Finding The Line Equation Perpendicular To Plane 2x+2y-3z=6

In the realm of three-dimensional geometry, understanding the relationships between planes and lines is crucial. This article delves into the process of determining the equation of a line that is perpendicular to a given plane and passes through a specific point. This is a fundamental concept in linear algebra and multivariable calculus, with applications in various fields such as computer graphics, physics, and engineering. Let's consider the specific problem of finding the equation of the line perpendicular to the plane defined by the equation 2x + 2y - 3z = 6 and passing through the point A(-2, 3, 4). This exploration will not only provide a step-by-step solution but also illuminate the underlying principles and geometrical interpretations.

Before diving into the solution, it's essential to grasp the concept of a plane in three-dimensional space and its normal vector. A plane is a flat, two-dimensional surface that extends infinitely far. It can be uniquely defined by a linear equation of the form Ax + By + Cz = D, where A, B, and C are the coefficients of the variables x, y, and z, respectively, and D is a constant. The vector n = (A, B, C) is called the normal vector to the plane. This vector is perpendicular to the plane, meaning it forms a right angle with any vector lying in the plane. In our specific case, the plane is given by the equation 2x + 2y - 3z = 6. Therefore, the normal vector to this plane is n = (2, 2, -3). This normal vector plays a crucial role in determining the direction of the line we seek, as the line perpendicular to the plane will be parallel to the normal vector. Visualizing this geometrically can be incredibly helpful. Imagine the plane as a flat surface and the normal vector as an arrow pointing directly away from the plane, perpendicular to its surface. Any line that we want to be perpendicular to the plane must align with this arrow.

The normal vector's components directly correspond to the coefficients in the plane's equation, making it straightforward to extract this crucial piece of information. This relationship stems from the dot product properties and the definition of a plane. The equation of the plane essentially states that the dot product of the normal vector with any vector lying in the plane is constant. This constant is related to the distance from the origin to the plane. Understanding the normal vector is paramount because it dictates the orientation of the plane in space, and consequently, it dictates the direction of any line perpendicular to it. The concept of a normal vector is not just limited to planes; it extends to surfaces in general, where the normal vector at a point on the surface indicates the direction of the steepest ascent. This concept is fundamental in fields like computer graphics, where surface normals are used for lighting calculations and determining surface orientation. In physics, normal vectors are used to calculate forces acting perpendicularly to surfaces, such as the normal force between two objects in contact. Therefore, mastering the concept of the normal vector is not just an exercise in mathematical manipulation but a key to unlocking a deeper understanding of three-dimensional space and its applications in various scientific and technological domains.

Now that we've identified the normal vector of the plane, we can use it to find the direction vector of the line. Since the line is perpendicular to the plane, its direction vector will be parallel to the plane's normal vector. This means that the direction vector of the line can be taken directly as the normal vector of the plane. In our case, the normal vector is n = (2, 2, -3), so the direction vector of the line, which we'll denote as v, is also v = (2, 2, -3). The direction vector essentially defines the orientation of the line in space. It's a vector that, when placed with its tail at any point on the line, points along the line's direction. The magnitude of the direction vector is not important; only its direction matters. Therefore, any scalar multiple of the direction vector will also serve as a valid direction vector for the same line. For instance, (4, 4, -6) or (-2, -2, 3) could also be used as direction vectors for the line, as they are simply scalar multiples of (2, 2, -3). This flexibility in choosing the direction vector allows for different parameterizations of the same line, which can be useful in different contexts.

The significance of using the normal vector as the direction vector lies in the geometric relationship between the plane and the line. The line, being perpendicular to the plane, must align its direction with the normal vector, which is, by definition, perpendicular to the plane. This alignment ensures that the line intersects the plane at a right angle. Visualizing this relationship helps solidify the understanding. Imagine the plane as a flat surface and the line as a rod piercing through the surface at a 90-degree angle. The direction of the rod is dictated by the normal vector, which points straight out from the surface. This concept is fundamental in various applications. For example, in computer graphics, when calculating how light reflects off a surface, the normal vector is used to determine the angle of incidence, which in turn dictates the angle of reflection. In robotics, when programming a robot to move perpendicularly to a surface, the normal vector is used to guide the robot's motion. Therefore, understanding how the normal vector of a plane dictates the direction of a perpendicular line is a crucial skill in many fields.

With the direction vector v = (2, 2, -3) and a point on the line A(-2, 3, 4) known, we can now construct the parametric equation of the line. The parametric equation of a line in three-dimensional space is given by:

r = r₀ + tv

where:

• r is the position vector of a general point (x, y, z) on the line.
• r₀ is the position vector of a known point on the line, which in our case is A(-2, 3, 4), so r₀ = (-2, 3, 4).
• v is the direction vector of the line, which we found to be v = (2, 2, -3).
• t is a parameter, a scalar that can take any real value.

Substituting the values we have, the parametric equation of the line becomes:

r = (-2, 3, 4) + t(2, 2, -3)

This equation represents all the points on the line. As the parameter t varies over the real numbers, the position vector r traces out the line in three-dimensional space. The parametric equation is a powerful tool because it allows us to represent a line using a single vector equation, which is often more convenient than using separate equations for x, y, and z. The parameter t acts as a