Finding the Equation of a Plane Perpendicular to the Y-Axis

Finding the Equation of a Plane Perpendicular to the Y-Axis

In this article, we will explore how to find the equation of a plane that passes through a specific point and is perpendicular to the y-axis. This involves understanding the role of the normal vector and using the point-normal form of the plane equation.

Understanding the Problem

We are given a plane that passes through the point ((-1, 2, 3)) and is perpendicular to the y-axis. This means that the plane is parallel to the xz-plane and cuts the y-axis at a specific value.

Identifying the Normal Vector

A plane perpendicular to the y-axis will have a normal vector pointing in the direction of the y-axis. Therefore, the normal vector can be represented as (langle 0, 1, 0 rangle).

Formulating the Plane Equation

The general equation of a plane is given by:

A x B y C z D

Here, ( langle A, B, C rangle ) is the normal vector of the plane. Since the normal vector is (langle 0, 1, 0 rangle), the equation simplifies to:

1 y 0 z D

Which further simplifies to:

(y D)

Determining the Value of D

To determine the value of (D), we substitute the coordinates of the given point ((-1, 2, 3)) into the equation:

y 2

This means that the value of (D) is 2.

Final Equation of the Plane

Substituting (D) back into the equation, we get:

(y 2)

This equation indicates that the plane is horizontal and is located at (y 2) in 3D space, extending infinitely in the x and z directions.

Verification Using the Dot Product

Another method to verify the equation is by using the dot product approach. The normal vector is (langle 0, 1, 0 rangle) and the plane passes through the point ((-1, 2, 3)). For a point ((x, y, z)) to lie on the plane, the following condition must be satisfied:

(langle 0, 1, 0 rangle cdot langle x - (-1), y - 2, z - 3 rangle 0)

Which simplifies to:

(y - 2 0)

Thus, the equation of the plane is:

(y 2)

This confirms our earlier result.

Conclusion

In summary, the equation of a plane that passes through the point ((-1, 2, 3)) and is perpendicular to the y-axis is (y 2). This plane is horizontal and extends infinitely in the x and z directions.

Further Reading

If you are interested in further concepts related to planes, normal vectors, and their equations, you can explore:

Plane Geometry: Explore the properties and equations of various geometric planes.

Vector Algebra: Dive into the algebraic operations involving vectors and their applications in 3D space.

Calculus in 3D: Examine how calculus concepts extend to three-dimensional space.