Find the parametric equation of a line that contains (2, -1, 1) and is parallel to the vector (3, 0.5, 0).
To find the parametric equation of a line, we need a point on the line and a vector parallel to the line.
Given that the line is parallel to the vector (3, 0.5, 0), we can use this vector as the direction vector for our line.
Let's start with the point (2, -1, 1) and the vector (3, 0.5, 0) to find the parametric equations.
The parametric equations for a line can be written in the form:
x = x₁ + at,
y = y₁ + bt,
z = z₁ + ct,
where (x₁, y₁, z₁) is a point on the line, (a, b, c) is the direction vector, and t is a parameter.
In this case, x₁ = 2, y₁ = -1, z₁ = 1, and (a, b, c) = (3, 0.5, 0).
So, the parametric equations for the line are:
x = 2 + 3t,
y = -1 + 0.5t,
z = 1 + 0t.
Simplifying, we find:
x = 2 + 3t,
y = -1 + 0.5t,
z = 1.
To find the parametric equation of a line, we need a point on the line and a direction vector. In this case, we have the point (2, -1, 1) and the direction vector (3, 0.5, 0), which is parallel to the line.
Let's call the coordinates of our point on the line (x, y, z), and the parameters of the direction vector a, b, and c. Then, the parametric equation of the line can be written as:
x = 2 + 3a
y = -1 + 0.5a
z = 1 + 0a
So, the parametric equation of the line that contains the point (2, -1, 1) and is parallel to the vector (3, 0.5, 0) is:
x = 2 + 3a
y = -1 + 0.5a
z = 1