Question

Find the symmetric equations of the line of intersection of the given pair of planes. \(x+y-z=2,3 x-2 y+z=3\)

Short answer

The symmetric equations of the line are \(\frac{x - \frac{3}{2}}{3} = \frac{y - 1}{4} = \frac{z - \frac{1}{2}}{-5}\).

Step-by-step solution

Identify the Normal Vectors of the Planes

The first plane, \(x+y-z=2\), has a normal vector \(\mathbf{N_1} = \langle 1, 1, -1 \rangle\). The second plane, \(3x-2y+z=3\), has a normal vector \(\mathbf{N_2} = \langle 3, -2, 1 \rangle\).

Compute the Direction Vector of the Line

The direction vector of the line of intersection of the two planes is given by the cross product of their normal vectors: \(\mathbf{D} = \mathbf{N_1} \times \mathbf{N_2}\). Calculate:\[\mathbf{D} = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\1 & 1 & -1\3 & -2 & 1\end{vmatrix} = \mathbf{i}(1 - (-2)) - \mathbf{j}(-1 - 3) + \mathbf{k}(-2 - 3) = 3\mathbf{i} + 4\mathbf{j} - 5\mathbf{k}\]Thus, \(\mathbf{D} = \langle 3, 4, -5 \rangle\).

Find a Point on the Line of Intersection

To find a point on the line of intersection, solve the system:\[\begin{cases}x + y - z = 2\3x - 2y + z = 3\end{cases}\]First, eliminate \(z\) by adding the two equations:\[4x - y = 5\]--> (1).Choose a value for \(y\), say \(y = 1\), then \(4x - 1 = 5\), which gives \(x = \frac{3}{2}\).Substitute \(x = \frac{3}{2}\) and \(y = 1\) into \(x + y - z = 2\):\[\frac{3}{2} + 1 - z = 2\rightarrow z = \frac{1}{2}\].Thus, a point on the line is \(\left(\frac{3}{2}, 1, \frac{1}{2}\right)\).

Write the Symmetric Equations of the Line

The symmetric equations for the line use the point \(\left(\frac{3}{2}, 1, \frac{1}{2}\right)\) and direction vector \(\langle 3, 4, -5 \rangle\):\[\frac{x - \frac{3}{2}}{3} = \frac{y - 1}{4} = \frac{z - \frac{1}{2}}{-5}\].These equations describe the line of intersection of the planes.

Key concepts

Symmetric Equations

Symmetric equations are a helpful way to describe a line in three-dimensional space. They allow you to express a line using a single statement, instead of multiple parametric equations. In the context of finding the line of intersection between two planes, symmetric equations present the relationship between each coordinate of points on the line and a known point, along with the direction ratios derived from the direction vector.
To form symmetric equations, we need a specific point on the line and a direction vector. You take the coordinates of the specific point, which is the intersection of the two planes. With the direction vector, each part of the symmetric equation corresponds to one coordinate (x, y, or z) and their respective direction ratios. For example, if the direction vector is \( \langle a, b, c \rangle \), the symmetric equations are:\[ \frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c} \]This set of equations clearly outlines how a line progresses through space using its specific direction.

Normal Vectors

Normal vectors are crucial in understanding planes and their intersections. These vectors are perpendicular to their respective planes and define the orientation or tilt of the plane in 3D space. When you're given a plane equation like \( ax + by + cz = d \), the coefficients \( \langle a, b, c \rangle \) form the normal vector.
In our problem, we worked with two planes represented by equations \( x+y-z=2 \) and \( 3x-2y+z=3 \). Their normal vectors are \( \mathbf{N_1} = \langle 1, 1, -1 \rangle \) and \( \mathbf{N_2} = \langle 3, -2, 1 \rangle \) respectively. These vectors help define the position and orientation of each plane. Knowing them is essential for calculating the line of intersection through operations like the cross product to find the direction vector.

Cross Product

The cross product is a mathematical operation that is invaluable when working with vectors in three dimensions. This operation takes two vectors and produces a third vector perpendicular to the plane formed by the initial vectors. This property makes the cross product very useful for finding direction vectors for lines that intersect planes.
To calculate the direction vector of the line formed by the intersection of our two planes, we use the cross product of their normal vectors. If you have vectors \( \mathbf{A} = \langle a_1, a_2, a_3 \rangle \) and \( \mathbf{B} = \langle b_1, b_2, b_3 \rangle \), their cross product \( \mathbf{C} \) is computed as:

• \( c_1 = a_2b_3 - a_3b_2 \)
• \( c_2 = a_3b_1 - a_1b_3 \)
• \( c_3 = a_1b_2 - a_2b_1 \)

In our example, the direction vector from \( \mathbf{N_1} \) and \( \mathbf{N_2} \) is \( \langle 3, 4, -5 \rangle \), indicating the direction of the line through space.

Direction Vector

The direction vector in the context of a line is a vector that indicates the direction in which the line extends through space. It is derived from the cross product of the normal vectors of intersecting planes and helps define the orientation of the line.
For the line where the two planes intersect, we discovered that the direction vector is \( \langle 3, 4, -5 \rangle \). This vector gives us the proportions in which the coordinates increase as you move along the line. It's an essential component of the symmetric equations because it ties the separate equations for x, y, and z together, maintaining the correct proportions for the line's direction.
The importance of the direction vector lies in its ability to describe a line through space via its direction ratios. These ratios are what you use in the symmetric equation to prescribe how each coordinate changes relative to one another, providing both consistency and clarity in plotting or understanding how these lines navigate through a 3D environment.