Posted by Anonymous on Wednesday, March 12, 2008 at 11:09am.
I'm stuck on these questions. Can you please show me a step by step solution?
20) For each of the following, find a normal vector, a direction vector, and a point of each line.
Note, normal vector means perpendicular to the direction vector.
a) 3x  6y = 14
TEXTBOOK ANSWER:
normal vector = (1 , 2)
direction vector = (2 , 1)
point on the line = [(14/3) , 0]

21) Find vector, parametric, and symmetric equations of the following lines.
a) 4x + 6y + 9 = 0
TEXTBOOK ANSWER:
Vector Equation > r = [0 , (3/2)] + t(3 , 2)
Parametric Equation > x = 3t and y = 2t  (3/2)
Symmetric Equation > (x/3) = [y + (3/2)] / 2

Discrete Math: Equations of Line in a Plane  Reiny, Wednesday, March 12, 2008 at 12:32pm
I will do the second one, since it looks a bit more difficult than the first
You are clearly working in only 2D here.
I assume you know that for any straight line
Ax + By + C = 0
[A,B] is a normal to the line which makes
[B,A] a direction vector
and the slope of the line is A/B
the vector equation of a line in 2D is
r = [any point on the line] + t[direction vector of the line], where t is a parameter.
so they started by letting x=0, then y = 9/6 = 3/2
so the point they used is (0,3/2)
and the direction of the line is [6,4] or [3,2]
so the vector equation is
r = [0,3/2] + t[3,2] as shown in your text.
if we "expand" that we get
r = [0+3t,3/2 + 2t]
from that you can see that the
x=3t and the
y=3/2 + 2t as shown in your book answer
now if we solve these last two equations for t we get
t = x/3 and t = (y+3/2)/2
equating t=t gives us their third answer of
x/3 = (y+3/2)/2
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