Final answer:
To find the equation of the plane through the point (1, -1, 1) and containing the line x=3y=4z, you need two direction vectors from the line and point, take their cross product for the normal vector, and use the point-normal form of a plane.
Step-by-step explanation:
To find the equation of a plane that goes through a specific point and contains a line, you need a point on the plane and a normal vector to the plane. The symmetric equations given, x = 3y = 4z, suggest that any point on the line can be written as (t, t/3, t/4), for a particular value of t. The point (1, -1, 1) is also on the plane, but not on the given line.
We can use the two points to find the direction vector of the line on the plane. Let's take t=3 (arbitrary choice) to get the second point (3, 1, 0.75) on the line, which is also on the plane. Now, we find the direction vector by subtracting the coordinates of the given point from this new point - direction vector = (3-1, 1-(-1), 0.75-1) = (2, 2, -0.25).
Since planes are defined by a point and a normal vector, we need to find a vector that is perpendicular to the direction vector. Here, we can use the cross product of two direction vectors if we find another direction vector on the plane. To do this, we may take another value for t - let's say t=4, which gives us the point (4, 4/3, 1) on the plane. Subtracting again from the given point, we get the second direction vector = (4-1, 4/3-(-1), 1-1) = (3, 7/3, 0).
The cross product of the two direction vectors will give us the normal vector: (2, 2, -0.25) x (3, 7/3, 0). Let's call the resulting normal vector (A, B, C). With the normal vector (A, B, C) and the point (1, -1, 1), the equation of the plane is given by A(x-1) + B(y+1) + C(z-1) = 0.