Problem 13 - APEX Calculus

Question

Write the vector, parametric and symmetric equations of the lines described. Passes through \(P=(1,1)\), parallel to \(\vec{d}=\langle 2,3\rangle\).

Step-by-Step Solution

Answer

Vector: \( \vec{r}(t) = \langle 1+2t, 1+3t \rangle \). Parametric: \( x = 1 + 2t \), \( y = 1 + 3t \). Symmetric: \( \frac{x-1}{2} = \frac{y-1}{3} \).

Step 1: Identify the Given Elements

First, identify the given point and direction vector. The line passes through point \( P = (1,1) \) and is parallel to the direction vector \( \vec{d} = \langle 2,3 \rangle \).

Step 2: Write the Vector Equation

Use the point and the direction vector to write the vector equation of the line. The vector equation is given by:\[\vec{r}(t) = \vec{r}_0 + t\vec{d}\]where \( \vec{r}_0 = \langle 1,1 \rangle \) is the position vector of point \( P \) and \( \vec{d} = \langle 2,3 \rangle \). Thus, the vector equation is:\[\vec{r}(t) = \langle 1,1 \rangle + t\langle 2,3 \rangle\]\[\vec{r}(t) = \langle 1+2t, 1+3t \rangle\]

Step 3: Write the Parametric Equations

The components of the vector equation give the parametric equations of the line:\[x = 1 + 2t\]\[y = 1 + 3t\]

Step 4: Write the Symmetric Equations

To derive the symmetric equations, eliminate the parameter \( t \) from the parametric equations. Solving for \( t \) yields:\[t = \frac{x-1}{2}\]\[t = \frac{y-1}{3}\]Hence, the symmetric equations are:\[\frac{x-1}{2} = \frac{y-1}{3}\]

Step 5: Verify the Equations

Check that all forms of the equation satisfy the conditions given in the exercise, including the point \( P = (1,1) \) and the direction vector \( \vec{d} = \langle 2,3 \rangle \).

Key Concepts

Vector Equation

In vector calculus, the vector equation of a line is a powerful way to describe a line in terms of vectors. It expresses the position of any point on the line using a mathematical formula. The basic form of a vector equation of a line passing through a point \( P \) and parallel to a direction vector \( \vec{d} \) is given by:

• \( \vec{r}(t) = \vec{r}_0 + t\vec{d} \)

where:

• \( \vec{r}(t) \) is the vector representing a point on the line.
• \( \vec{r}_0 \) is the position vector of the point \( P \).
• \( t \) is a scalar parameter, representing how far to move along the line.
• \( \vec{d} \) is the direction vector indicating the line’s direction.

For this exercise, with \( P = (1,1) \) and \( \vec{d} = \langle 2,3 \rangle \), we substitute into the vector equation to get:

• \( \vec{r}(t) = \langle 1,1 \rangle + t\langle 2,3 \rangle \)
• Which simplifies to \( \vec{r}(t) = \langle 1+2t, 1+3t \rangle \)

This formula gives us a clear and concise way to see how the line is formed by moving along the direction vector from the initial point.

Parametric Equations

Parametric equations are derived from the vector equation and represent a line using variables linked to a parameter, typically \( t \). These equations express each coordinate of a point on the line as a separate function of the parameter.From the vector equation \( \vec{r}(t) = \langle 1+2t, 1+3t \rangle \), we can see how the line's coordinates develop independently:

• The parametric equation for \( x \) is: \( x = 1 + 2t \)
• Similarly, for \( y \): \( y = 1 + 3t \)

In these equations:

• \( t \) controls the position along the line.
• As \( t \) changes, \( (x, y) \) traces all points on the line.

These parametric forms are particularly useful in computational applications, such as computer graphics and physics simulations, because they provide a convenient means to iterate over points on a line.

Symmetric Equations

Symmetric equations eliminate the parameter entirely, making them particularly useful for various mathematical analysis scenarios. They express the relationship between \( x \), \( y \) (and \( z \), if in 3D) directly, without a parameter, and are often used to determine if a point lies on a given line.We derive the symmetric equations from the parametric versions \( x = 1 + 2t \) and \( y = 1 + 3t \). By solving for \( t \) in each:

• From \( x = 1 + 2t \), solve for \( t \): \( t = \frac{x-1}{2} \)
• From \( y = 1 + 3t \), solve for \( t \): \( t = \frac{y-1}{3} \)

Setting these equal for consistency yields the symmetric equation:

• \( \frac{x-1}{2} = \frac{y-1}{3} \)

This form is valuable in quickly determining the specific characteristics of the line's alignment. It provides a straightforward check if a certain point \((x, y)\) is on the line, by verifying this equation holds true.