Problem 29 - Algebra and Trigonometry

Question

A description of a line is given. Find parametric equations for the line. The line perpendicular to the \(x z\) -plane that contains the point \((2,-1,5) .\)

Step-by-Step Solution

Answer

The parametric equations are \(x = 2\), \(y = -1 + t\), \(z = 5\).

Step 1: Understand the Problem

We need to find the parametric equations of a line that is perpendicular to the \(x z\)-plane and passes through the point \((2, -1, 5)\). Being perpendicular to the \(x z\)-plane means the line is parallel to the \(y\)-axis.

Step 2: Identify the Direction Vector

Since the line is perpendicular to the \(x z\)-plane, it runs parallel to the \(y\)-axis. The direction vector for a line parallel to the \(y\)-axis is \(\vec{d} = (0, 1, 0)\).

Step 3: Write the Parametric Equations

A line passing through a point \((x_0, y_0, z_0)\) with direction vector \(\vec{d} = (a, b, c)\) can be described by the parametric equations: \[ x = x_0 + at \]\[ y = y_0 + bt \]\[ z = z_0 + ct \]Substituting \((x_0, y_0, z_0) = (2, -1, 5)\) and \(\vec{d} = (0, 1, 0)\), we get:\[ x = 2 \]\[ y = -1 + t \]\[ z = 5 \]

Step 4: Verify the Solution

The parametric equations \(x = 2\), \(y = -1 + t\), \(z = 5\) correctly describe a line that goes through the point \((2, -1, 5)\) and remains constant in the \(x\) and \(z\) components, making it perpendicular to the \(x z\)-plane.

Key Concepts

Perpendicularity to Plane

When we talk about a line being perpendicular to a plane, we mean that the line intersects the plane at a right angle, or 90 degrees. In this problem, the line is meant to be perpendicular to the \(xz\)-plane. The \(xz\)-plane is simply a plane where the \(y\)-axis varies, and both \(x\) and \(z\) remain constant.
For a line to be perpendicular to the \(xz\)-plane, it must have no component in either the \(x\) or \(z\) directions. This means the direction of the line is completely in the \(y\)-direction. Hence, it is parallel to the \(y\)-axis. In simpler terms:

• The line’s direction is purely along \(y\).
• It's not tilted towards \(x\) or \(z\).

Understanding this concept is crucial as it dictates the specific path the line will follow through space, given any initial point it passes through.

Direction Vector

The direction vector of a line is a vector that shows the line's direction in space. It is a crucial component when forming the parametric equations of the line.
For a line parallel to the \(y\)-axis and perpendicular to the \(xz\)-plane, the direction vector is \(\vec{d} = (0, 1, 0)\). This specific direction vector tells us:

• The line moves only in the \(y\) direction.
• No movement occurs in the \(x\) and \(z\) directions.

This choice of direction vector ensures that the line maintains its perpendicular relationship with the given plane throughout its path. By adopting \(\vec{d} = (0, 1, 0)\), we establish a clear and direct direction which our line follows without deviation from the \(y\)-axis.

Three-Dimensional Geometry

Three-dimensional geometry involves understanding and visualizing objects in a space that has three dimensions: \(x\), \(y\), and \(z\). It's about how points, lines, and planes interact within this 3D space.
In this exercise, the line needs to be expressed through parametric equations, which are particularly helpful in three-dimensional space. Unlike regular equations where each variable has a direct relationship, parametric equations express coordinates as functions of a parameter \(t\):

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

This system allows each coordinate of the point on the line to vary as the parameter \(t\) varies and is especially useful for describing a line in 3D geometry. Here, each change in \(t\) translates movement in \(y\) while \(x\) and \(z\) remain constant, reinforcing our understanding of the line's perpendicularity and direction. Understanding such parametric systems is fundamental in analyzing movements and shapes in three-dimensional spaces.