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
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
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 2-D 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 2-D 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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