2. Determine the Cartesian equation of the plane
passing through the point P(2, -1,2) and
containing the line r = (2, -1, 4) + f(0, 3, -5).
To determine the Cartesian equation of the plane passing through the point P(2, -1, 2) and containing the line r = (2, -1, 4) + f(0, 3, -5), we can use the following steps:
Step 1: Find a vector normal to the plane.
Since the plane is defined by a point and a line, the vector normal to the plane can be found by taking the cross product of the direction vector of the line and a vector formed by two points on the plane.
The direction vector of the line is (0, 3, -5).
To find the vector formed by two points on the plane, we can use the point P(2, -1, 2) and another point on the line. Let's choose P(2, -1, 2) and P(2, -1, 4).
Vector formed by two points on the plane = P(2, -1, 4) - P(2, -1, 2) = (0, 0, 2)
Now we can take the cross product of the direction vector of the line and the vector formed by two points on the plane:
Normal vector = (0, 3, -5) x (0, 0, 2) = (6, 0, 0)
Step 2: Write the equation of the plane using the normal vector and the coordinates of a point on the plane.
The equation of a plane can be written as Ax + By + Cz + D = 0, where (A, B, C) is the normal vector and (x, y, z) are the coordinates of a point on the plane.
Using the normal vector (6, 0, 0) and the coordinates of the point P(2, -1, 2), we can write the equation of the plane as:
6x + 0y + 0z + D = 0
Simplifying the equation gives us:
6x + D = 0
As we have not been given any information about the value of D, we cannot determine the exact Cartesian equation of the plane. However, we can say that the equation of the plane is of the form 6x + D = 0, where D is an unknown constant.