PART I: APPLICATION OF VECTORS AND VECTOR-VALUED FUNCTIONS Lines in Space Vectors that connect any pair of points on a line l are k. Vector description of a line Consider the locus of the terminal points of the vectors (first in 2D) ~ tA −∞<t<∞ It is a straight line. This line can be displaced to ~ describe any straight line k to A. The general vector expression for a line in space is: ~ ~r = ~r0 + tA −∞<t<∞ ~ is a where ~r0 is an arbitrary point on the line and A vector k to the line. Example Find a vector equation of the line that contains (−1, 3, 0) and is parallel to 2¯i − 3¯j − k¯ Example Find a vector equation of the line that passes through the two points: (1, 2, 3) and (4, 5, 6). Note ~r(t) gives the location of a point on the line. It ~ is. Namely, if P ~ is not parallel to the line. A ~ are any two different points on the line, and Q ~ then P~Q k A. Parametric equations x = x0 + tA1 y = y0 + tA2 z = z0 + tA3 Example Find a set of parametric equations for the line of the previous example. Analytic geometry (AG) description of a line a1 x + b1 y + c1 z = d1 a2 x + b2 y + c2 z = d2 Example Find an AG description of the line in the previous example. Degrees of Freedom / Dimensions ≡ number of free parameters used to describe the geometrical object ~ be a vector in space, the object tA ~ Example Let A (−∞ < t < ∞) has one free parameter, therefore the object has dimension 1. ~ and B ~ are not parallel, and both 6= ~0, Example If A what is the object ~ + sB ~ −∞ < t < ∞ −∞ < s < ∞ ? tA Planes in 3D Space Basic geometrical facts (i) Two different intersecting lines l1 , l2 (directions given ~ B ~ respectively) define a plane through them. by A, (ii) There is a unique line l3 that passes through the point of intersection and is ⊥ to both lines. A vector ~ k l3 defines a normal to the plane that contains N ~ ·A ~=N ~ ·B ~ = 0) l1 , l2 . (N (iii) There is only one plane that contains a given point and is ⊥ to a given nonzero vector. (i) can be viewed as the definition of a plane; (ii) & (iii) come from the three-dimensional nature of space. Vector description of a plane A plane through two intersecting lines l1 and l2 can be generated as: ~ + sB ~ ~r = ~r0 + tA − ∞ < t < ∞, −∞ < s < ∞ ~ B ~ where ~r0 is the point of intersection of l1 , l2 , and A, are vectors along l1 , l2 respectively. A normal is perpendicular to all directed line seg~ · (tA ~ + sB) ~ = 0. ments on the plane as N Recall (ii) above for the definition of a normal. A vector is a normal to a plane iff it is perpendicular to all vectors associated with the directed line segments on the plane. Parametric description of a plane: Decompose the vector equation into components, as (linear) functions of the parameters. x = x0 + tA1 + sB1 y = y0 + tA2 + sB2 z = z0 + tA3 + sB3 This approach is not used much for planes, but very useful for general surfaces. The coordinates of a point on a surface can be specified as: x = f1 (t, s), y = f2 (t, s), z = f3 (t, s). These are two-variable functions which make components of a function f : R2 → R3 . AG description of a plane: Let P be a fixed point on the plane, Q be any point ~ be a normal. Then on the plane, and N ~ · P~Q = 0 N ~ = (x, y, z), and N ~ = Writing P~ = (x0 , y0 , z0 ), Q (a, b, c), one has (a, b, c)· [(x, y, z) − (x0 , y0 , z0 )] = a(x − x0 ) + b(y − y0 ) + c(z − z0 ) = 0. The resultant equation is of the form ax + by + cz = d. A normal to the plane can be found easily as ~ = a~i + b~j + c~k. N Example: Find an equation of the plane that contains the point (−2, 4, 5) and has normal vector 7~i − 6~k Example: Find a unit normal to the plane x + y + z = 1. Example: Find the plane through the three points (0, 0, 1), (1, 0, 1), (0, 1, 1). Quadric Surfaces Definition: A quadric surface is a surface which contains points that satisfies the 2nd degree polynomial equation (AG description) Ax2 +By 2 +Cz 2 +Dxy+Eyz+F zx+Gx+Hy+Iz+J = 0 where A, ..., J are constants. Quadric surfaces fall into nine major classes. Examples of each are provided as following (a, b, c > 0): 1. 2. 3. 4. 5. 6. 7. 8. 9. x2 y2 z2 Ellipsoid + 2 + 2 =1 2 a b c x2 y2 z2 Hyperboloid of one sheet + 2 − 2 =1 a2 b c y2 z2 x2 − 2 − 2 =1 Hyperboloid of two sheets 2 a b c x2 y2 z2 Elliptic double cone + 2 − 2 =0 2 a b c x2 y2 z Elliptic paraboloid + 2 = a2 b c x2 y2 z Hyperbolic paraboloid − = a2 b2 c y2 x2 + 2 =1 Elliptic cylinder a2 b x2 y2 Hyperbolic cylinder − 2 =1 2 a b z Parabolic cylinder x2 = c —– Problem Set 1 —–

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