Problem 5 - Advanced Engineering Mathematics

Question

In Problems, find a vector equation for the line through the given points. $$ (1,1,-1),(-4,1,-1) $$

Step-by-Step Solution

Answer

The vector equation is \( \mathbf{r} = (1 - 5t, 1, -1) \).

Step 1: Identify Given Points

We are given two points through which the line passes: \((1,1,-1)\) and \((-4,1,-1)\). These points will help us find the direction vector for the line.

Step 2: Find the Direction Vector

The direction vector \( \mathbf{d} \) can be found by subtracting the coordinates of the first point from the second point. So, \( \mathbf{d} = (-4 - 1, 1 - 1, -1 - (-1)) = (-5, 0, 0) \).

Step 3: Use the Vector Equation of a Line

The vector equation of a line can be written as \( \mathbf{r} = \mathbf{r}_0 + t \mathbf{d} \), where \( \mathbf{r}_0 \) is a position vector of a point on the line and \( t \) is a parameter. Let's use the first point as \( \mathbf{r}_0 = (1,1,-1) \).

Step 4: Write the Vector Equation

Substitute \( \mathbf{r}_0 = (1,1,-1) \) and \( \mathbf{d} = (-5, 0, 0) \) into the line equation: \( \mathbf{r} = (1,1,-1) + t(-5, 0, 0) \).

Step 5: Simplify the Equation

Expanding the equation, we have \( \mathbf{r} = (1 - 5t, 1, -1) \). This is the vector equation of the line through the given points.

Key Concepts

Direction Vector

A direction vector is essential when you're working with the vector equation of a line. It shows the line's orientation in space. Imagine it as the arrow directing the flow or path of the line. The direction vector is not unique — any non-zero scalar multiple of a direction vector can also work.

To find this vector, use the differences in the coordinates of two points on the line. This involves vector subtraction. For example, if you have points \( (x_1, y_1, z_1) \) and \( (x_2, y_2, z_2) \), the direction vector \( \mathbf{d} \) is calculated as:

• \( x_2 - x_1 \)
• \( y_2 - y_1 \)
• \( z_2 - z_1 \)

In our original exercise, this results in the direction vector of \( (-5, 0, 0) \). This tells us the line moves 5 units backward along the x-axis, while y and z remain constant.

Position Vector

A position vector is one of the two main ingredients in the vector equation of a line. It pinpoints a specific location on the line, serving as the line's anchor point.

Usually, you select one of the given points on the line as the position vector. For example, if our given points are \((1,1,-1)\) and \((-4,1,-1)\), we can choose either as the position vector, but let's go with \(\mathbf{r}_0 = (1,1,-1)\). Choosing the initial point makes calculations consistent and straightforward. This vector tells us exactly where the line starts or anchors in a 3D space.

Vector Subtraction

Vector subtraction is important in calculating the direction vector. Think of it as finding the 'difference' between two vectors. The process involves subtracting corresponding components from two vectors.

Consider two points forming vectors \( \mathbf{A} = (A_x, A_y, A_z) \) and \( \mathbf{B} = (B_x, B_y, B_z) \). When you subtract them to find \( \mathbf{B} - \mathbf{A} \), you're effectively creating a vector that points from the coordinates of \( \mathbf{A} \) to \( \mathbf{B} \):

• \( B_x - A_x \)
• \( B_y - A_y \)
• \( B_z - A_z \)

In the exercise, subtracting \((1,1,-1)\) from \((-4,1,-1)\) gives us the direction vector \((-5, 0, 0)\). This vector tells us how to 'travel' from \(\mathbf{A}\) to \(\mathbf{B}\).

Parameterization of Lines

Parameterizing the line gives a flexible equation that covers all points on it. It's like mapping out the entire path in one go.

The equation:\[ \mathbf{r} = \mathbf{r}_0 + t\mathbf{d} \]lets us describe every possible point along that line using a parameter \( t \). This parameter acts like a 'slider' that moves the set position along the line.

• The position vector \( \mathbf{r}_0 \) gives us a starting or reference point.
• The direction vector \( \mathbf{d} \) indicates which way we're going.

Adjust \( t \) to yield different points on the line:

• When \( t = 0 \), we're at the position vector \( \mathbf{r}_0 \).
• When \( t > 0 \), we move in the direction of the vector.
• When \( t < 0 \), we move opposite to the direction vector.

In our problem, using \( \mathbf{r}_0 = (1,1,-1) \) and \( \mathbf{d} = (-5, 0, 0) \), the line's equation becomes:\[ \mathbf{r} = (1 - 5t, 1, -1) \] This lets you effortlessly find any point on the line by selecting a value for \( t \).