8.++Equations+of+Lines+and+Planes

**Tuesday, May 22, 2012 - Sophie Yung**

 * EQUATIONS OF LINES**

Intro: From a dot on a line, you can arrive at any other dot along the line using a scalar multiple of a direction vector

[x, y, z] = [ a1, a2, a3] + t[m1, m2, m3]

r = r0 + tm where… r = all possible points on the line r0 = original point on the line t = belongs to all real numbers m = direction vector

E.g. r0 = (-1, 2.1, 0.5) m = (7, 3.8, -0.8)

r = r0 + tm r = (-1, 2.1, 0.5) + t(7, 3.8, -0.8)


 * Parametric Equations**

x = a1 + m1t y = a2 + m2t z = a3 + m3t

E.g.

x = -1 + 7t y = 2.1 + 3.8t z = 0.5 = 0.8t


 * Symmetric Equations**

Rearrange for t:

((x-a1)/m1) = ((y-a2)/m2) = ((z-a3)/m3) m1, m2, m3 ≠ 0

E.g. ((x+1)/7) = ((y-2.1)/3.8) = ((z-0.5)/-0.8)

Given r = [2, 3, -3] + t[4, 0, -1] a) Find two points on the line other than (2, 3, -3) b) Is the point (-10, 3, 0) on the line? c) Write the parametric and symmetric equations

a) Let t = 1, (6, 3, -4) Let t = 2, (10, 3, -5) Let t = -1, (-2, 3, -2)

b) -10 = 2 + 4t -10 – 2 = 4t -12 = 4t t = -3 Check: 3 + (-3)(0) = 3 and -3 + (-3)(-1) = -3 + 3

Therefore, (-10, 3, 0) is on the line.

c) Parametric: x = 2+4t y = 3 z = -3-t

Symmetric: ((x-2)/4) = ((z+3)/-1), y = 3


 * Equations of Lines in 2-Space**

Ax + By + C = 0 x – 9y + 5 = 0

y = mx + b y = (1/9)x + (5/9)

y – y0 = m(x – x0) y – 1 = (1/9)(x – 4) r = (a1, a2) + t(m1, m2) r = (4, 1) + t(9, 1) or r = (0, (5/9)) + t(9, 1)

n = normal (vector perpendicular to line)

x = 4 + 9t y = 1 + t

((x-4)/9) = ((y-1)/1)

Find the parametric equation of the line 3x – 2y + 1 = 0 x = a1 + tm1 y = a2 + tm2

m = (2, 3) n = (3, -2) nm = 0

let x = -1, (-1, -1) or (1, 2), etc. x = -1 + 2s y = -1 + 3s

Given r = (-3, 4) + t(-4, 1), find the equation of the line in slope intercept form y = (-1/4)x + b 4 = (-1/4)(-3) + b (16/4) – (3/4) = b (13/4) = b

y = (-1/4)x + (13/4)

Homework: txb - p. 141# 4, 5 11, 13 - p. 132# 7-11, 13, 14

Link: http://www.netcomuk.co.uk/~jenolive/vect3.html Video: http://www.youtube.com/watch?v=CpA6OP3THwc


 * __Wednesday, May 23, 2012 - Charles Lin__**
 * __Intersection of Lines in 3-Space__**

Lesson summary:
 * In 3-space, a pair of lines can interact in four separate ways (see video)
 * they can be parallel (and thus never intersect)
 * they can be not parallel and intersect
 * they can be not parallel and not intersect (they are skewed)
 * to find the point of intersection of two lines
 * convert each line to their parametric equations
 * set the x equations, y equations, and z equations equal to each other
 * use two equations to solve for your variable, and then plug it back into your third equation
 * If the variable works for all three equations, then the lines intersect at the x, y and z coordinates given by those equations
 * If the variable does not work for all three equations, the lines do not intersect

E.g. find the intersection between

r = [-1, 2, 3] + t[3, 1, 1]

and

(x + 2) / 3 = (y - 1) / 2 = z / 3

//convert both lines to their parametric equations//

x = -1 + 3t y = 2 + t z = 3 + t

x = -2 + 3s y = 1 + 2s z = 3s

//set the equations equal to each other//

x: -1 + 3t = -2 + 3s y: 2 + t = 1 + 2s z: 3 + t = 3s

//solve for s and t, using two equations//

subtracting y from z: 3 + t - (2 + t) = 3s - (1 + 2s) 1 = s - 1 s = 2 t = 3

//sub s and t back into the third equation//

using x: LS: -1 + 3t = 8

RS: -2 + 3s = 4

LS =! RS, therefore the lines are skewed.

Homework: p. 133 #15, 16; p. 142 #14, 15

Resource: []

**Monday, May 28 - Dhairya Patel**

 * Tuesday, May 29 - Chanice Hazlitt**


 * Two planes can be parallel, intersecting or the same plane
 * If they are the same plane they will have the same scalar equation
 * If they are parallel their normal vectors will be scalar multiples
 * If they are not parallel they intersect in a line

E.g. Find the intersection between the following planes.

2x - y + z + 3 = 0 x + 3y - z + 5 = 0

Since the normals are not scalar multiples the planes are not parallel.

Add the equations:

3x + 2y + 8 = 0 3x = -2y - 8 x = -8/3 - (2/3)y

let y = s

x = -8/3 - (2/3)s

Substitute into one of the original equations and solve for z.

-8/3 - (2/3)s + 3s - z + 5 = 0 z = 7/3 + (7/3)s

The equation of the line is

x = -8/3 - (2/3)s y = s z = 7/3 + (7/3)s

or

r = (-8/3, 0, 7/3) + s(-2/3, 1, 7/3)

Homework: p. 168 #5, 8, 16

Resource: __[]__

Intersection of 3 Planes - Charlie Zhao Wednesday 30/05/2012

The Intersection of Two Planes

Finding Distances: June 1st (Olivia Weller)

Review & Test Prep: June 4th (Ayesha Tasneem) No lesson today. Review on pages: 204-207. Self-Test- Pg 208. Helpful sites with examples: [] [|http://www.math.washington.edu/~king/coursedir/m445w04/notes/vector/equations.html] []

__Dhairya__ __&__ __Ren-David__ Below are the answers to the latest test. Sorry for the quality, my scanner is out of order.
 * __TEST ANSWERS:__**