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Find an equation of the line of intersection of the planes \(Q\) and \(R\). $$Q: x+2 y-z=1 ; R: x+y+z=1$$

Short Answer

Expert verified
Answer: The parametric equation of the line of intersection of the two planes Q and R is given by: $$ \begin{cases} x = 1 - 3t + 3s \\ y = 2t - 2s \\ z = t - s \end{cases} $$ where \(s\) is the parameter for the line.

Step by step solution

01

Find the normal vectors of the planes

The normal vectors are the coefficients of the variables in the plane equations. For plane Q, the normal vector is \(\vec{n_Q} = \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}\) and for plane R, the normal vector is \(\vec{n_R} = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}\).
02

Find the direction vector of the line

The direction vector of the line of intersection is the cross product of the normal vectors of the two planes: $$\vec{v} = \vec{n_Q} \times \vec{n_R} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & -1 \\ 1 & 1 & 1 \end{vmatrix} = (3)\mathbf{i} + (-2)\mathbf{j} + (-1) \mathbf{k} = \begin{pmatrix} 3 \\ -2 \\ -1 \end{pmatrix}$$. So, the direction vector of the line \(\vec{v} = \begin{pmatrix} 3 \\ -2 \\ -1 \end{pmatrix}\).
03

Find a point on both planes

To find a point that lies on both planes, we can solve the system of equations formed by the plane equations: $$ \begin{cases} x+2y-z=1 \\ x+y+z=1 \end{cases} $$ We can solve this system by considering z as a free parameter. Let \(z=t\). Then we have: $$ \begin{cases} x+2y = 1+t \\ x+y = 1-t \end{cases} $$ Subtracting the second equation from the first equation, we get \(y=2t\). Substituting for \(y\) in the second equation, we get \(x=1-3t\). Thus, the coordinates of the point are \((1-3t, 2t, t)\).
04

Write the parametric equation of the line

Using the point from Step 3 and the direction vector from Step 2, we can write the parametric equation of the line as follows: $$ \begin{cases} x = 1 - 3t + 3s \\ y = 2t - 2s \\ z = t - s \end{cases} $$ where \(s\) is the parameter for the line. This parametric equation represents the line of intersection of the two planes Q and R.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Cross Product
The cross product is an essential operation in vector mathematics, particularly useful in three-dimensional space. In the context of finding the line of intersection between two planes, the cross product helps us determine the direction vector of the line. When you calculate the cross product of two vectors, you essentially find a vector that is perpendicular to both. Here, we take the normal vectors of the two planes, \( \vec{n_Q} = \begin{pmatrix} 1 \ 2 \ -1 \end{pmatrix} \) and \( \vec{n_R} = \begin{pmatrix} 1 \ 1 \ 1 \end{pmatrix} \), and perform the cross product to find the direction of the line.
  • Calculate using the determinant of a 3x3 matrix with \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) as unit vectors.
  • Result: Direction vector \( \vec{v} = \begin{pmatrix} 3 \ -2 \ -1 \end{pmatrix} \).
The resulting vector is essential in forming the parametric equation of the line.
Normal Vectors
Normal vectors play a crucial role in describing planes in 3D space. They are vectors that are perpendicular to a given plane and consist of the coefficients of \(x, y, \) and \(z\) from the plane's equation. For instance, the plane equation \(x + 2y - z = 1\) corresponds to the normal vector \(\vec{n_Q} = \begin{pmatrix} 1 \ 2 \ -1 \end{pmatrix}\). Similarly, for \(x + y + z = 1\), the normal vector is \(\vec{n_R} = \begin{pmatrix} 1 \ 1 \ 1 \end{pmatrix}\).
  • Normal vectors are essential for determining the orientation of a plane.
  • They help find the line of intersection by providing direction information through their cross product.
Understanding these vectors allows us to better comprehend the geometry of space and how planes can interact.
Parametric Equations
Parametric equations are a way to express the coordinates of the points that make up a geometric object, such as a line, using one or more parameters. When finding the line of intersection between two planes, parametric equations describe this line in terms of a parameter, often denoted \(s\). From the calculations, the point on the line is expressed in terms of \(t\), and we formulate the equation as:\[ \begin{cases}x = 1 - 3t + 3s \y = 2t - 2s \z = t - s\end{cases} \]
  • Each parameter increment represents movement along the line.
  • Parametric representation gives flexibility in dealing with 3D geometric configurations.
Parametric equations thus offer a clear and structured way to understand the positioning and direction of the line derived from intersecting planes.
System of Equations
The concept of solving a system of equations is fundamental when dealing with intersections in geometry. Finding a common point on both planes involves setting up equations from their plane equations and solving simultaneously. The given plane equations are:\[\begin{cases}x+2y-z=1 \x+y+z=1\end{cases}\]To solve, you choose a free variable, typically \(z = t\), and solve for the others:
  • Subtract the second from the first to find \(y\).
  • Substitute into any equation to find \(x\).
This process allows us to determine a specific point that satisfies both equations, resulting in the expression \((1-3t, 2t, t)\). The system of equations thus helps in ensuring the correctness of mathematical modeling and finding intersections.

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