Lines and planes in space (Sect. 12.5) Review: Lines on a plane. Lines in space (Today). Planes in space (Next class). Equation of a line


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1 Lines and planes in space (Sect. 2.5) Lines in space (Toda). Review: Lines on a plane. The equations of lines in space: Vector equation. arametric equation. Distance from a point to a line. lanes in space (Net class). Equations of planes in space. Vector equation. Components equation. The line of intersection of two planes. arallel planes and angle between planes. Distance from a point to a plane. Review: Lines on a plane Equation of a line The equation of a line with slope m and vertical intercept b is given b = m + b. Vector equation of a line The equation of the line b the point = (0, b) parallel to the vector v =, m is given b r(t) = r 0 + t v, where r 0 = O = 0, b. b m b r 0 V m
2 Review: Lines on a plane Find the vector equation of a line = + 3. Solution: The vertical intercept is at the point = (0, 3). A vector tangent to the line is v =,, since the point = (, 2) belongs to the line, which implies that v = = ( 0), (2 3) =,. The vector equation for the line is r(t) = 0, 3 + t,. 0 3 r = O = ( 0, 3 ) V O 3 Review: Lines on a plane We verif the result above: That the line = + 3 is indeed r(t) = 0, 3 + t,, (Vector equation of the line.) If r(t) = (t), (t), then (t), (t) = (0 + t), (3 t). That is, (t) = t, (t) = 3 t. (arametric equation of the line.) (The parameter is t.) Replacing t b is the second equation above we obtain () = + 3.
3 Review: Lines on a plane Vector equation of a line The equation of the line b the point = (0, b) parallel to the vector v =, m is given b r(t) = r 0 + t v, where r 0 = O = 0, b. arametric equation of a line A line with vector equation r(t) = r 0 + t v, where r 0 = 0, b and v =, m can also be written as follows (t), (t) = (0 + t), (b + tm), that is, b r 0 V m (t) = t (t) = b + mt. Lines and planes in space (Sect. 2.5) Lines in space Review: Lines on a plane. The equations of lines in space: Vector equation. arametric equation. Distance from a point to a line.
4 A line is specified b a point and a tangent vector Vector equation of a line Definition Fi Cartesian coordinates in R 3 with origin at a point O. Given a point and a vector v in R 3, the line b parallel to v is the set of terminal points of the vectors O r 0 V line r(t) r(t) = r 0 + t v, t R, where r 0 = O. We refer to a line to mean both the set of vectors r(t) and the set of terminal points of these vectors. Vector equation of a line. Find the vector equation of the line b the point = (, 2, ) tangent to the vector v =, 2, 3. Solution: The vector r 0 = O =, 2,, therefore, the formula r(t) = r 0 + t v implies r(t) =, 2, + t, 2, 3.
5 Lines and planes in space (Sect. 2.5) Lines in space Review: Lines on a plane. The equations of lines in space: Vector equation. arametric equation. Distance from a point to a line. arametric equation of a line. Definition The parametric equations of a line b = ( 0, 0, 0 ) tangent to v = v, v, v are given b (t) = 0 + t v, (t) = 0 + t v, (t) = 0 + t v. Remark: It is simple to obtain the parametric equations form the vector equation, and viceversa, noticing the relation r(t) = r 0 + t v (t), (t), (t) = 0, 0, 0 + t v, v, v = ( 0 + t v ), ( 0 + t v ), ( 0 + t v ).
6 arametric equation of a line. Find the parametric equations of the line with vector equation r(t) =, 2, + t, 2, 3. Solution: Rewrite the vector equation in vector components, (t), (t), (t) = ( + t), ( 2 + 2t), ( + 3t). We conclude that (t) = + t, (t) = 2 + 2t, (t) = + 3t. arametric equation of a line. Find both the vector equation and the parametric equation of the line containing the points = (, 2, 3) and Q = (3, 2, ). Solution: A vector tangent to the line is v = Q, which is given b v = (3 ), ( 2 2), ( + 3) v = 2, 4, 4. We can use either or Q to epress the vector equation for the line. If we use, then the vector equation of the line is r(t) =, 2, 3 + t 2, 4, 4. If we choose Q, the vector equation of the line is r(s) = 3, 2, + s 2, 4, 4. We use s to do not confuse it with the t above.
7 arametric equation of a line. Find both the vector equation and the parametric equation of the line containing the points = (, 2, 3) and Q = (3, 2, ). Solution: The parametric equation of the line is simple to obtain once the vector equation is known. Since r(t) =, 2, 3 + t 2, 4, 4, then (t), (t), (t) = ( + 2t), (2 4t), ( 3 + 4t). Then, the parametric equations of the line are given b (t) = + 2t, (t) = 2 4t, (t) = 3 + 4t. arallel lines, perpendicular lines, intersections Definition The lines r(t) = r 0 + t v and ˆr(t) = ˆr 0 + t ˆv are parallel iff their tangent vectors v and ˆv are parallel; the are perpendicular iff v and ˆv are perpendicular; and the lines intersect iff the have a common point. V V V V erpendicular lines in space ma not intersect. Nonparallel lines in space ma not intersect.
8 arallel lines, perpendicular lines, intersections Find the line through = (,, ) and parallel to the line ˆr(t) =, 2, 3 + t 2,, V V Solution: We need to find r 0 and v such that r(t) = r 0 + t v. The vector r 0 is simple to find: r 0 = O =,,. The vector v is simple to find too: v = 2,,. We conclude: r(t) =,, + t 2,,. Find the line through = (,, ) perpendicular to and intersecting the line ˆr(t) =, 2, 3 + t 2,, S 0 V = S 0 V Solution: Find a point S on the intersection such that S is perpendicular to ˆv = 2,,. Writing S t = ˆr(t) = ( + 2t), (2 t), (3 + t), S t = 2t, ( t), (2 + t) ˆv = 2,, St ˆv = 0. 0 = St ˆv = 4t + ( + t) + (2 + t) = 6t + t 0 = 6. S 0 = 2 ( 6, + ) (, 2 ) 6 6 S 0 = 2, 7,. 6 r(t) = O + t S 0 r(t) =,, + t 6 2, 7,.
9 Lines and planes in space (Sect. 2.5) Lines in space Review: Lines on a plane. The equations of lines in space: Vector equation. arametric equation. Distance from a point to a line. Distance from a point to a line. Theorem The distance from a point S in space to a line through the point with tangent vector v is given b d = S v. v S d = S sin( 0 ) S line 0 V
10 Distance from a point to a line. S d = S sin( 0 ) S 0 line V roof. The distance from the point S to the line passing b the point with tangent vector v is given b d = S sin(θ). Recalling that S v = S v sin(θ), we conclude that d = S v. v Distance from a point to a line. Find the distance from the point S = (, 2, ) to the line S 0 S d = S sin( 0 ) line = 2 t, = + 2t, = 2 + 2t. V Solution: First we need to compute the vector equation of the line above. This line has tangent vector v =, 2, 2. (The vector components are the numbers that multipl t.) This line contains the vector = (2,, 2). (Just evaluate the line above at t = 0.) Therefore, S =, 3,.
11 Find the distance from the point S = (, 2, ) to the line S 0 S d = S sin( 0 ) line = 2 t, = + 2t, = 2 + 2t. V Solution: So far: = (2,, 2), v =, 2, 2, and S =, 3,. Since d = S v / v, we need to compute: i j k S v = 3 = (6 + 2) i ( 2 ) j + ( 2 + 3) k, 2 2 that is, S v = 8, 3,. We then compute the lengths: S v = = 74, v = = 3. The distance from S to the line is d = 74/3. Eercise Consider the lines (t) = + t, (s) = 2s, (t) = t, (s) = + s, (t) = t, (s) = 2 + 4s. Are the lines parallel? Do the intersect? Answer: The lines are not parallel. The lines intersect at = (, 3 2, 0 ).
12 Lines and planes in space (Sect. 2.5) lanes in space. Equations of planes in space. Vector equation. Components equation. The line of intersection of two planes. arallel planes and angle between planes. Distance from a point to a plane. A point an a vector determine a plane. Definition Given a point 0 and a nonero vector n in R 3, the plane b 0 perpendicular to n is the set of points solution of the equation ( 0 ) n = 0. n n 0
13 A point an a vector determine a plane. Does the point = (, 2, 3) belong to the plane containing 0 = (3,, 2) and perpendicular to n =,,? 0 n n Solution: We need to know if the vector 0 is perpendicular to n. We first compute 0 as follows, 0 = ( 3), (2 ), (3 2) 0 = 2,,. This vector is orthogonal to n, since ( 0 ) n = = 0. We conclude that belongs to the plane. Lines and planes in space (Sect. 2.5) lanes in space. Equations of planes in space. Vector equation. Components equation. The line of intersection of two planes. arallel planes and angle between planes. Distance from a point to a plane.
14 Equation of a plane in Cartesian coordinates Theorem Given an Cartesian coordinate sstem, the point = (,, ) belongs to the plane b 0 = ( 0, 0, 0 ) perpendicular to n = n, n, n iff holds ( 0 )n + ( 0 )n + ( 0 )n = 0. Furthermore, the equation above can be written as n + n + n = d, d = n 0 + n 0 + n 0. n n 0 O Equation of a plane in Cartesian coordinates Theorem Given an Cartesian coordinate sstem, the point = (,, ) belongs to the plane b 0 = ( 0, 0, 0 ) perpendicular to n = n, n, n iff holds ( 0 )n + ( 0 )n + ( 0 )n = 0. Furthermore, the equation above can be written as n + n + n = d, d = n 0 + n 0 + n 0. roof. In Cartesian coordinates 0 = ( 0 ), ( 0 ), ( 0 ). Therefore, the equation of the plane is 0 = ( 0 ) n = ( 0 )n + ( 0 )n + ( 0 )n.
15 Equation of a plane in Cartesian coordinates Find the equation of a plane containing 0 = (, 2, 3) and perpendicular to n =,, 2. Solution: The point = (,, ) belongs to the plane above iff ( 0 ) n = 0, that is, ( ), ( 2), ( 3),, 2 = 0. Computing the dot product above we get ( ) ( 2) + 2( 3) = 0. The equation of the plane can be also written as + 2 = 5. Equation of a plane in Cartesian coordinates Find a point 0 and the perpendicular vector n to the plane = 3. Solution: We know that the general equation of a plane is n + n + n = d. The components of the vector n, called normal vector, are the coefficients that multipl the variables, and. Therefore, n = 2, 4,. A point 0 on the plane is simple to find. Just look for the intersection of the plane with one of the coordinate ais. For eample: set = 0, = 0 and find from the equation of the plane: 2 = 3, that is = 3/2. Therefore, 0 = (3/2, 0, 0).
16 Equation of a plane in Cartesian coordinates Find the equation of the plane containing the points = (2, 0, 0), Q = (0, 2, ), R = (, 0, 3). R Q Solution: Find two tangent vectors to the plane, for eample, Q = 2, 2, and R =, 0, 3. R Q Equation of a plane in Cartesian coordinates Solution: Find two tangent vectors to the plane, for eample, Q = 2, 2, and R =, 0, 3. A normal vector n to the plane tangent to Q and R can be obtained using the cross product: n = Q R. That is, n = Q i j k R = 2 2 = (6 0) i ( 6 + ) j + (0 + 2) k. 0 3 The result is: n = Q R = 6, 5, 2. Choose an point on the plane, sa = (2, 0, 0). Then, the equation of the plane is: 6( 2) = 0. R Q
17 Lines and planes in space (Sect. 2.5) lanes in space. Equations of planes in space. Vector equation. Components equation. The line of intersection of two planes. arallel planes and angle between planes. Distance from a point to a plane. The line of intersection of two planes. Find a vector tangent to the line of intersection of the planes = 2 and + 2 =. Solution: We need to find a vector perpendicular to both normal vectors n = 2,, 3 and N =, 2,. We choose v = N n. That is, i j k v = N n = 2 = ( 6 + ) i (3 + 2) j + ( 4) k 2 3 Result: v = 5, 5, 5. A simpler choice is v =,,. n n N N V n n N N
18 Lines and planes in space (Sect. 2.5) lanes in space. Equations of planes in space. Vector equation. Components equation. The line of intersection of two planes. arallel planes and angle between planes. Distance from a point to a plane. arallel planes and angle between planes Definition Two planes are parallel if their normal vectors are parallel. The angle between two nonparallel planes is the smaller angle between their normal vectors. n N N n N n
19 arallel planes and angle between planes Find the angle between the planes = 2 and + 2 =. Solution: We need to find the angle between the normal vectors n = 2,, 3 and N =, 2,. We use the dot product: cos(θ) = n N n N. The numbers we need are: n N = = 3, n = = 4, N = = 6 Therefore, cos(θ) = 3/ 84. We conclude that θ = Lines and planes in space (Sect. 2.5) lanes in space. Equations of planes in space. Vector equation. Components equation. The line of intersection of two planes. arallel planes and angle between planes. Distance from a point to a plane.
20 Distance formula from a point to a plane Theorem The distance d from a point to a plane containing 0 with normal vector n is the shortest distance from to an point in the plane, and is given b the epression d = ( 0 ) n. n n n d 0 Distance formula from a point to a plane roof. We need to proof the distance formula d = ( 0 ) n. n n d n From the picture we see that d = 0 cos(θ), where θ is the angle between 0 and n, where the absolute value are needed since the distance is a nonnegative number. Recall: ( 0 ) n = 0 n cos(θ) 0 cos(θ) = 0 ( 0 ) n. n Take the absolute value above, and that is the formula for d.
21 Distance formula from a point to a plane Find the distance from the point = (, 2, 3) to the plane = 4. Solution: We need to find a point 0 on the plane and its normal vector n. Then use the formula d = ( 0 ) n / n. A point on the plane is simple to find: Choose a point that intersects one of the ais, for eample = 0, = 0, and = 4. That is, 0 = (4, 0, 0). The normal vector is in the plane equation: n =, 3, 2. We now compute 0 = 3, 2, 3. Then, d = d = 3 4. Clinders and quadratic surfaces (Sect. 2.6). Clinders. Quadratic surfaces: Spheres, 2 r r 2 Ellipsoids, 2 a + 2 Cones, 2 a + 2 Hperboloids, 2 a r 2 =. + 2 c 2 =. 2 c 2 = 0. araboloids, 2 a b 2 c = 0. Saddles, 2 a 2 2 b 2 c = 0. 2 c =, 2 2 a c 2 =.
22 Clinders. Definition Given a curve on a plane, called the generating curve, a clinder is a surface in space generating b moving along the generating curve a straight line perpendicular to the plane containing the generating curve. A circular clinder is the particular case when the generating curve is a circle. In the picture, the generating curve lies on the plane. r Find the equation of the clinder given in the picture. r Solution: The intersection of the clinder with the = 0 plane is a circle with radius r, hence points of the form (,, 0) belong to the clinder iff = r 2 and = 0. For 0, the intersection of horiontal planes of constant with the clinder again are circles of radius r, hence points of the form (,, ) belong to the clinder iff = r 2 and constant. Summariing, the equation of the clinder is = r 2. We do not mention the coordinate, since the equation above holds for ever value of R.
23 Find the equation of the clinder given in the picture. r Solution: The generating curve is a circle, but this time on the plane = 0. Hence point of the form (, 0, ) belong to the clinder iff = r 2. We conclude that the equation of the clinder above is = r 2. We do not mention the coordinate, since the equation above holds for ever value of R. Find the equation of the clinder given in the picture. 4 2 parabola Solution: The generating curve is a parabola on planes with constant. This parabola contains the points (0, 0, 0), (, 0, ), and (2, 0, 4). Since three points determine a unique parabola and = 2 contains these points, then at = 0 the generating curve is = 2. The clinder equation does not contain the coordinate. Hence, = 2, R.
24 Clinders and quadratic surfaces (Sect. 2.6). Clinders. Quadratic surfaces: Spheres. Ellipsoids. Cones. Hperboloids. araboloids. Saddles. Quadratic surfaces. Definition Given constants a i, b i and c, with i =, 2, 3, a quadratic surface in space is the set of points (,, ) solutions of the equation Remark: a 2 + a a b + b 2 + b 3 + c = 0. The coefficients b, b 2, b 3 pla a role moving around the surface in space. We stud onl quadratic equations of the form: a 2 + a a b 3 = c 2. () The surfaces below are rotations of the one in Eq. (), a 2 + a a b 3 = c 2, a 2 + a a b 3 = c 2.
25 Clinders and quadratic surfaces (Sect. 2.6). Clinders. Quadratic surfaces: Spheres. 2 r r 2 Ellipsoids. Cones. Hperboloids. araboloids. Saddles. + 2 r 2 =. Spheres. Recall: We stud onl quadratic equations of the form: a 2 + a a b 3 = c 2. A sphere is a simple quadratic surface, the one in the picture has the equation 2 r r r 2 =. (a = a 2 = a 3 = /r 2, b 3 = 0 and c 2 =.) Equivalentl, = r 2. r
26 Spheres. Recall: Linear terms move the surface around in space. Graph the surface given b the equation = 0. Solution: Complete the square: 2 + [ ( 4 ( 4 2 ] ( 4 ) ) 2) 2 = 0. 2 Therefore, 2 + ( + 4 2) = 4. This is the equation of a sphere centered at 0 = (0, 2, 0) and with radius r = 2. 2 Clinders and quadratic surfaces (Sect. 2.6). Clinders. Quadratic surfaces: Spheres, 2 r r 2 Ellipsoids, 2 a + 2 araboloids. Cones. Hperboloids. Saddles. + 2 r 2 =. + 2 c 2 =.
27 Ellipsoids. Definition Given positive constants a, b, c, an ellipsoid centered at the origin is the set of point solution to the equation 2 a b c 2 =. Graph the ellipsoid, =. 2 3 Graph the ellipsoid, Solution: =. On the plane = 0 we have the ellipse =. 2 3 On the plane = 0, with 2 ( < 0 < 2 we have the ellipse = ). 2 2 Denoting c = (0 2 /4), then 0 < c <, and 2 c c =. 2 3
28 Clinders and quadratic surfaces (Sect. 2.6). Clinders. Quadratic surfaces: Spheres, 2 r r 2 Ellipsoids, 2 a + 2 Cones, 2 a + 2 Hperboloids. araboloids. Saddles. + 2 r 2 =. + 2 c 2 =. 2 c 2 = 0. Cones. Definition Given positive constants a, b, a cone centered at the origin is the set of point solution to the equation 2 = ± a b 2. Graph the cone, =
29 Graph the cone, = Solution: On the plane = we have the ellipse =. 2 2 On the plane = 0 > 0 we have the ellipse = 0 2, that is, =. 2 2 Clinders and quadratic surfaces (Sect. 2.6). Clinders. Quadratic surfaces: Spheres, 2 r r 2 Ellipsoids, 2 a + 2 Cones, 2 a + 2 Hperboloids, 2 a + 2 araboloids. Saddles. + 2 r 2 =. + 2 c 2 =. 2 c 2 = 0. 2 c =, 2 2 a c 2 =.
30 Hperboloids. Definition Given positive constants a, b, c, a one sheet hperboloid centered at the origin is the set of point solution to the equation (One negative sign, one sheet.) 2 a b 2 2 c 2 =. ellipse Graph the hperboloid, =. hperbola 2 hperbola hperbolas Hperboloids. ellipse Graph the hperboloid =. hperbola 2 hperbola hperbolas Solution: Find the intersection of the surface with horiontal and vertical planes. Then combine all these results into a qualitative graph. On horiontal planes, = 0, we obtain ellipses = On vertical planes, = 0, we obtain hperbolas 2 2 = On vertical planes, = 0, we obtain hperbolas = 2 0.
31 Hperboloids. Definition Given positive constants a, b, c, a two sheet hperboloid centered at the origin is the set of point solution to the equation (Two negative signs, two sheets.) 2 a 2 2 b c 2 =. Graph the hperboloid, =. ellipse 2 hperbola hperbola Hperboloids. Graph the hperboloid =. Solution: Find the intersection of the surface with horiontal and vertical planes. Then combine all these results into a qualitative graph. ellipse 2 hperbola On horiontal planes, = 0, with 0 >, we obtain ellipses = On vertical planes, = 0, we obtain hperbolas = On vertical planes, = 0, we obtain hperbolas = hperbola
32 Clinders and quadratic surfaces (Sect. 2.6). Clinders. Quadratic surfaces: Spheres, 2 r r 2 Ellipsoids, 2 a + 2 Cones, 2 a + 2 Hperboloids, 2 a r 2 =. + 2 c 2 =. 2 c 2 = 0. araboloids, 2 a b 2 c = 0. Saddles. 2 c =, 2 2 a c 2 =. araboloids. Definition Given positive constants a, b, a paraboloid centered at the origin is the set of point solution to the equation = 2 a b 2. Graph the paraboloid, = parabola 2 ellipse
33 araboloids. Graph the paraboloid = parabola 2 Solution: Find the intersection of the surface with horiontal and vertical planes. Then combine all these results into a qualitative graph. On horiontal planes, = 0, with 0 > 0, we obtain ellipses = 0. On vertical planes, = 0, we obtain parabolas = On vertical planes, = 0, we obtain parabolas = ellipse Clinders and quadratic surfaces (Sect. 2.6). Clinders. Quadratic surfaces: Spheres, 2 r r 2 Ellipsoids, 2 a + 2 Cones, 2 a + 2 Hperboloids, 2 a r 2 =. + 2 c 2 =. 2 c 2 = 0. araboloids, 2 a b 2 c = 0. Saddles, 2 a 2 2 b 2 c = 0. 2 c =, 2 2 a c 2 =.
34 Saddles, or hperbolic paraboloids. Definition Given positive constants a, b, c, a saddle centered at the origin is the set of point solution to the equation = 2 a 2 2 b 2. Graph the paraboloid, = Saddles. Graph the saddle = parabola parabola hperbola Solution: Find the intersection of the surface with horiontal and vertical planes. Then combine all these results into a qualitative graph. On planes, = 0, we obtain hperbolas = 0. On planes, = 0, we obtain parabolas = On planes, = 0, we obtain parabolas =
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