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
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
Sure! Let's go through each question step by step.
20) To find the normal vector, direction vector, and a point on the line, we can rearrange the given equation into the form Ax + By + C = 0.
The given equation is 3x - 6y = 14.
By rearranging it, we get: 3x - 6y - 14 = 0.
Comparing this equation with the general form Ax + By + C = 0, we can identify A = 3, B = -6, and C = -14.
Now, the normal vector of the line is given by (A, B). So, the normal vector is (3, -6).
To find the direction vector, we can take any convenient nonzero values for x and y in our initial equation and solve for the corresponding variable.
Let's take x = 0. When we substitute x = 0 into the equation 3x - 6y = 14, we get -6y = 14, which implies y = -14/6 = -7/3.
This gives us the point (0, -7/3) on the line.
Now, let's find the direction vector by taking another point on the line and subtracting the coordinates of the two points. Let's choose x = 1. When we substitute x = 1 into the equation 3x - 6y = 14, we get 3 - 6y = 14, which implies y = (3-14)/(-6) = -11/6.
This gives us another point (1, -11/6) on the line.
The direction vector is obtained by subtracting the coordinates of the two points: (1 - 0, -11/6 - (-7/3)) = (1, -11/6 + 7/3) = (1, -5/6).
So, the direction vector is (2, 1).
The point on the line can be found by rearranging the equation and solving for x.
Starting with the initial equation, 3x - 6y = 14, we can isolate x as follows:
3x = 14 + 6y
x = (14 + 6y)/3
x = 14/3 + 2y
Therefore, the point on the line is (14/3, 0).
In summary,
Normal vector = (1, -2)
Direction vector = (2, 1)
Point on the line = (14/3, 0)
21) To find the vector, parametric, and symmetric equations of the given line, we need to rearrange the equation into different forms.
The given equation is -4x + 6y + 9 = 0.
To find the vector equation, we can rewrite the equation in the form r = a + t*b, where r is the position vector, a is a known vector, t is a scalar parameter, and b is the direction vector.
From the given equation, we can rearrange it as follows:
-4x + 6y = -9
6y = 4x - 9
y = (4/6)x - (9/6)
y = (2/3)x - (3/2)
Now, we can write the vector equation as:
r = [0, (-3/2)] + t[3, 2]
To find the parametric equations, we can express x and y in terms of a parameter, usually denoted as t.
From the original equation, we have:
-4x + 6y + 9 = 0
Rearranging it, we get:
-4x = -6y - 9
x = (-6y - 9)/-4
x = (3/2)y + (9/4)
Therefore, the parametric equations are:
x = (3/2)t + (9/4)
y = t
Lastly, to find the symmetric equation, we can express x and y in terms of each other.
Starting with the equation: y = (2/3)x - (3/2)
We can multiply both sides by 3:
3y = 2x - 9/2
Rearranging, we get:
-2x + 3y = -9/2
Dividing by -2, we have:
x - (3/2)y = 9/4
Therefore, the symmetric equation is:
(x/3) = [y + (3/2)] / 2
In summary,
Vector equation: r = [0, (-3/2)] + t[3, 2]
Parametric equations: x = (3/2)t + (9/4), y = t
Symmetric equation: (x/3) = [y + (3/2)] / 2