Finding the Equation of a Plane Parallel to the Line of Intersection of Two Planes

Introduction to Analytic Geometry and Planes: In the realm of analytic geometry, we often deal with various geometric shapes and their equations. One such concept involves finding a plane that is parallel to the line of intersection of two given planes. This article will explore the steps to determine the equation of the plane passing through a specific point and parallel to the line of intersection of two planes.

Given Planes and Points

Consider two planes given by their respective equations:

$$p_1: 3x - 2y - z 0$$ $$p_2: 4x - y - 3z 0$$

These planes intersect to form a line. The problem at hand is to find an equation of a plane that passes through the point $(1, -1, 0)$ and is parallel to this line of intersection.

The Normal Vectors of the Given Planes

The normal vectors $mathbf{n_1}$ and $mathbf{n_2}$ of the planes (p_1) and (p_2) respectively can be found from their coefficients:

$$mathbf{n_1} begin{bmatrix}3-2-1end{bmatrix}$$ $$mathbf{n_2} begin{bmatrix}4-1-3end{bmatrix}$$

The cross product of these normal vectors,$mathbf{n_1} times mathbf{n_2}$, gives the direction vector of the line of intersection of the two planes. The calculation is as follows:

$$mathbf{E} mathbf{n_1} times mathbf{n_2} begin{vmatrix}mathbf{i} mathbf{j} mathbf{k}3 -2 -14 -1 -3end{vmatrix}$$

This evaluates to:

$$mathbf{E} begin{vmatrix}-2(-3) - (-1)(-1) 3(-3) - (-1)(4) 3(-1) - (-2)(4)end{vmatrix}$$ $$mathbf{E} begin{vmatrix}6 - 1 -9 4 -3 8end{vmatrix} begin{vmatrix}5 -5 5end{vmatrix}$$ $$mathbf{E} 5mathbf{i} - 5mathbf{j} 5mathbf{k}$$

Direction Vector of the Desired Plane

The direction vector of the line of intersection is (mathbf{E} 5mathbf{i} - 5mathbf{j} 5mathbf{k} begin{bmatrix}5-55end{bmatrix}).

Deriving the Normal Vector to the Desired Plane

The desired plane is parallel to the line of intersection, meaning its normal vector should be perpendicular to (mathbf{E}). Since the plane also passes through the point ((1, -1, 0)), we can use the point-normal form of the plane equation. The normal vector (mathbf{N}) to the desired plane can be taken as (mathbf{D}), where (mathbf{D} -mathbf{E} -5mathbf{i} 5mathbf{j} - 5mathbf{k} begin{bmatrix}-55-5end{bmatrix}).

Equation of the Desired Plane

The equation of the plane passing through ((1, -1, 0)) with normal vector (mathbf{N} begin{bmatrix}-55-5end{bmatrix}) is:

$$-5(x - 1) 5(y 1) - 5z 0$$

Expanding and simplifying this, we get:

$$-5x 5 5y 5 - 5z 0$$ $$-5x 5y - 5z -10$$

Dividing the entire equation by -5, the final equation of the desired plane is:

$$x - y z 2$$

Conclusion

In summary, the equation of the plane that passes through the point ((1, -1, 0)) and is parallel to the line of intersection of the planes (3x - 2y - z 0) and (4x - y - 3z 0) is: