Equation of a Parallel Line Passing Through a Given Point and Parallel to a Plane
When dealing with geometry in three-dimensional space, it's important to understand how to derive the equation of a line that is parallel to a given plane and passes through a specific point. This involves understanding the relationship between the plane's normal vector and the direction vector of the line.
Understanding the Problem
Given the equation of a plane in the form of (3x 2y - z 4), we need to find a line that is parallel to this plane and passes through the point ( (2, -11) ). The key is recognizing that any line parallel to the plane must have a direction vector that is orthogonal to the normal vector of the plane.
Identifying the Normal Vector
The normal vector to the plane (3x 2y - z 4) can be directly derived from the coefficients of (x), (y), and (z) in the plane's equation. Therefore, the normal vector is (langle 3, 2, -1 rangle).
Deriving the Line Equation
The vector equation of a line in 3-dimensional space can be written as (vec{r} vec{a} tvec{b}), where (vec{a}) is a point on the line and (vec{b}) is the direction vector of the line. Given the point ((2, -11)), we can write the line equation as (vec{r} langle 2, -11 rangle t langle l, m, n rangle).
For the line to be parallel to the given plane, the direction vector (langle l, m, n rangle) must satisfy the condition that it is orthogonal to the normal vector of the plane. This means the dot product of (langle l, m, n rangle) and (langle 3, 2, -1 rangle) must be zero.
Mathematically, this is expressed as:
(3l 2m - n 0)
Constructing the Line Equation
Using the condition (3l 2m - n 0), we can express (n) in terms of (l) and (m). For instance, let (l 1) and (m -1). Then we can solve for (n):
(3(1) 2(-1) - n 0)
(3 - 2 - n 0)
(1 - n 0)
(n 1)
This gives us a possible direction vector (langle 1, -1, 1 rangle). Therefore, the line equation in parametric form is:
[ begin{cases} x 2 t y -11 - t z 1 t end{cases} ]Thus, the equation of a line parallel to the plane (3x 2y - z 4) and passing through the point ((2, -11)) is:
[ begin{cases} x 2 t y -11 - t z 1 t end{cases} ]Note that this is just one possible solution. There are infinitely many lines parallel to the given plane passing through the point ((2, -11)) as long as their direction vectors satisfy the condition (3l 2m - n 0).
Conclusion
Understanding how to derive the equation of a line that is parallel to a given plane is a crucial concept in three-dimensional geometry. By leveraging the relationship between the normal vector of the plane and the direction vector of the line, we can construct the line equation with ease. For further exploration, consider experimenting with different values of (l) and (m) to find different direction vectors and corresponding line equations.