Chapter 11: Problem 6
Write both the parametric equations and the symmetric equations for the line through the given point parallel to the given vector. \((-1,3,-6),\langle-2,0,5\rangle\)
Short Answer
Expert verified
Parametric: \(x = -1 - 2t\), \(y = 3\), \(z = -6 + 5t\); Symmetric: \(\frac{x + 1}{-2} = \frac{z + 6}{5},\; y = 3\).
Step by step solution
01
Understand the Problem
We need to find both the parametric and the symmetric equations of a line. The line should pass through the point \((-1, 3, -6)\) and be parallel to the vector \(\langle -2, 0, 5 \rangle\).
02
Write the Parametric Equations
The parametric equations of a line can be written using a point \((x_0, y_0, z_0)\) and a direction vector \(\langle a, b, c \rangle\) as follows:\[\begin{align*}x &= x_0 + at, \y &= y_0 + bt, \z &= z_0 + ct,\end{align*}\]where \(t\) is the parameter. Substituting \((-1, 3, -6)\) as the point and \(\langle -2, 0, 5 \rangle\) as the direction vector, we get:\[\begin{align*}x &= -1 - 2t, \y &= 3 + 0t, \z &= -6 + 5t.\end{align*}\]
03
Write the Symmetric Equations
The symmetric equations can be derived from the parametric equations by solving for \(t\) in each equation. From the parametric equations:\[\begin{align*}x &= -1 - 2t y &= 3 \z &= -6 + 5t\end{align*}\]Solving for \(t\), we get:\[\begin{align*}t &= \frac{x + 1}{-2}, \y &= 3, \t &= \frac{z + 6}{5}.\end{align*}\]The symmetric equations are:\[\frac{x + 1}{-2} = \frac{z + 6}{5}.\]Since \(y = 3\) doesn't contain \(t\), it's written separately.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Symmetric Equations of a Line
Symmetric equations offer an elegant way to represent a line in space. They eliminate the parameter in parametric equations, providing a relation between the coordinates of the points lying on the line. For example, from the parametric equations \[ x = -1 - 2t, \, y = 3, \, z = -6 + 5t \] we can solve for the parameter \( t \) in terms of \( x \) and \( z \).
- For \( x \): \( t = \frac{x + 1}{-2} \).
- For \( z \): \( t = \frac{z + 6}{5} \).
Direction Vector
A direction vector literally points the way for a line in three-dimensional space. It's crucial for defining both the parametric and symmetric equations of a line. The direction vector \( \langle a, b, c \rangle \), in this scenario, is \( \langle -2, 0, 5 \rangle \). This vector controls the line's trajectory:
- The \( -2 \) indicates the line's direction in the negative \( x \)-axis.
- The \( 0 \) shows there's no movement along the \( y \)-axis.
- The \( 5 \) points the line towards the positive \( z \)-axis.
Parameterization
Parameterization translates the essence of complex geometric shapes, such as lines, into more manageable mathematical forms. It uses a parameter, typically denoted as \( t \), to express each coordinate of a line:
- \( x = x_0 + at \)
- \( y = y_0 + bt \)
- \( z = z_0 + ct \)
- \( x = -1 - 2t \)
- \( y = 3 \)
- \( z = -6 + 5t \)
Vectors and Equations
Vectors and equations are integral components of mathematical representations in three-dimensional geometry. The line in question is described using both a point and a direction vector, which originates from vector theory. Here's why they're important:
- The **point** \((-1, 3, -6)\) identifies a specific location on the line.
- The **direction vector** \( \langle -2, 0, 5 \rangle \) provides directional cues on where the line moves.