Introduction

Understanding the equation of a plane in three-dimensional space is a fundamental aspect of both high school and college-level mathematics. A plane is a two-dimensional surface that extends infinitely in all directions, and it can be described in several ways. One way to define a plane is through a point on the plane and a line that either lies on the plane or intersects it. In this article, we will walk through the steps to find the equation of the plane that contains a given point and a line described by parametric equations.

Identifying a Point on the Line

A line in three-dimensional space can be represented by parametric equations. For the given line:

x t, y t 1, z -32t

We can find a point on the line by setting t 0. When t 0, the coordinates of the point are:

(0, 1, -3)

Determining the Direction Vector of the Line

The direction vector of the line can be found by examining the coefficients of t in the parametric equations:

mathbf{d} (1, 1, -32)

Vector from the Given Point to a Point on the Line

Given a point on the plane (1, -1, 2) and a point on the line (0, 1, -3), the vector from the given point to the point on the line is:

mathbf{v} (0 - 1, 1 - (-1), -3 - 2) (-1, 2, -5)

Finding the Normal Vector to the Plane

The normal vector to the plane can be found by taking the cross product of the direction vector of the line and the vector mathbf{v}. The cross product can be calculated using the determinant of a matrix:

mathbf{n} mathbf{d} times mathbf{v} (1, 1, -32) times (-1, 2, -5)

Calculating this determinant, we have:

mathbf{n} mathbf{i} begin{vmatrix} 1, 2, -32 end{vmatrix} - mathbf{j} begin{vmatrix} 1, 2, -32 end{vmatrix} mathbf{k} begin{vmatrix} 1, 1, -32 end{vmatrix}

Evaluating the 2x2 determinants:

begin{vmatrix} 1, 2, -32 end{vmatrix} (1)(-5) - (2)(-32) -5 64 59

begin{vmatrix} 1, 2, -32 end{vmatrix} (1)(-5) - (2)(-1) -5 2 -3

begin{vmatrix} 1, 1, -32 end{vmatrix} (1)(-5) - (1)(-1) -5 1 -4

Thus, we have:

mathbf{n} 59, -3, -4

The Equation of the Plane

The equation of a plane can be expressed in the form:

n_x(x - x_0) n_y(y - y_0) n_z(z - z_0) 0

where (x_0, y_0, z_0) is a point on the plane and n_x, n_y, n_z is the normal vector. Using the point (1, -1, 2) and the normal vector (-9, 3, 3), the equation is:

-9(x - 1) 3(y 1) 3(z - 2) 0

Simplifying this, we get:

-9x 9 3y 3 3z - 6 0

Rearranging the equation of the plane, we obtain:

-9x 3y 3z -6

The equation of the plane is:

9x - 3y - 3z 6

Conclusion

In this article, we have seen how to find the equation of a plane that contains a given point and a line described by parametric equations. We used the direction vector of the line, the vector from the given point to a point on the line, and the cross product to find the normal vector to the plane. Finally, we wrote the equation of the plane in standard form.

Keywords

equation of a plane, parametric equations, cross product