Find a linear function whose graph is the plane that intersects the xy-plane along the line y=2x+2 and contains the point (1,2,2).
To find a linear function whose graph is the plane that intersects the xy-plane along the line y=2x+2 and contains the point (1,2,2), we need to follow these steps:
Step 1: Understand the problem.
The given line, y=2x+2, represents the intersection of the plane with the xy-plane. The equation of the line is in the form of a linear function, y = mx + b, where m is the slope and b is the y-intercept. The plane to be found needs to pass through the given point (1,2,2) and also contain the line.
Step 2: Find the normal vector of the plane.
Since we need to find a linear function (equivalent to finding a plane), we can recognize that the normal vector of the plane should be orthogonal (perpendicular) to the xy-plane. The xy-plane has a normal vector pointing in the direction (0, 0, 1) since it is parallel to the z-axis.
Step 3: Find the direction vector of the line.
To find the direction vector of the line y=2x+2, we can observe that the coefficients of x and y give the values of the direction vector. Thus, the direction vector of the line is (1, 2).
Step 4: Find the cross product of the normal vector and the direction vector.
The cross product of the normal vector and the direction vector will give a vector that is orthogonal to both vectors. This vector will lie on the plane we are looking for. Let's calculate the cross product:
(0, 0, 1) x (1, 2, 0) = (2, -1, 0)
Step 5: Find the equation of the plane.
We know that the plane passes through the point (1, 2, 2). Using the equation of a plane, Ax + By + Cz = D, we can substitute the values to find the equation:
(2)(x - 1) + (-1)(y - 2) + (0)(z - 2) = 0
Simplifying, we get:
2x - 2 - y + 2 = 0
This simplifies to:
2x - y = 0
Thus, the linear function (equation of the plane) is 2x - y = 0.