1 . Three Dimensional Curves and Surfaces.. Parametric Equation of a Line An line in two- or three-dimensional space can be uniquel specified b a point on the line and a vector parallel to the line. The line then is the line parallel to the vector v = (a, b, c) passing through the point P (,, ). In particular, if we view the vector as having its initial point at P, then we can move awa from P along the line b adding multiples of v. In other L.5. P,,.5. v a,b,c Figure : Parametric equation of a line. words, we can move tv along the line. In other words, the position of point Q(,, ) on the line is given b (,, ) = (,, ) + t(a, b, c) = ( + at, + bt, + ct). () This gives us the parametric equations of a line in 3-space for a line passing through P (,, ) and parallel to v = ai + bj + ck : = + at, = + bt, = + ct. ()
2 Here t is the parameter that determines how far along the line ou have moved. Eample: Find the parametric equation of the line passing through P (, 3, ), parallel to i j + 3k. Solution: The line is given b parametric equations = + t, = 3 t, = + 3t. Note that these lines are infinite. If we want onl a line segment we must restrict the parameter. Thus, the line given in the eample can be restricted to the line segment joining P (, 3, ) to Q(3,, 5) if we restrict t to the interval t. Finall, two lines that are not parallel and do not cross are called skew lines.... Representing the parametric equation using vectors If instead of an initial point P (,, ), we define a position vector r = (,, ) to give an initial position on a line and r = (,, ) for an point on the line parallel to v = (a, b, c) through P, then equation () can be represented b the vector equation of a line as r = r + tv. (3) This is shown in Figure. Note that all these vectors are defined with initial points at the origin. This is entirel equivalent to the equations in (), but the notation is obviousl briefer, and the interpretation is purel in terms of vectors. If ou need to think about this more intuitivel, think of r as ordinar position vectors in space, r as a velocit vector and t as time. This then looks like one of the usual equations for linear motion in vector form. However, we can use this more generall for an parameter t and appropriate vector v. Let s return to the previous eample: Eample: Find the vector equation of the line passing through P (, 3, ), parallel to i j + 3k.. Solution: The initial position is given b the vector r = (, 3, ),
3 ..5 L P,..5 r r v t v Figure : Vector equation of a line r = r + vt. and the vector equation is r = (, 3, )+t(,, 3) = (+t, 3 t, +3t) = (+t)i+(3 t)j+(+3t)k. It is important to be comfortable with all these was of writing parametric equations of a line... Planes The first tpe of planes we might think of are the coordinate planes. The -plane, for eample, is the set of a and with =. B etension, we can imagine planes parallel to the coordinate planes. For eample, the plane = a is the set of all and such that = a. These tpes of planes are shown in Figure 3. 3
4 a b c Figure 3: Planes parallel to the coordinate aes.... Planes specified b a point and normal vector An plane in 3-space can be uniquel determined b giving a point on the plane and a vector perpendicular to the plane, called a normal vector. Suppose we want to find an equation of the plane passing through P (,, ) and perpendicular to n = (a, b, c). Let us define the vector r = (,, ) pointing to P, and r = (,, ) pointing to another point Q. Since n is perpendicular to the plane, n r = n r =, and therefore which in component form is written which results in the equation n (r r ) =, (4) (a, b, c) (,, ) =, (5) a( ) + b( ) + c( ) =. (6) This is known as the point-normal form of the equation of a plane. Equation (4) is the vector form of this equation. Eample: Find the equation of the plane passing though (,, ) and with normal vector n = (, 3, ). Solution: We use equation (6) to give us ( ) + 3( ) + ( + ) =, 4
5 which can be simplified to =. In fact, this final form can be generalised to a + b + c + d =, (7) which is the equation with graph that is a plane with n = (a, b, c) as a normal. This is called the general form of the equation of a plane. d is determined b knowing a point that lies in the plane and substituting into the equation to find d. Other was to determine a plane are a point and two vectors parallel to the plane, or three points in the plane. Note that these allow us to find a normal vector and so the equation of the plane. The cross-product of two vectors in the plane will give a vector normal to the plane. Three points P, P and P 3 allow us to define two vectors P P and P P 3, which again allow us to find the normal vector.... Intersecting planes Two distinct planes have an acute angle of intersection, θ π/. The angle is the same as either the angle between n and n or between n and n depending on the direction of the normal vectors. However, in either case the angle is given b cos θ = n n n n. (8) The absolute value ensures that regardless of the sign of the normal vectors we alwas have an acute angle. Eample: Find the angle between the planes + = 3, =. Solution: The normal vectors are given b n = (,, ) and n = (, 4, 4), and the angle is found from cos θ = = 8 (3)(6) = 4 9, 5
6 n n. Θ 5 Θ Figure 4: Intersecting planes. which gives the angle θ = cos 4 9 =.4 rad = 63.6o....3 Distance problems involving planes There are three distance problems we will be concerned with: the distance between a point and a plane; the distance between two parallel planes; find the distance between two skew lines. Theorem: The distance D between a point P (,, ) and the plane a + b + c + d = is D = a + b + c + d a + b + c. (9) Proof: Let Q(,, ) be a point in the plane, and n = (a, b, c) the normal vector with its initial point is at Q. Consider Figure 5. The distance D is equal to the length of the orthogonal projection of QP onto n. Therefore, recalling proj b v = v b b b, () 6
7 P,, proj n QP D Q,, Figure 5: Projection onto normal vector: distance from plane. we get However, we can write D = proj n QP = QP n n QP = (,, ) QP na( ) + b( ) + c( ), n = QP n. () n () and which together give us n = a + b + c, (3) D = a( ) + b( ) + c( ) a + b + c. (4) Moreover, since Q lies in the plane, it satisfies the equation a + b + c + d =, (5) which allows us to find d, d = a b c, (6) and therefore we find the result. 7
8 Eample: Find the distance between the point P (4, 4, ) and the plane =. Solution: Using () we get D = ()(4) + ( )(4) + (4)() = 6 6 =. (7) To compute the distance between two parallel planes, compute the distance between one plane and an point in the other plane. To find the distance between skew lines, define two parallel planes each of which contains one of the skew lines. Then the distance between the planes gives the distance between the skew lines. Eample: Find the distance between the skew lines L : = + 4t = 5 4t, = + 5t, L : = + 8t = 4 3t, = 5 + t. Solution: Let P and P be parallel planes containing L and L respectivel. We can find a point on each line and hence in each plane b setting t =, giving Q (, 5, ) and Q (, 4, 5), see Figure 6. Let s use Q and find the L Q D L Q Figure 6: Distance between skew lines. equation of the plane P. Since the planes are parallel, the vectors used to define the parametric equations of the lines u = (4, 4, 5) and u = (8, 3, ) are both parallel to P. Hence i j k n = u u = = i + 36j + k, 8 3 8
9 is normal to P and P. With this normal vector and the point Q, we find the equation of P : (c ) + 36( 4) + ( 5) =, which can be written in the general form =. Therefore the distance from P to the point Q (, 5, ) is D = ()() + (36)(5) + ()( ) = 95 87, which in turn is the distance from L to L, since the lie in the parallel planes...3 Quadric Surfaces The generalisation to the general quadratic equation?? which from which conic sections were derived in -space is the second-degree equation in, and, A + B + C + D + E + F + G + H + Ik + J =. (8) The graphs of this famil of equations are called the quadric surfaces. There are si common tpes of quadrics,. Ellipsoid:. Hperboloid of One Sheet: 3. Hperboloid of Two Sheets: a + b + c = a + b c = c a b = 4. Elliptic Cone: = a + b 5. Elliptic Paraboloid: = a + b (9) 6. Hperbolic Paraboloid: = b a where we assume that a, b, c >. These are shown in Figure 7. The have the following traces 9
10 . Ellipsoid: Traces in the coordinate planes are ellipses.. Hperboloid of One Sheet: The trace in the -plane are ellipses, and traces in the - and -planes are hperbolas. 3. Hperboloid of Two Sheets: There is no trace in the -plane, although the traces in planes parallel to the -plane are ellipses provided there is a trace, and traces in the - and -planes are hperbolas. 4. Elliptic Cone: The trace in the -plane is a point and in planes parallel to the -plane the traces are ellipses, and traces in the - and planes are pairs of intersecting lines. 5. Elliptic Paraboloid: The trace in the -plane is a point and in planes parallel to and above the -plane the traces are ellipses, and traces in the - and -planes are parabolas. 6. Hperbolic Paraboloid: The trace in the -plane is a pair of intersecting lines, and traces in planes parallel to the -plane are hperbolas, which open in the -direction when above the -plane and open in the - direction when below the -plane. Traces in the - and -planes are parabolas. These of course can also appear in other orientations along different coordinate aes, or indeed with cross-product terms which would result in other orientations. If the elliptic cross-section of an elliptic cone or elliptic paraboloid is circular the are called a circular cone and a circular paraboloid respectivel. Of course an ellipsoid with all the cross-sections circular is a sphere, i.e. a = b = c. You are not epected to accuratel draw an of these surfaces. If asked for a sketch, draw the traces on the planes, and join these to give a rough idea of the shape. As with the conic sections we can translate a quadric surface b moving awa from the origin to (a, b, c), which will result in the change (,, ) ( a, b, c) in the equations for the surfaces. However, this will not be required for this course.
11 Figure 7: Quadric surfaces.
12 ..4 Clindrical and Spherical Coordinates We have alread met polar coordinates in -space. Now we introduce two coordinate sstems that are often useful when rectangular coordinates are awkward in 3-space. Clindrical coordinates (ρ, θ, ): In terms of clindrical coordinates the rectangular coordinates are written = ρ cos θ, = ρ sin θ,, () where ρ <, θ < π. This is shown in Figure 8 If we wish to go Θ Ρ Figure 8: Clindrical coordinates (ρ, θ, φ). from rectangular to clindrical coordinates directl, we can use the relations ρ = +, tan θ =, =. () Spherical coordinates: In terms of spherical coordinates the rectangular coordinates are written = ρ sin φ cos θ, = ρ sin φ sin θ, = ρ cos φ, () where ρ <, θ < π and φ π. Figure 9 shows spherical coordinates. If we wish to go from rectangular to spherical coordinates directl,
13 Ρ Φ.5 Θ Figure 9: Spherical coordinates. we can use the relations ρ = + +, tan θ =, cos φ = + +. (3) Eample: Change + + = 9 to spherical polar coordinates. Solution: The coordinates for this surface (a sphere) are for ρ = 3 and so = 3 sin φ cos θ, = 3 sin φ sin θ, = 3 cos φ, (4) where θ and φ are the parameters. 3