane x z z xy plane y y xz plane x x

  6. Plotting Points in Space Every position or point in 3-dimensional space is identified by an ordered triple, (x, y, z). Here is one example of plotting points in 3-dimensional space: z P (3, 4, 2) y The point is 3 units in front of the yz plane, 4 points in front of the xz plane and 2 units up from the xy plane. x

  7. Here is another example of plotting points in space. In plotting the point Q (-3,4,-5) you will need to go back from the yz plane 3 units, out from the xz plane 4 units and down from the xy plane 5 units. z y Q (-3, 4, -5) x As you can see it is more difficult to visualize points in 3 dimensions.

  8. Distance Between Two Points in Space The distance between two points in space is given by the formula: Take a look at the next two slides to see how we come up with this formula.

  9. Consider finding the distance between the two points, . It is helpful to think of a rectangular solid with P in the bottom back corner and Q in the upper front corner with R below it at . Using two letters to represent the distance between the points, we know from the Pythagorean Theorem that PQ² = PR²+ RQ² Q P R Using the Pythagorean Theorem again we can show that PR² = Note that RQ is .

  10. Starting with PQ² = PR²+ RQ² Make the substitutions: PR² = and RQ = Thus, PQ² = Or the distance from P to Q, PQ = Q P R That’s how we get the formula for the distance between any two points in space.

  11. We will look at example problems related to the three-dimensional coordinate system as we look at the different topics. Example 1: Find the distance between the points P(2, 3, 1) and Q(-3,4,2). Solution: Plugging into the distance formula:

  12. Example 2: Find the lengths of the sides of triangle with vertices (0, 0, 0), (5, 4, 1) and (4, -2, 3). Then determine if the triangle is a right triangle, an isosceles triangle or neither. Solution: First find the length of each side of the triangle by finding the distance between each pair of vertices. (0, 0, 0) and (5, 4, 1) (0, 0, 0) and (4, -2, 3) (5, 4, 1) and (4, -2, 3) These are the lengths of the sides of the triangle. Since none of them are equal we know that it is not an isosceles triangle and since we know it is not a right triangle. Thus it is neither.

  13. The Midpoint Between Two Points in Space The midpoint between two points, is given by: Each coordinate in the midpoint is simply the average of the coordinates in P and Q. Example 3: Find the midpoint of the points P(2, 3, 0) and Q(-4,4,2).

  14. Equation of a Sphere A sphere is the collection of all points equal distance from a center point. To come up with the equation of a sphere, keep in mind that the distance from any point (x, y, z) on the sphere to the center of the sphere, is the constant r which is the radius of the sphere. Using the two points (x, y, z), and r, the radius in the distance formula, we get: If we square both sides of this equation we get: The standard equation of a sphere is where r is the radius and is the center.

  15. Example 4: Find the equation of the sphere with radius, r = 5 and center, (2, -3, 1). Solution: Just plugging into the standard equation of a sphere we get: Example 5: Find the equation of the sphere with endpoints of a diameter (4, 3, 1) and (-2, 5, 7). Solution: Using the midpoint formula we can find the center and using the distance formula we can find the radius. Thus the equation is:

  16. Example 6: Find the center and radius of the sphere, . Solution: To find the center and the radius we simply need to write the equation of the sphere in standard form, . Then we can easily identify the center, and the radius, r. To do this we will need to complete the square on each variable. Thus the center is (2, -3, -4) and the radius is 6.

  17. Vectors in Three-Dimensional Space Now that we have an understanding of the three-dimensional system, we are ready to discuss vectors in the three-dimensional system. All the information you learned about vectors in the previous lesson will apply, only now we will add in the third component. Vectors in component form in three dimensions are written as ordered triples, in other words, now a vector in component form is . In three dimensions the zero vector is O = < 0, 0, 0> and the standard unit vectors are . z Each of the unit vectors represents one unit of change in the direction of each of their respective positive axes. y x

  18. Given the initial point, and the terminal point, , the component form of the vector can be found the same way it was on the Cartesian Plane. Component form of a vector Be sure to subtract the initial point’s coordinates from the terminal point’s coordinates. The same vector can be written as a combination of the unit vectors. Standard Unit Vector Notation We will look at examples using both forms.

  19. More on Vectors in Three-Dimensions Let and let c be a scalar. • Vector Equality: • Magnitude or Length of a Vector: • Vector Addition: • Scalar Multiplication: • Unit Vector in the Direction of : Note: This is simply the vector multiplied by the reciprocal of its magnitude.

  20. Let’s look at some example problems involving vectors. Example 1: Sketch the vector with initial point P(2, 1, 0) and terminal point Q(3, 5, 4). Then find the component form of the vector, the standard unit vector form and a unit vector in the same direction. Solution: First draw a 3D system and plot P and Q. The vector connects P to Q. Q P

  21. Example 1 Continued: Second, find the component form of the vector. Do this by subtracting the initial point’s coordinates from the terminal point’s coordinates. Component form Standard Unit Vector Form Last, find a unit vector in the same direction. Do this by multiplying the vector by the reciprocal of the magnitude. Note: You can verify it’s a unit vector by finding its magnitude.

  22. Example 2: Given the vectors find the following: a. b. c. Solution: a. b. c.

  23. Parallel Vectors You may recall from the previous section that a nonzero scalar multiple of a vector has the same direction as the vector (positive scalar) or the opposite direction as the vector (negative scalar). Since this is the case, any nonzero scalar multiple of a vector is considered a parallel vector. In other words, if two vectors, , are parallel, then there exists some scalar, c such that . The zero vector does not have direction so it cannot be parallel. To get the idea, look at these vectors on the Cartesian Plane. y x

  24. Example 3: Determine if the vector with initial point, P(3,2,-2) and terminal point, Q(7,5,1-3) is parallel to the vector . Solution: First find the component form of the vector from P to Q. Second, if the two vectors are parallel, then there exists some scalar, c, such that Then –12 = 4c c = -3 And -9 = 3c c = -3 And 3 = -c c = -3 For the two vectors to be parallel, c would have to be the same for each coordinate. Since it is, the two vectors are parallel.

  25. Example 4: Determine whether the points A(2,3,-1), B(0,1,3) and C(-3,-2,8) are collinear. Solution: We need to find two vector between the three points and determine if they are parallel. If the two vectors are parallel and pass through a common point then the three points must be in the same line. The vector from A to B is Now we need to find the vector from A to C or B to C. The vector from A to C is To be parallel: -2 = -5c c = 2/5 -2 = -5c c = 2/5 4 = 9c c = 4/9 Since c is not the same in each case, the vectors are not parallel and the points are not collinear.

  26. Example 5: Find a vector parallel to the vector with magnitude 5. Solution: Be careful. We might quickly assume that all we need to do is to multiply the vector by 5. This would be fine if we were dealing with a unit vector. Since we are not, we need to multiply by the reciprocal of the magnitude first to get a unit vector and then multiply by 5.

  27. Solution to Example 5 Continued: You can verify the new vector is parallel if you look at the form: Obviously the scalar multiple is . You can verify the magnitude is 5 by finding the magnitude of the form:

  28. Example 6: The weight of an 80lb. chandelier hanging 2.5 feet from the ceiling is distributed over 3 chains. If the chains are located as shown below, represent the force exerted on each chain with a vector. (-1,-1,0) 1 ft 1 ft 1 ft (0,1,0) (1,-1,0) 1 ft (0,0,-2.5)

  29. Solution to Example 6: First find the vectors from the chandelier to the three points on the ceiling. Each force is a multiple of the vector since we can find the direction, but we don’t know the magnitude. (-1,-1,0) 1 ft 1 ft 1 ft (0,1,0) (1,-1,0) 1 ft (0,0,-2.5)

  30. Solution to Example 6 Continued: The sum of the three forces must negate the downward force of the chandelier from its weight. So, This gives us a system of three equations in three unknowns, a,b and c. Solving the system, you get a = 16, b = 8 and c = 8. Thus the three forces are

  31. You can find your practice problems for this lesson in Blackboard under the Assignments button under Lesson 7.2.

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