4. A plane has Cartetiar equation
4x - 3y + 6z - 12 =-- 0.
a. Determine vector and p.iramqtric conations for
this plane.
b. Determine the equation of a line that lies in
the plane.
To determine the vector and parametric equations for the plane with the Cartesian equation 4x - 3y + 6z - 12 = 0, we need to isolate one variable in terms of the other two variables.
a. Vector equation for the plane:
Let's isolate z:
6z = 12 - 4x + 3y
Divide both sides by 6:
z = (12 - 4x + 3y)/6
The vector equation for the plane is [x, y, z] = [x, y, (12 - 4x + 3y)/6], where x and y can take any real values.
Parametric equations for the plane:
From the vector equation, we can express x, y, and z in terms of a parameter, typically denoted by t.
Let's assume t = y, then we have:
x = t
y = t
z = (12 - 4t + 3t)/6 = (12 - t)/6
So, the parametric equations for the plane are:
x = t
y = t
z = (12 - t)/6
b. Equation of a line in the plane:
Since the parametric equations represent a line, we can choose any two values for t and calculate the corresponding values of x, y, and z to get points on the line. Let's choose t = 0 and t = 1.
For t = 0:
x = 0
y = 0
z = (12 - 0)/6 = 2
For t = 1:
x = 1
y = 1
z = (12 - 1)/6 = 11/6
So, the points on the line are (0, 0, 2) and (1, 1, 11/6). Using these two points, we can find the direction vector of the line.
Direction vector of the line:
Let's find the difference vector between the two points:
[1 - 0, 1 - 0, 11/6 - 2] = [1, 1, -1/6]
The equation of the line can be written as:
[x, y, z] = [0, 0, 2] + t[1, 1, -1/6], where t is a real parameter.