Question

Find the symmetric equations and the parametric equations of a line, and/or the equation of the plane satisfying the following given conditions. Line through (4,-1,3) and parallel to $\mathbf{i}-2\mathbf{k}$

Short answer

Parametric: $x=4+ty=-1z=3-2t$; Symmetric: \frac{x - 4}{1} = \frac{z - 3}{-2}\ and y = -1.

Step-by-step solution

Identify the given point

The given point through which the line passes is $4,-1,3$. This point will be used as the initial point in the parametric equations.

Identify the direction vector

The direction vector for the line is given as \mathbf{i} - 2 \mathbf{k}\. This vector can be written in component form as $1,0,-2$. This will be used to determine the direction of the line.

Write the parametric equations

The parametric equations can be written using the initial point $4,-1,3$ and the direction vector $1,0,-2$. The parametric equations are: \x = 4 + t\, y = -1 + 0t = -1\, z = 3 - 2t\.

Write the symmetric equations

To find the symmetric equations, solve the parametric equations for $t$. From the parametric equations: \x = 4 + t\ implies \t = x - 4\. Similarly, \z = 3 - 2t\ implies \t = \frac{3 - z}{2}\. Since \y = -1\ is a constant, the symmetric equations are: \frac{x - 4}{1} = \frac{z - 3}{-2}\ and y = -1.

Key concepts

symmetric equations

Symmetric equations of a line are a way of describing the line in a form that reveals the relationship between the coordinates of the points on the line without involving a parameter like $t$. Starting from the parametric equations of a line, which are expressed as: $x=x_0+at$, $y=y_0+bt$, and $z=z_0+ct$, where $(x_0,y_0,z_0)$ is an initial point and $(a,b,c)$ is the direction vector. The symmetric equations are derived by eliminating the parameter $t$.

The process involves solving each parametric equation for $t$ and then setting them equal to each other. For example:
- From $x=4+t$, we get $t=x-4$.
- From $z=3-2t$, we get $t=\frac{3-z}{2}$.

Therefore, substituting back gives the symmetric form $\frac{x-4}{1}=\frac{3-z}{2}$. And because $y$ remains constant at $-1$, this factors into the final symmetric equations, incorporating both parts: $\frac{x-4}{1}=\frac{3-z}{2}$, $y=-1$. These equations help in identifying any point on the line and understanding its structure in space.

direction vector

The direction vector of a line is a vector that gives the direction in which the line extends. It's crucial for both parametric and symmetric equations because it determines the variation of the line in 3D space. For example, in our problem, the direction vector is given by $\mathbf{i}-2\mathbf{k}$, which translates to the vector $(1,0,-2)$.

To fully comprehend its significance:

• The direction vector indicates the movement from any point on the line. In this case, adding the vector to any point on the line moves you one unit along the x-axis while subtracting two units on the z-axis without changing the y-axis.
• The components of the direction vector directly influence the coefficients in the parametric equations and the numerators in the symmetric equations.

Using the direction vector in conjunction with the initial point enables us to form parametric and symmetric equations, which describe all points on the line in a structured way.

initial point

An initial point on a line is a specific point through which the line passes. It is essential for writing the equations of the line, as it provides a reference position for the line. For our problem, the initial point is given as $(4,-1,3)$.

Understanding the significance of the initial point:

• In parametric equations, the initial point $(x_0,y_0,z_0)$ sets the starting coordinates: $x=x_0+at$, $y=y_0+bt$, $z=z_0+ct$.
• It ensures the line passes through this specific point, giving meaning to the parameter $t$ as the 'travel distance' along the direction vector from this initial point.

In our specific solution:

• The parametric equations are formed as $x=4+t$, $y=-1$, and $z=3-2t$, originating from $(4,-1,3)$.
• This helps in maintaining consistency and making sure all points derived from the equations lie on the same line.

Thus, the initial point serves as an anchor for constructing the line's equations.