Problem 5 - Algebra and Trigonometry

Question

Find parametric equations for the line that passes through the point \(P\) and is parallel to the vector \(\mathbf{v} .\) $$ P(3,2,1), \quad \mathbf{v}=\langle 0,-4,2\rangle $$

Step-by-Step Solution

Answer

The parametric equations are \( x = 3 \), \( y = 2 - 4t \), \( z = 1 + 2t \).

Step 1: Understand the problem

We need to find parametric equations for a line in 3-dimensional space. The line should pass through a given point \( P = (3,2,1) \) and be parallel to a given direction vector \( \mathbf{v} = \langle 0, -4, 2 \rangle \).

Step 2: Recall the formula for parametric equations

A line in 3-dimensional space can be described by the parametric equations using a point \((x_0, y_0, z_0)\) on the line and a direction vector \(\langle a, b, c \rangle \) as follows:\[x = x_0 + at\]\[y = y_0 + bt\]\[z = z_0 + ct\] where \(t\) is a parameter.

Step 3: Substitute point and vector values

Substitute the coordinates of point \(P = (3, 2, 1)\) and the direction vector \(\mathbf{v} = \langle 0, -4, 2 \rangle \) into the parametric equation formulas. This gives us:\[ x = 3 + 0t \]\[ y = 2 - 4t \]\[ z = 1 + 2t \].

Step 4: Simplify the equations

Since the first equation contains \(0t\), it simplifies to \( x = 3 \). Thus, the parametric equations for the line are:\[ x = 3 \]\[ y = 2 - 4t \]\[ z = 1 + 2t \].

Key Concepts

3-Dimensional Space

Welcome to the fascinating world of 3-dimensional space! Unlike the flat 2-dimensional world, a 3-dimensional (or 3D) space has depth, making it more complex and exciting. In 3D, you can use three coordinates, typically labeled as \(x, y, z\), to indicate the position of any point.

Consider a room. You are able to move left-right (x-axis), forward-backward (y-axis), and up-down (z-axis). This three-axis combination gives 3-dimensional space its depth and complexity, allowing for interesting geometric representations like spheres, cubes, and more.

• The x-coordinate measures horizontal position.
• The y-coordinate measures vertical position.
• The z-coordinate measures depth, adding the third dimension.

Understanding this basic concept helps in visualizing problems like the one in this exercise, where we find a line in 3-dimensional space using coordinates and vectors.

Direction Vector

A direction vector is a key tool for understanding how objects, such as lines, orient themselves within space. In our exercise, the vector \(\mathbf{v} = \langle 0, -4, 2 \rangle \) gives the direction in which the line moves.

Think of a direction vector as a set of instructions that tell you how to "travel" from one point to another. Each component of the vector shows movement along one of the three axes:

• 0 along the x-axis: no movement left or right.
• -4 along the y-axis: points downward, indicating movement downwards.
• 2 along the z-axis: moves upwards in the depth direction.

This example helps us find lines in 3-dimensional space that are parallel, as they have the same direction vector. It doesn't matter where the line is positioned, if the vector is the same, the line will point in the same direction.

Line Equations

Line equations in 3D help describe exactly where a line is in space, and we use parametric equations to express these lines. A parametric line equation involves a point on the line and a direction vector, and it takes the form:

• \( x = x_0 + at \)
• \( y = y_0 + bt \)
• \( z = z_0 + ct \)

Here, \(x_0, y_0, z_0\) are the coordinates of the starting point of the line, and \(\langle a, b, c \rangle\) is the direction vector. The letter \(t\) is a parameter which you can think of as a 'time' variable, showing different positions the line passes through.

In the problem, we have:

• \( x = 3 \)
• \( y = 2 - 4t \)
• \( z = 1 + 2t \)

These equations clearly describe how the line moves in space, and allow for easy location of any point on it simply by picking different values for \(t\).

Parameterization

Parameterization is all about using a parameter to represent a line's path in a neat and organized way. In 3D line equations, the role of the parameter \(t\) is crucial because it determines the line's exact position at any point.

Imagine this parameter as a slider which changes the position on the line from one point to another. By altering \(t\), you can find every possible point on the line: at \(t = 0\), you're at the starting point, and as \(t\) progresses, you move along in space.

• For negative \(t\), you move in the opposite direction of the vector.
• For positive \(t\), you move in the vector's direction.

This flexibility in formulating line equations makes it a handy tool for various applications in geometry and physics. In our exercise, parameterization allows us to visualize and mathematically express a line through a point and parallel to a vector!