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]
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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
Sure, I can help you with that.
20) To find the normal vector of the line, we need to rearrange the equation of the line in the form Ax + By = C, where A, B, and C are constants.
Given the equation 3x - 6y = 14, we can rewrite it as -6y = -3x + 14.
Dividing both sides of the equation by -6, we get y = (1/2)x - (7/3).
Now we can compare this equation with the general form Ax + By = C. We can see that A = 1, B = -2, and C = 0.
So the normal vector of the line is (1, -2).
To find the direction vector of the line, we can rearrange the equation in slope-intercept form y = mx + b, where m is the slope.
From the equation y = (1/2)x - (7/3), we can see that the slope is 1/2.
Therefore, the direction vector of the line is (2, 1).
To find a point on the line, we can set y = 0 in the original equation and solve for x.
In 3x - 6y = 14, if y = 0, then we have 3x = 14, and solving for x, we get x = 14/3.
So a point on the line is (14/3, 0).
Therefore, the normal vector is (1, -2), the direction vector is (2, 1), and a point on the line is (14/3, 0).
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21) To find the vector equation of the line, we need a direction vector and a point on the line.
Given the equation -4x + 6y + 9 = 0, we can rewrite it as 6y = 4x - 9.
Dividing both sides of the equation by 6, we get y = (2/3)x - (3/2).
Comparing this with the slope-intercept form, we can see that the slope is 2/3.
Therefore, the direction vector of the line is (3, 2).
To find a point on the line, we can set x = 0 in the original equation and solve for y.
In -4x + 6y + 9 = 0, if x = 0, then we have 6y = -9, and solving for y, we get y = -3/2.
So a point on the line is (0, -3/2).
Therefore, the vector equation of the line is r = [0, (-3/2)] + t(3, 2).
To find the parametric equations of the line, we can write x = x0 + at and y = y0 + bt, where (x0, y0) is a point on the line.
From the vector equation, we can see that x0 = 0 and y0 = -3/2.
Also, the coefficients of t in the vector equation represent a and b.
So the parametric equations of the line are x = 3t and y = 2t - (3/2).
To find the symmetric equation of the line, we can set x = 3t and y = 2t - (3/2) in the original equation and solve for t.
Substituting these values in -4x + 6y + 9 = 0, we get -4(3t) + 6(2t - (3/2)) + 9 = 0.
Simplifying the equation, we have -12t + 12t - 9 + 9 = 0, which simplifies to 0 = 0.
Since the equation is always satisfied, any value of t is a solution.
So, the symmetric equation of the line is (x/3) = [y + (3/2)] / 2.
Therefore, the vector equation of the line is r = [0, (-3/2)] + t(3, 2), the parametric equations are x = 3t and y = 2t - (3/2), and the symmetric equation is (x/3) = [y + (3/2)] / 2.