The Three-Dimensional Coordinate System. We represent any point in space by an ordered triple of real numbers.

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1 Study Sheet (12.1) The Three-Dimensional Coordinate System To locate a point in space, numbers are required. We represent any point in space by an ordered triple of real numbers. In order to represent points in space, we first choose: A fixed point O (the origin) Three directed lines through O that are perpendicular to each other. Coordinate axes They are labeled: x-axis, y-axis, and z-axis The x- and y-axes as being horizontal The z-axis as being vertical Right-hand rule The direction of the z-axis is determined by the right-hand rule. Curl the fingers of your right hand around the z-axis in the direction of a 90 counterclockwise rotation from the positive x-axis to the positive y-axis. Then, your thumb points in the positive direction of the z-axis. Coordinate Planes The three coordinate axes determine the three coordinate planes. The xy-plane contains the x- and y-axes. The yz-plane contains the y- and z-axes. The xz-plane contains the x- and z-axes. Octants These three coordinate planes divide space into eight parts, called octants. The octants (quadrants) are numbered by rotating counterclockwise around the positive z-axis. There are 8 octants.

2 Plotting Points in Space We represent the point P by the ordered triple of real numbers (a, b, c). We call a, b, and c the coordinates of P. a is the x-coordinate. b is the y-coordinate. c is the z-coordinate. Example: Plot the points: (2,-3,3), (-2,6,2), (1,4,0), and (2,2,-3). Draw a sideways x, then put a perpendicular line through the origin. The point P(a, b, c) determines a rectangular box. If we drop a perpendicular from P to the xy-plane, we get a point Q with coordinates (a, b, 0). This is called the projection of P on the xy-plane. Similarly, R(0, b, c) and S(a, 0, c) are the projections of P on the yz-plane and xz-plane, respectively. Cartesian Product R x R x R = {(x, y, z) x, y, z R} is the set of all ordered triples of real numbers and is denoted by R 3. Three Dimensional Rectangular Coordinate System: Ordered triples (a, b, c) in R 3 In 3-D analytic geometry, an equation in x, y, and z represents a surface in R 3. Note: When an equation is given, we must understand from the context whether it represents either: A curve in R 2 A surface in R 3 Example 1: What surfaces in R 3 are represented by the following equations? (a) z = 3 (b) y = 5

3 Note: In general, if k is a constant, then x = k represents a plane parallel to the yz-plane. y = k is a plane parallel to the xz-plane. z = k is a plane parallel to the xy-plane. Example 2: (a) Which points (x, y, z) satisfy the equations x 2 + y 2 =1 and z = 3 (b) What does the equation x 2 + y 2 =1represent as a surface in 3 R? Example 2: Describe and sketch the surface in 3 R represented by the equation y = x. Distance Formuls In Three Dimensions Recall the distance formula in R 2. Given two points x, y ) and ( x, y ), d = ( It looks the same in space as it did before except with a third coordinate: d =

4 Example 3: Find the distance between (1, 0, 2) and (2, 4,-3). Midpoint Formula In Three Dimensions So the midpoint is just the average of the x s, y s, and z s. Example 4: Find the midpoint of the line segment joining (5, -2, 3) and (0, 4, 4). Equation of a Sphere Recall: The equation of a circle is x 2 + y 2 = r If the center is not at the origin, then the equation is ( x h) + ( y k) = r. The equation of a sphere whose center is at ( h, k, l) with radius r is Finding the Equation of a Sphere Example 5: Find the standard equation of a sphere with center (2,4,3) and radius 3 Example 6: Find the center and radius of the sphere given by x 2 + y 2 + z 2 2x + 4y 6z +8 = 0 Example 7: Finding the Center and Radius of a Sphere (a) x 2 + y 2 + z 2 2x + 4y 6z +8 = 0 (b) (x-1) 2 + (y+2) 2 + (z-3) 2 = 6 (c) The center is (1,-2,3) and the radius is 6. (d)

5 Study Sheet (12.2) Vectors The term vector is used by scientists to indicate a quantity (such as displacement or velocity or force) that has both. A vector is represented by, which is just a line segment with an arrow indicating a direction. (Vectors are used for representing quantities such as.) We denote a vector by either: or. The length of the arrow represents of the vector. The arrow points in the direction of the vector. Quantities that we measure that have magnitude but not direction are called. Example: Consider driving down the road in your car. The speedometer gives the (scalar) speed that you are driving at any instant. But also, you are driving in a particular direction. So in any instant, your movement down the road is described by both your speed and your direction. v = length of v = speed of car Vectors are equal if they have the same. Quantities that we measure that have magnitude but not direction are called. The zero vector, denoted by 0, has length 0. (It is the only vector with no specific direction.) Example: Suppose a particle moves along a line segment from point A to point B. The corresponding displacement vector v has initial point A (the tail) and terminal point B (the tip). We indicate this by writing v = AB. Standard Position: A vector is in standard position if the initial point is at the origin. Component Form: The component form of this vector is: v = v, v. 1 2 Magnitude: The magnitude (length) of v = v1, v2 is:. Unit Vector:. Zero Vector: The zero vector, denoted by, has length.

6 Resulting Displacement Vector: The resulting displacement vector AC is called the sum of AB and BC. We write: AC = AB + BC Vector Addition: The definition of vector addition is illustrated here. Triangle Law: Parallelogram Law: Example 1: Draw the sum of the vectors a and b shown here. Multiplying Scalars: We multiply a vector by a scalar as follows. If c is a scalar and v is a vector, the scalar multiple cv is: The vector whose length is c times the length of v and whose direction is the same as v if c > 0 and is opposite to v if c < 0. If c = 0 or v = 0, then cv = 0. Note: Two nonzero vectors are parallel if they are scalar multiples of one another. The vector v = ( 1)v has the same length as v but points in the opposite direction. We call it the negative of v. Difference of two vectors: By the difference u v of two vectors, we mean:. By the Parallelogram Law, we can construct u v by first drawing the negative of v, v, and then adding it to u. We also could construct u by means of the Triangle Law. Example 2: If a and b are the vectors shown here, draw a 2b.

7 Components in Two- or Three-dimensional: Consider the vectors shown here. What do they have in common?!!!" The particular representation OP from the origin to the point P(3, 2) is called of the point P. Position Vector: In three dimensions, the vector!!!" a = OP = a 1, a 2, a 3 is the position vector of the point P(a 1, a 2, a 3 ). Given the points A(x 1, y 1, z 1 ) and B(x 2, y 2, z 2 ), the vector a with representation a =!!!" AB is: Example 3: Find the vector represented by the directed line segment with initial point A(2, 3, 4) and terminal point B( 2, 1, 1).

8 Length of Vector: The magnitude or length of the vector v is the length of any of its representations. The length of the two-dimensional (2-D) vector a = a 1, a 2 is: Unit Vectors: A unit vector is a vector whose length is. 2-D Algebraic Vectors: If a = a 1, a 2 and b = b 1, b 2, then Properties of Vectors: Standard Basis Vectors: Two vectors i, and j are called the standard basis vectors. They have length 1 and point in the directions of the positive x- and y-axes. And i and j are unit vectors. If a = a 1, a 2,, then we can write: i = j = In two dimensions, we can write: a = Example 4: If a = i + 2j and b = 4i + 7j, express the vector 2a + 3b in terms of I and j.

9 In general, if a 0, then the unit vector that has the same direction as a is: u = In order to verify this, we let c = 1/ a. Then, u = ca and c is a positive scalar; so, u has the same direction as a. Also, u = Example 5: Find the unit vector in the direction of the vector 2i j. Direction Angles If u is a unit vector such that θ is the angle (measured counterclockwise) from the positive x-axis to u, the terminal point of u lies on the unit circle and you have u = x, y = cosθ, sinθ = (cosθ )i + (sinθ )j The angle θ is the direction angle of the vector u. Given Unit Vector u = <cosθ, sinθ> = cosθ i + sinθ j, Vector v = is a scalar multiple of u v = v <cosθ, sinθ> = v (cosθ i + sinθ j) = v cosθ i + v sinθ j If v = <v 1, v 2 > tanθ = sin cos θ θ => v v sin cos θ θ = v v 2 1

10 Notation: We denote: V 2 as the set of all 2-D vectors V 3 as the set of all 3-D vectors V n as the set of all n-dimensional vectors: An n-dimensional vector is an ordered n-tuple a = a 1, a 2,, a n where a 1, a 2,, a n are real numbers that are called the components of a. (Addition and scalar multiplication are defined in terms of components just as for the cases n = 2 and n = 3.) 3-D Algebraic Vectors: If a = a 1, a 2, a 3 and b = b 1, b 2, b 3, then Example 6: If a = 4, 0, 3 and b = 2, 1, 5, find: a and the vectors a + b, a b, 3b, 2a + 5b Three vectors in V 3 play a special role. Standard Basis Vectors: These vectors i, j, and k are called the standard basis vectors. They have length 1 and point in the directions of the positive x-, y-, and z-axes. And i, j, and k are all unit vectors. If a = a 1, a 2, a 3, then we can write: a = Example 6: If a = i + 2j 3k and b = 4i + 7k, express the vector 2a + 3b in terms of i, j, and k. Example 7: Find the unit vector in the direction of the vector 2i j 2k.

11 Study Sheet (12.3) Dot Product In 1773, Joseph Lagrange was working on a problem involving tetrahedrons, and he came up with two ways to multiply vectors. We call these two methods the dot product and the cross product. The dot product is useful for several things. One of the important uses is in a formula for finding the angle between two vectors that have the same initial point. Definition of Dot Product: If a = a 1, a 2, a 3 and b = b 1, b 2, b 3, then the dot product of a and b is the number a b given by: To find the dot product of a and b, we multiply corresponding components and add. The result is not. It is a real number, that is,. For this reason, the dot product is sometimes called. The dot product of two-dimensional vectors: a 1, a 2 b 1, b 2 = a 1 b 1 + a 2 b 2 Example 1: 2, 4 3, 1 = 1, 7, 4 6, 2, ½ = (i + 2j 3k) (2j k) = PROPERTIES OF DOT PRODUCT If a, b, and c are vectors in V 3 and c is a scalar, then Geometric Interpretation: The dot product a b can be given a geometric interpretation in terms of the angle θ between a and b. This is defined to be the angle between the representations of a and b that start at the origin, where 0 θ π. θ is the angle between the line segments OA and OB. Note: If a and b are parallel vectors, then θ = 0 or θ = π.

12 Consider two non-zero vectors. We can use their dot product and their magnitudes to calculate the angle between the two vectors. From the Law of Cosines where c is the side opposite the angle theta: c 2 = a 2 + b 2 2abcosθ a b 2 = a 2 + b 2 2 a b cosθ From the properties that we just developed for dot product! 2!! we have that w = w w. It follows that : a b 2 = ( a b) ( a b) = = = From the previous slide a b 2 = a 2 + b 2 2 a b cosθ Substituting from above a 2 2a b + b 2 = a 2 + b 2 2 a b cosθ Simplifying 2a b = Theorem : If θ is the angle between the vectors a and b, then Example 2: If the vectors a and b have lengths 4 and 6, and the angle between them is π/3, find a b. Alternative Form of Dot Product: Example 3: Find the angle between the vectors a = 2, 2, 1 and b = 5, 3, 2.

13 True or False? Whenever two non-zero vectors are perpendicular, their dot product is 0. Since the two vectors are perpendicular, the angle between them will be 90! Whenever two non-zero vectors are perpendicular, their dot product is 0. cos π 2 = 0 and cos π 2 = u v u v u v u v = 0 u v = 0 or π 2. Orthogonal Vectors: Two nonzero vectors a and b are called perpendicular or orthogonal if the angle between them is. Then, the theorem gives:. Conversely, if a b = 0, then cos θ = 0; so, θ = π/2. Zero Vectors: The zero vector 0 is considered to be to all vectors. ORTHOGONAL VECTORS: Two vectors a and b are orthogonal if and only if Example 4: Show that 2i + 2j k is perpendicular to 5i 4j + 2k. As cos θ > 0 if 0 θ < π/2 and cos θ < 0 if π/2 < θ π, we see that a b is for θ < π/2 and for θ > π/2. We can think of a b as measuring the extent to which a and b point in the same direction. The dot product a b is: Positive, if a and b point in the same general direction Zero, if they are perpendicular Negative, if they point in generally opposite directions Extreme case: In the case where a and b point in exactly the same direction, we have. So, cos θ = 1 and. If a and b point in exactly opposite directions, then. So, cos θ = 1 and.

14 Direction Angles: The direction angles of a nonzero vector a are the angles α, β, and γ (in the interval [0, π]) that a makes with the positive x-, y-, and z-axes. The cosines of these direction angles cos α, cos β, and cos γ are called the direction cosines of the vector a. Direction Cosines: cos α = cos β = cosγ = By squaring the above expressions and adding, we see that:. We can also use above equations to write: a = a 1, a 2, a 3 = a cos α, a cos β, a cos γ = a cos α, cos β, cos γ 1 a a = This states that the direction cosines of a are the components of the unit vector in the direction of a. Example 5: Find the direction angles of the vector a = 1, 2, 3. An Application of the Dot Product: Projection Consider the example of a heavy box being dragged across the floor by a rope. If the box weighs 250 pounds and the angle between the rope and the horizontal is 25 degrees, how much force does the tractor have to exert to move the box?

15 The force being exerted by the tractor can be interpreted as a vector with direction of 25 degrees. So we need to find the magnitude. We are going to look at the force vector as the sum of its vertical component vector and its horizontal component. The work of moving the box across the floor is done by the horizontal component. (For maximum accuracy, don t round off until the end of the problem. In this case, we left the value of the cosine in the calculator and did not round off until the end.) The problem gets more complicated when the direction in which the object moves is not horizontal or vertical. In the sketch, u = w 1 + w 2 so that w 1 is parallel to v and w 2 is orthogonal to w 1. If a force represented by u is moving an object at the origin in the direction of v, then all of the work is being done by the component w 1. w 1 is called and is denoted as. w 2 is called. A Vocabulary Tip: When two vectors are perpendicular we say that they are orthogonal. When a vector is perpendicular to a line or a plane we say that the vector is normal to the line or plane. If u and v are nonzero vectors, then the projection of u onto v can be calculated as follows:

16 proj v u = The quantity inside the parenthesis is a of vector v. so the projection of u onto v is a multiple WORK Work is traditionally defined as follows: where F is the constant force acting along the line of motion and D is the distance traveled along the line of motion. Example: An object is pulled 12 feet across the floor using a force of 100 pounds. Find the work done if the force is applied at an angle of 50 degrees above the horizontal. Solution A: using W = FD we use the projection of F in the x direction. W = Solution B: Note that the force vector in coordinate form is, and the direction vector is,. Note also that. Two Ways to Calculate Work Method A: Method B: W = W = Example: Find the work done by a force of 20 lb acting in the direction N50W while moving an object 4 ft due west.!!!" uuur Projections: The figure shows representations PQand PR of two vectors a and b with the same initial point P.

17 Let S be the foot of the perpendicular from R to the line containing PQ. Then, the vector with representation PS is called the vector projection of b onto a and is denoted by proj a b. (You can think of it as a shadow of b.) Scalar Projection: The scalar projection of b onto a (also called the component of b along a) is defined to be the signed magnitude of the vector projection. This is the number b cos θ, where θ is the angle between a and b. This is denoted by comp a b. Observe: It is negative if π/2 < θ π. The equation a b = a b cos θ = a ( b cos θ) shows that: The dot product of a and b can be interpreted as the length of a times the scalar projection of b onto a. Since a b a b cosθ = = b, the component of b along a can be computed by taking the dot product a a of b with the unit vector in the direction of a. Scalar projection of b onto a: Vector projection of b onto a: Notice that the vector projection is the scalar projection times the unit vector in the direction of a. Example 6: Find the scalar and vector projections of: b = 1, 1, 2 onto a = 2, 3, 1

18 Calculating Work: In Section 6.4, we defined the work done by a constant force F in moving an object through a distance d as: W = Fd This, however, applies only when the force is directed along the line of motion of the object. However, suppose that the constant force is a vector pointing in some other direction, as shown. If the force moves the object from P to Q,then the displacement vector is D =!!!" PQ. The work done by this force is defined to be the product of the component of the force along D and the distance moved: W = ( F cos θ) D Equation 12: W = F =!!!" PR Therefore, the work done by a constant force F is the dot product F D, where D is the displacement vector. Unit: Force is measured in Newtons or pounds. Distance is measured in meters, or feet. Work is measured in Joules or foot-pounds. Example 7: A force is given by a vector F = 3i + 4 j + 5k and moves a particle from the point P(2,1, 0) to the point Q(4, 6, 2). Find the work done.

19 Study Sheet (12.4) Cross Product The cross product u x v of two vectors u and v, unlike the dot product, is. For this reason, it is also called the vector product. Note that u x v is defined only when u and v are three-dimensional (3-D) vectors. The reason for the particular form of the definition is that the cross product defined in this way has many useful properties. In particular, we will show that the vector a x b is perpendicular to both a and b. Determinant of Order 2: A determinant of order 2 is defined by: Determinant of Order 3: A determinant of order 3 can be defined in terms of second-order determinants as follows: a 1 a 2 a 3 a 1 a 2 a 3 a 1 a 2 a 3 a 1 a 2 a 3 b 1 b 2 b 3 = a 1 b 1 b 2 b 3 a 2 b 1 b 2 b 3 + a 3 b 1 b 2 b 3 c 1 c 2 c 3 c 1 c 2 c 3 c 1 c 2 c 3 c 1 c 2 c 3 = a 1 a 2 + a 3 Observe that: Each term on the right side of the equation involves a number a i in the first row of the determinant. This is multiplied by the second-order determinant obtained from the left side by deleting the row and column in which it appears. Notice also the minus sign in the second term.

20 Now, let s rewrite the definition using second-order determinants and the standard basis vectors i, j, and k. i j k i j k i j k i j k a b = = i j + k = i j + k = ( ) i ( ) j + ( ) k We see that the cross product of the vectors a = a 1 i +a 2 j + a 3 k and b = b 1 i +b 2 j + b 3 k is: a b = ( ) i ( ) j + ( ) k Example 1: If a = <1, 3, 4> and b = <2, 7, 5>, then Example 2: Show that a x a = 0 for any vector a in V 3.

21 One of the most important properties of the cross product is given by the following theorem. Theorem: The vector a x b is orthogonal to both and. Let a and b be represented by directed line segments with the same initial point, as shown. Then, Theorem states that the cross product a x b points in a direction perpendicular to the plane through a and b. It turns out that the direction of a x b is given by the right-hand rule, as follows. RIGHT-HAND RULE: If the fingers of your right hand curl in the direction of a rotation (through an angle less than 180 ) from a to b, then your thumb points in the direction of a x b. We know the direction of the vector a x b. The remaining thing we need to complete its geometric description is its length a x b. This is given by the following theorem. Theorem : If θ is the angle between a and b (so 0 θ π), then a b =. A vector is completely determined by its magnitude and direction. Thus, we can now say that a x b is the vector that is perpendicular to both a and b, whose: Orientation is determined by the right-hand rule Length is a b sin θ Two nonzero vectors a and b are parallel if and only if. The geometric interpretation of Theorem 9 can be seen from this figure. If a and b are represented by directed line segments with the same initial point, then they determine a parallelogram with base a, altitude b sin θ, and area A = a ( b sin θ) = a x b

22 Interpreting the magnitude of a cross product: CROSS PRODUCT MAGNITUDE: The length of the cross product a x b is equal to determined by a and b. Example 3: Find a vector perpendicular to the plane that passes through the points P(1, 4, 6), Q(-2, 5, -1), R(1, -1, 1) Example 4: Find the area of the triangle with vertices P(1, 4, 6), Q(-2, 5, -1), R(1, -1, 1) If we apply Theorems 8 and 9 to the standard basis vectors i, j, and k using θ = π/2, we obtain: i x j =. j x k =. k x i =. j x i =. k x j =. i x k =. Observe that: i x j j x i Thus, the cross product is not commutative. Also, i x (i x j) = =. However, (i x i) x j = =. So, the associative law for multiplication does not usually hold. That is, in general, (a x b) x c a x (b x c)

23 However, some of the usual laws of algebra do hold for cross products. Theorem: CROSS PRODUCT PROPERTIES If a, b, and c are vectors and c is a scalar, then 1. a x b = (b x a) 2. (ca) x b = c(a x b) = a x (cb) 3. a x (b + c) = a x b + a x c 4. (a + b) x c = a x c + b x c 5. a (b x c) = (a x b) c 6. a x (b x c) = (a c)b (a b)c These properties can be proved by writing the vectors in terms of their components and using the definition of a cross product. We give the proof of Property 5 and leave the remaining proofs as exercises. SCALAR TRIPLE PRODUCTS Notice from the equation, a b = a b sinθ, that we can write the scalar triple product as a determinant: SCALAR TRIPLE PRODUCTS: The geometric significance of the scalar triple product can be seen by considering the parallelepiped determined by the vectors a, b, and c. The area of the base parallelogram is: A = b x c If θ is the angle between a and b x c, then the height h of the parallelepiped is: h = a cos θ We must use cos θ instead of cos θ in case θ > π/2. Hence, the volume of the parallelepiped is: V = Ah = b x c a cos θ = a (b x c) Thus, we have proved the following formula. Formula: The volume of the parallelepiped determined by the vectors a, b, and c is the magnitude of their scalar triple product:

24 COPLANAR VECTORS: If we use the formula and discover that the volume of the parallelepiped determined by a, b, and c is 0, then the vectors must lie in the same plane. That is, they are coplanar. Example 5: Use the scalar triple product to show that the vectors a = <1, 4, -7>, b = <2, -1, 4>, c = <0, -9, 18> are coplanar. VECTOR TRIPLE PRODUCT: The product a x (b x c) that occurs in Property 6 is called the vector triple product of a, b, and c. CROSS PRODUCT IN PHYSICS: The idea of a cross product occurs often in physics. In particular, we consider a force F acting on a rigid body at a point given by a position vector r. For instance, if we tighten a bolt by applying a force to a wrench, we produce a turning effect. TORQUE: The torque τ (relative to the origin) is defined to be the cross product of the position and force vectors τ = r x F It measures the tendency of the body to rotate about the origin. The direction of the torque vector indicates the axis of rotation. According to Theorem 9, the magnitude of the torque vector is τ = r x F = r F sin θ where θ is the angle between the position and force vectors. Observe that the only component of F that can cause a rotation is the one perpendicular to r that is, F sin θ. The magnitude of the torque is equal to the area of the parallelogram determined by r and F. Example 6: A bolt is tightened by applying a 40-N force to a 0.25-m wrench, as shown. Find the magnitude of the torque about the center of the bolt.

25 Study Sheet (12.5) EQUATIONS OF LINES AND PLANES EQUATIONS OF LINE A line L in three-dimensional (3-D) space is determined when we know: A point P 0 (x 0, y 0, z 0 ) on L The direction of L In three dimensions, the direction of a line is conveniently described by a vector. So, we let v be a vector parallel to L. Let P(x, y, z) be an arbitrary point on L. Let r 0 and r be the position vectors of P 0 and P. That is, they have representations OP 0 and OP. In three dimensions, the direction of a line is conveniently described by a vector. So, we let v be a vector parallel to L, and a vector v = <a, b, c> is called the and numbers a, b, and c are called of L. If a is the vector with representation P 0 P, then the Triangle Law for vector addition gives: r = r 0 + a To determine the equation of the line passing through the point P 0 (x 0, y 0, z 0 ) and parallel to the direction vector, v = <a, b, c>, we will use our knowledge that parallel vectors are scalar multiples. Thus, the vector through P and any other point P(x, y, z) on the line is a scalar multiple of the direction vector, v = <a, b, c>. In other words, P 0 P = t a, b, c, where t is any real number (scalar), or x x 0, y y 0, z z 0 = at, bt,ct Equate the respective components and there are three equations. or x x 0 = at, y y 0 = bt, and z z 0 = ct Parametric Equations of a Line in Space: These equations are called the parametric equations of the line L through the point P 0 (x 0, y 0, z 0 ) and parallel to the vector v = <a, b, c>. Each value of the parameter t gives a point (x, y, z) on L.

26 Example 1: (a) Find a vector equation and parametric equations for the line that passes through the point (5, 1, 3) and is parallel to the vector i + 4 j 2 k. (b) Find two other points on the line. The vector equation and parametric equations of a line are not unique. If we change the point or the parameter or choose a different parallel vector, then the equations change. For instance, if, instead of (5, 1, 3), we choose the point (6, 5, 1) in Example 1, the parametric equations of the line become: x = 6 + t y = 5 + 4t z = 1 2t Alternatively, if we stay with the point (5, 1, 3) but choose the parallel vector 2 i + 8 j 4 k, we arrive at: x = 5 + 2t y = 1 + 8t z = 3 4t DIRECTION NUMBERS In general, if a vector v = <a, b, c> is used to describe the direction of a line L, then the numbers a, b, and c are called direction numbers of L. Any vector parallel to v could also be used. Thus, we see that any three numbers proportional to a, b, and c could also be used as a set of direction numbers for L. If the components of the direction vector are all nonzero, each equation can be solved for the parameter t and then the three can be set equal. Symmetric Equations of a Line in Space: These equations are called the symmetric equations of the line. Notice that the numbers a, b, and c that appear in the denominators of the equations are direction numbers of L. That is, they are components of a vector parallel to L. If one of a, b, or c is 0, we can still eliminate t. For instance, if a = 0, we could write the equations of L as: x = x 0, This means that L lies in the vertical plane x = x 0. y y 0 b = z z 0 c

27 Example 2: (a) Find parametric equations and symmetric equations of the line that passes through the points A(2, 4, 3) and B(3, 1, 1). (b) At what point does this line intersect the xy-plane? In general, the procedure of Example 2 shows that direction numbers of the line L through the points P 0 (x 0, y 0, z 0 ) and P 1 (x 1, y 1, z 1 ) are: x 1 x 0, y 1 y 0, and z 1 z 0. So, symmetric equations of L are: x x0 y y0 z z0 = = x x y y z z EQUATIONS OF LINE SEGMENTS Often, we need a description, not of an entire line, but of just a line segment. If we put t = 0 in the parametric equations in Example 2 a, we get the point (2, 4, 3). If we put t = 1, we get (3, 1, 1). So, the line segment AB is described by either: The parametric equations x = 2 + t y = 4 5t z = 3 + 4t where 0 t 1 The corresponding vector equation r(t) = <2 + t, 4 5t, 3 + 4t> where 0 t 1 The vector equation of a line through the (tip of the) vector r 0 in the direction of a vector v is: r = r 0 + t v If the line also passes through (the tip of) r 1, then we can take v = r 1 r 0. So, its vector equation is: r = r 0 + t(r 1 r 0 ) = (1 t)r 0 + t r 1 The line segment from r 0 to r 1 is given by the parameter interval 0 t 1. EQUATIONS OF LINE SEGMENTS The line segment from r 0 to r 1 is given by the vector equation Example 3: Show that the lines L 1 and L 2 with parametric equations x = 1 + t y = 2 + 3t z = 4 t x = 2s y = 3 + s z = 3 + 4s are skew lines; that is, they do not intersect and are not parallel, and therefore do not lie in the same plane.

28 PLANES IN SPACE In previous sections we have looked at planes in space. For example, we looked at the xy-plane, the yz-plane and the xz-plane when we first introduced 3-dimensional space. Now we are going to examine the equation for a plane. In the figure below P(x 0, y 0, z 0 ) is a point in the highlighted plane and n = a, b, c is the vector normal to the highlighted plane. Q P For any point Q(x, y, z) in the plane, the vector from P to Q, PQ = t x x 1, y y 1, z z 1 is also in the plane. Since the vector from P to Q is in the plane, PQ and n are perpendicular and their dot product must equal zero. n PQ = 0 a, b,c x x 0, y y 0, z z 0 = 0 ( ) + b( y y 0 ) + c( z z 0 ) = 0 a x x 0 This last equation is the equation of the highlighted plane. So the equation of any plane can be found from a point in the plane and a vector normal to the plane. The standard equation of a plane containing the point P(x 0, y 0, z 0 ) and having normal vector, n = a, b, c is SCALAR EQUATION of a Plane in Space: Note: The equation can be simplified by using the distributive property and collecting like terms. This results in the Linear Equation in x, y, and z. LINEAR EQUATION: where d = (ax 0 + by 0 + cz 0 ). Conversely, it can be shown that, if a, b, and c are not all 0, then the linear Equation 8 represents a plane with normal vector <a, b, c>.

29 NORMAL VECTOR This orthogonal vector n is called a normal vector. Let P(x, y, z) be an arbitrary point in the plane. Let r 0 and r 1 be the position vectors of P 0 and P. Then, the vector r r 0 is represented by The normal vector n is orthogonal to every vector in the given plane. In particular, n is orthogonal to r r 0 and so we have Vector Equation:!!!" P0 P And the equation can be rewritten as Vector Equation: Either equation is called a vector equation of the plane. Example 4: Find an equation of the plane through the point (2, 4, 1) with normal vector n = <2, 3, 4>. Find the intercepts and sketch the plane. Example 5: Find an equation of the plane that passes through the points P(1, 3, 2), Q(3, 1, 6), R(5, 2, 0) Example 6: Find the point at which the line with parametric equations x = 2 + 3t y = 4t z = 5 + t intersects the plane 4x + 5y 2z = 18.

30 PARALLEL PLANES Two planes are parallel if their normal vectors are parallel. For instance, the planes x + 2y 3z = 4 and 2x + 4y 6z = 3 are parallel because: Their normal vectors are n 1 = <1, 2, 3> and n 2 = <2, 4, 6> and n 2 = 2n 1. NONPARALLEL PLANES If two planes are not parallel, then They intersect in a straight line. The angle between the two planes is defined as the acute angle between their normal vectors. Example 7: (a) Find the angle between the planes x + y + z = 1 and x 2y + 3z = 1 (b) Find symmetric equations for the line of intersection L of these two planes. NOTE: A linear equation in x, y, and z represents a plane. Also, two nonparallel planes intersect in a line. It follows that two linear equations can represent a line. The points (x, y, z) that satisfy both a 1 x + b 1 y + c 1 z + d 1 = 0 and a 2 x + b 2 y + c 2 z + d 2 = 0 lie on both of these planes. So, the pair of linear equations represents the line of intersection of the planes (if they are not parallel). For instance, in Example 7, the line L was given as the line of intersection of the planes x + y + z = 1 and x 2y + 3z = 1 1 The symmetric equations that we found for L could be written as: x y and y z = = They exhibit L as the line of intersection of the planes x 1 5 = y 2 and y 2 = z 3

31 In general, when we write the equations of a line in the symmetric form x x y y z z = = a b c we can regard the line as the line of intersection of the two planes x x0 y y0 y y z z and = a b b c = 0 0 Example 8: Find a formula for the distance D from a point P 1 (x 1, y 1, z 1 ) to the plane ax + by + cz + d = 0. The formula for the distance D from a point P 1 to the plane is: D = Example 9: Find the distance between the parallel planes 10x + 2y 2z = 5 and 5x + y z = 1 Example 10: In Example 3, we showed that the lines L 1 : x = 1 + t y = 2 + 3t z = 4 t L 2 : x = 2s y = 3 + s z = 3 + 4s are skew. Find the distance between them.

32 Study Sheet (12.6) Cylinders and Quadric Surfaces TRACES: To sketch the graph of a surface, it is useful to determine the curves of intersection of the surface with planes parallel to the coordinate planes. These curves are called traces (or cross-sections) of the surface. CYLINDER: A cylinder is a surface that consists of all lines (called rulings) that are parallel to a given line and pass through a given plane curve. Example 1: Sketch the graph of the surface z = x 2 Let C be a curve in a plane and let L be a line not parallel to that plane. Then the set of points on lines parallel to L that intersect C is called a cylinder. The straight lines that make up the cylinder are called the rulings of the cylinder. In the sketch, we have the generating curve, a parabola in the xz-plane: In Example 1, we noticed the variable y is missing from the equation of the cylinder. This is typical of a surface whose rulings are parallel to one of the coordinate axes. If one of the variables x, y, or z is missing from the equation of a surface, then the surface is a cylinder. Now consider the cylinder generated by the equation (y 1) 2 + (z 1) 2 =1. For a point on this surface, the x-coordinate can take on any value as long as the y- and z- coordinates satisfy the equation (y 1) 2 + (z 1) 2 =1. z y x An effective way to visualize this surface is to move about 5 units down the x-axis and place a copy of the circle in its corresponding position. Then draw lines parallel to the x-axis (rulings) that connect corresponding points on the two circles and you have a pretty good idea what the cylinder looks like.

33 Example 2: Identify and sketch the surfaces. (a) x 2 + y 2 = 1 (b) y 2 + z 2 = 1 Note: When you are dealing with surfaces, it is important to recognize that an equation like x 2 +y 2 = 1 represents a cylinder and not a circle. The trace of the cylinder x 2 + y 2 = 1 in the xy-plane is the circle with equations x 2 + y 2 = 1, z = 0 QUADRIC SURFACE: A quadric surface is the graph of a second-degree equation in three variables x, y, and z. The most general such equation is: Ax 2 + By 2 + Cz 2 + Dxy + Eyz + Fxz + Gx + Hy + Iz + J = 0 where A, B, C,, J are constants. However, by translation and rotation, it can be brought into one of the two standard forms: Ax 2 + By 2 + Cz 2 + J = 0 Ax 2 + By 2 + Iz = 0 In this section, we will be working with equations where D=E=F=0. Quadric surfaces are the counterparts in three dimensions of the conic sections in the plane. ELLIPSOID: A football is an ellipsoid. Planet Earth is also an ellipsoid. The standard form of an ellipsoid is: d = Example 3: Use traces to sketch the quadric surface with equation x 2 + y2 9 + z2 4 =1. It is very helpful to examine the intersection of the quadric surface with the coordinate planes or even with planes that are parallel to the coordinate planes. Consider the table below for the ellipsoid. Plane Equation Trace xy-plane (z=0) xz-plane (y=0) yz-plane (x=0)

34 The trace in the xz-plane is the ellipse: The trace in the yz-plane is the ellipse:.. The trace in the xy-plane is the ellipse:. ELLIPTIC PARABOLOID: For z > 0, traces are ellipses. Planes parallel to the xz- and yz planes are parabolas. Horizontal traces are ellipses. Vertical traces are parabolas. The standard form of an elliptic paraboloid is: d = Example 4: Use traces to sketch the surface z = 4x 2 + y 2 Plane Equation Trace xy-plane (z=0) xz-plane (y=0) yz-plane (x=0) HYPERBOLIC PARABOLOID: Traces parallel to the xy-plane are hyperbolas. Traces parallel to the xz- and yz-planes: parabolas. The standard form of an hyperbolic paraboloid is: Example 5: Sketch the surface z = y 2 x 2 d = Plane Equation Trace xy-plane (z=0) xz-plane (y=0) yz-plane (x=0)

35 The traces in the vertical planes x = k are the parabolas z = y 2 k 2, which open upward. The traces in y = k are the parabolas z = x 2 + k 2, which open downward. The horizontal traces are y 2 x 2 = k, a family of hyperbolas. All traces are labeled with the value of k. Here, we show how the traces appear when placed in their correct planes. Here, we fit together the traces from the previous figure to form the surface z = y 2 x 2, a hyperbolic paraboloid. Notice that the shape of the surface near the origin resembles that of a saddle. HYPERBOLOID OF ONE SHEET: Traces parallel to the xy-plane are ellipses. Traces parallel to the xz- and yz-planes are hyperbolas. The standard form of hyperboloid of one sheet is: d =

36 Example 6: Sketch the surface x2 4 + y2 z2 4 =1 Plane Equation Trace xy-plane (z=0) xz-plane (y=0) yz-plane (x=0) GRAPHING SOFTWARE: The idea of using traces to draw a surface is employed in threedimensional (3-D) graphing software for computers. In most such software, Traces in the vertical planes x = k and y = k are drawn for equally spaced values of k. Parts of the graph are eliminated using hidden line removal. HYPERBOLOID OF TWO SHEETS: Traces parallel to the xy-plane are ellipses. Traces parallel to the xz- and yz-planes are hyperbolas. The standard form of hyperboloid of two sheets is: d = Example 7: Identify and sketch the surface 4x 2 y 2 + 2z 2 +4 = 0 Plane Equation Trace xy-plane (z=0) xz-plane (y=0) yz-plane (x=0) Example 8: Classify the quadric surface x 2 + 2z 2 6x y + 10 = 0

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