Problem 10 - Essential Calculus Early Transcendentals

Question

\(6-10=\) Find parametric equations and symmetric equations for the line. The line of intersection of the planes \(x+2 y+3 z=1\) and \(x-y+z=1\)

Step-by-Step Solution

Answer

Parametric: \(x=\frac{3}{5}+5t\), \(y=\frac{1}{5}-2t\), \(z=-3t\); Symmetric: \(\frac{x-\frac{3}{5}}{5}=\frac{y-\frac{1}{5}}{-2}=\frac{z}{-3}\).

Step 1: Find the Direction Vector

To find the direction vector of the line of intersection, we need to calculate the cross product of the normal vectors of the given planes. The normal vector of the first plane, \(x + 2y + 3z = 1\), is \(\langle 1, 2, 3 \rangle\). The normal vector of the second plane, \(x - y + z = 1\), is \(\langle 1, -1, 1 \rangle\). The cross product \(\mathbf{n}_1 \times \mathbf{n}_2\) is given by the determinant:\[\mathbf{d} = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \1 & 2 & 3 \1 & -1 & 1 \\end{vmatrix} = \mathbf{i}(2 \cdot 1 - 3 \cdot (-1)) - \mathbf{j}(1 \cdot 1 - 3 \cdot 1) + \mathbf{k}(1 \cdot (-1) - 2 \cdot 1)\]. This simplifies to \(\mathbf{d} = \langle 5, -2, -3 \rangle\).

Step 2: Find a Point on the Intersection Line

To find a point on the line of intersection, we solve the system of equations given by the plane equations simultaneously. Choosing \(z = 0\), we substitute into the equations:\[\begin{align*}x + 2y + 3(0) &= 1,\x - y + 0 &= 1.\end{align*}\]Solving these gives us \(x = \frac{3}{5}, y = \frac{1}{5}\), thus a point \((\frac{3}{5}, \frac{1}{5}, 0)\) lies on the line.

Step 3: Write the Parametric Equations

With the direction vector \(\langle 5, -2, -3 \rangle\) and the point \((\frac{3}{5}, \frac{1}{5}, 0)\), we can express the parametric equations of the line:\[\begin{align*}x &= \frac{3}{5} + 5t,\y &= \frac{1}{5} - 2t,\z &= -3t.\end{align*}\]

Step 4: Write the Symmetric Equations

The symmetric equations derived from the parametric form are obtained by solving for \(t\) in each parametric equation:\[\begin{align*}\frac{x - \frac{3}{5}}{5} &= \frac{y - \frac{1}{5}}{-2} = \frac{z}{-3}.\end{align*}\]These are the symmetric equations of the line of intersection.

Key Concepts

Direction Vector

To discover a line in three-dimensional space, first, we need to identify its direction vector. This can be achieved through the cross product of the normal vectors of the two intersecting planes.- **Normal Vector**: Each plane can be represented by a normal vector, which is perpendicular to the plane. For the plane equation \(x + 2y + 3z = 1\), the normal vector is \(\langle 1, 2, 3 \rangle\). For the plane equation \(x - y + z = 1\), it's \(\langle 1, -1, 1 \rangle\).- **Cross Product**: The direction vector of the intersection line is the cross product of these normal vectors. Calculate it using the determinant of a matrix formed by these vectors:\[\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 1 & 2 & 3 \ 1 & -1 & 1 \end{vmatrix} = \mathbf{i}(2 \cdot 1 - 3 \cdot (-1)) - \mathbf{j}(1 \cdot 1 - 3 \cdot 1) + \mathbf{k}(1 \cdot (-1) - 2 \cdot 1),\]which results in \(\langle 5, -2, -3 \rangle\). This vector provides the direction of the line of intersection.

Parametric Equations

Finding a point on the line is crucial to express it with parametric equations, alongside the direction vector.- **Choosing a Point**: Solving the given plane equations, setting \(z = 0\) for simplicity, yields the point \((\frac{3}{5}, \frac{1}{5}, 0)\).- **Parametric Form**: With the point and the direction vector \(\langle 5, -2, -3 \rangle\), the parametric equations, which describe every point \((x, y, z)\) on the line as a function of a parameter \(t\), are: \[ \begin{align*} x &= \frac{3}{5} + 5t, \ y &= \frac{1}{5} - 2t, \ z &= -3t. \end{align*} \]These equations enable us to generate any point on the line by choosing different values for \(t\).

Symmetric Equations

Converting parametric equations to symmetric form can offer a more succinct way to represent a line.- **Transition to Symmetric**: Symmetric equations are derived by isolating the parameter \(t\) in each parametric equation: \[ \frac{x - \frac{3}{5}}{5} = \frac{y - \frac{1}{5}}{-2} = \frac{z}{-3} \]- **Uniform Representation**: This form interrelates \(x\), \(y\), and \(z\) directly without explicitly showing the parameter \(t\). It provides a more geometric perspective of the line's alignment in space. Symmetric equations help in visualizing and verifying the alignment with other lines or planes.