Def.: a, b, and c are called the for the line L. x = y = z =

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1 Bob Brown, CCBC Dundalk Math 253 Calculus 3, Chapter Section 5 Completed Lines in Space Eercise : Consider the vector v = Sketch and describe the following set: t v ta, tb, tc : t a, b, c. Let P =,,. Sketch and describe the following set: ta tb, tc : t, Def.: The vector v is called the for the line L. Def.: a, b, and c are called the for the line L. Parametric Equations and Smmetric Equations of a Line The line L (in the second graph) parallel to the vector v that passes through the point P =, is represented b the following set of parametric equations:, = = = If the direction numbers are all nonero, then ou can solve for the parameter to obtain the following smmetric equations of the line without the parameter:

2 Bob Brown, CCBC Dundalk Math 253 Calculus 3, Chapter Section 5 Completed 2 Eercise 2: Determine a direction vector, parametric equations, and smmetric equations of the line that passes through the points P = (2, 0, 2) and Q = (, 4, -3). direction vector: PQ = parametric equations: Choose either one of the points; sa, P = (2, 0, 2). Then write = = = Note: t = 0 and t = smmetric equations: = 2 t 2 = -t = t = 0 + 4t = t = 2 5t = t Thus, smmetric equations are Note: We have seen that an equation of a line in space can be obtained from a and a

3 Bob Brown, CCBC Dundalk Math 253 Calculus 3, Chapter Section 5 Completed 3 Planes in Space We will now see that an equation of a plane in space can be obtained from a and a Let P =,, be a point in the plane, and let n = a, b, c be a nonero normal vector to the plane. The plane consists of all points Q =,, for which the vector PQ is Equations of a Plane Let P =,, be a point in the plane, and let n = a, b, c be a nonero normal vector to the plane. The following are equations for the plane. Standard Form General Form Eercise 3: Determine an equation of the plane that passes through the points (2, 3, -2), (3, 4, 2), and (, -, 0). u = v = normal: n = = * using the point (2, 3, -2): 8( 2) 6( 3) 3( + 2) = 0 * using the point (3, 4, 2): 8( 3) 6( 4) 3( 2) = 0

4 Bob Brown, CCBC Dundalk Math 253 Calculus 3, Chapter Section 5 Completed 4 Eercise 4a (Section.5 #66): Sketch a graph of the plane 2 + = 4 b determining the traces. -trace ( = 0) -trace ( = 0) -trace ( = 0) Note: The plane is not a triangle; the plane contains this triangle. Eercise 4b (Section.5 #66): Sketch a graph of the plane 2 + = 4 b determining the intercepts. -intercept: (, 0, 0) -intercept: (0,, 0) -intercept: (0, 0, )

5 Bob Brown, CCBC Dundalk Math 253 Calculus 3, Chapter Section 5 Completed 5 Angle Between Planes Two distinct planes in three-dimensional space either are parallel or intersect in a line. If the planes intersect in a line, ou can determine the angle between the planes b determining the angle between their normal vectors. The value of the angle between two intersecting planes is equivalent to the value of the angle between their respective normal vectors and calculating the angle between two vectors in three-dimensional space is something that we alread know how to do. (See Handout.3, top of page 3.) Theorem: If n and n 2 are normal vectors to two intersecting planes, then the angle θ between the normal vectors is equal to the angle between the two planes, and the angle is given implicitl b The two planes are if The two planes are if Of course, it is possible that the two planes are neither perpendicular nor parallel. Eercise 5 (Section.5 #60): Determine the angle between the following two planes = 7 and = 0 Hint: Normal vectors for these planes are, respectivel, and

6 Bob Brown, CCBC Dundalk Math 253 Calculus 3, Chapter Section 5 Completed 6 Verification of the Hint in Eercise 5 You ma have noticed in Eercise 5 the formulaic similarit between the equation of a plane and the components of its normal vector. That is no coincidence. How can we obtain a normal vector from the equation of the plane? Consider the plane = 7. First, since three points determine a plane, find three points in the plane. Secondl, those three points determine two vectors in the plane. Finall, the cross product of those two vectors is perpendicular to those two vectors and is, thus, normal to the plane. Note: Note that n = 3 i 2 j k is not the onl normal to the plane. An nonero scalar multiple of n is also normal to the plane, like, for eample,

7 Bob Brown, CCBC Dundalk Math 253 Calculus 3, Chapter Section 5 Completed 7 Distance Between a Point and a Plane Theorem: Let P be a point in the plane and n a normal vector to the plane. The distance between the plane and a point Q not in the plane is Eercise 6: (Compare with the work and solution found in Eample 5 on page 788 in the tetbook.) Determine the distance between Q = (, 5, -4) and the plane = 6. We know that n = The point P = is a normal vector to the plane. is a point in the plane because if ou (arbitraril) pick Thus, PQ = and, so, PQ n = 0,6, 5 3,, 2 = Also, n = Therefore, the distance between Q and the plane is

8 Bob Brown, CCBC Dundalk Math 253 Calculus 3, Chapter Section 5 Completed 8 Distance Between a Point and a Line in Space sin(θ) = ( * ) d = Recall Handout.4, page 2, Geometric Propert 3: u v = Multipl both sides of ( * ) b u : d = PQ sin(θ) d u = d = Theorem: Let P be a point on a line L and let u be a direction vector for L. The distance between Q and L is

9 Bob Brown, CCBC Dundalk Math 253 Calculus 3, Chapter Section 5 Completed 9 Eercise 7 (Section.5 #90): Determine the distance between Q = (, -2, 4) and the line given b the parametric equations = 2t, = t 3, = 2t + 2. A direction vector for the line is u = and, so, u = Note that the point Q is not on the line because To determine a point P on the line, take, for eample, t = P = (,, ) = Thus, PQ = Net, PQ u = 3,2,4 2,, 2 = Therefore, PQ u = Finall, the distance between Q and the line is