Answer:
To find the vector equation, parametric equations, and symmetric equations for the line that goes through the points (-8, 1, 4) and (3, -2, 4), we can follow these steps:
1. Find the direction vector of the line.
2. Choose one of the points to serve as the initial point (a point on the line).
Let's start with step 1:
1. Find the direction vector of the line:
To find the direction vector, subtract one point from the other. Let's use (-8, 1, 4) as the initial point (P1) and (3, -2, 4) as the terminal point (P2):
Direction vector (D) = P2 - P1
D = (3, -2, 4) - (-8, 1, 4)
D = (3, -2, 4) + (8, -1, -4)
D = (11, -3, 0)
So, the direction vector of the line is D = (11, -3, 0).
Now, let's move on to step 2:
2. Choose one of the points as the initial point. Let's use (-8, 1, 4) as the initial point (P1).
Now, we can write the vector equation, parametric equations, and symmetric equations for the line:
Vector Equation:
The vector equation of the line is given by:
R(t) = P1 + t * D
where:
- R(t) is a vector representing any point on the line.
- P1 = (-8, 1, 4) is the initial point on the line.
- D = (11, -3, 0) is the direction vector.
- t is a parameter that varies along the line.
Parametric Equations:
The parametric equations of the line are given by:
x(t) = -8 + 11t
y(t) = 1 - 3t
z(t) = 4
Symmetric Equations:
The symmetric equations of the line are given by:
(x + 8) / 11 = (y - 1) / (-3) = (z - 4) / 0
Please note that the third equation, (z - 4) / 0, doesn't have a meaningful solution for z, as the denominator is zero. This means that the line lies in the plane z = 4 and does not change in the z-direction. So, the line is essentially a horizontal line in the plane z = 4.
Explanation: