Introduction

In this article, we will delve into the process of finding the equation of a plane that is perpendicular to another given plane and contains a specific line. The problem is defined as: Find the equation of a plane that contains the line (mathbf{x} 1 - 3t, mathbf{y} 2t, mathbf{z} 13t) and is perpendicular to the plane (2x y - 3z 4).

Identifying the Direction Vector of the Line

To begin, we need to identify the direction vector of the given line. The parametric equations of the line are:

The direction vector of this line can be obtained from the coefficients of t in the parametric equations:

mathbf{d} langle -3, 2, 3 rangle

Normal Vector of the Given Plane

The normal vector of the given plane (2x y - 3z 4) can be directly identified from the coefficients of x, y, and z:

mathbf{n} langle 2, 1, -3 rangle

Normal Vector of the New Plane

Since the new plane is perpendicular to the given plane, its normal vector can be found by taking the cross product of the direction vector of the line (mathbf{d}) and the normal vector of the given plane (mathbf{n}).

Cross Product Calculation

The cross product is calculated using the determinant of a matrix:

mathbf{n} begin{vmatrix} mathbf{i} mathbf{j} mathbf{k} -3 2 3 2 1 -3 end{vmatrix}

Expanding the determinant:

mathbf{n} mathbf{i}(2 cdot -3 - 3 cdot 1) - mathbf{j}(-3 cdot -3 - 3 cdot 2) mathbf{k}(-3 cdot 1 - 2 cdot 2)

mathbf{n} mathbf{i}(-6 - 3) - mathbf{j}(9 - 6) mathbf{k}(-3 - 4)

mathbf{n} langle -9, -3, -7 rangle

Using a Point on the Line

To determine the equation of the plane, we use the point on the line when (t 0). This point is ((1, 0, 1)).

Equation of the Plane

The general equation of a plane can be written as:

(n_1x - x_0 n_2y - y_0 n_3z - z_0 0)

Substituting the normal vector (langle -9, -3, -7 rangle) and the point ((1, 0, 1)):

(-9x - 1 - 3y - 0 - 7z - 1 0)

Simplifying this equation:

(-9x 9 - 3y - 7z - 7 0)

(-9x - 3y - 7z - 16 0)

Rearranging the terms gives:

(9x 3y 7z 16)

Thus, the equation of the plane is:

(boxed{9x 3y 7z 16})