Problem 62 - Advanced Engineering Mathematics

Question

Find parametric equations for the line that contains \((-4,1,7)\) and is perpendicular to the plane \(-7 x+2 y+3 z=1\).

Step-by-Step Solution

Answer

Parametric equations are: \\(x(t) = -4 - 7t, \\ y(t) = 1 + 2t, \\ z(t) = 7 + 3t.\\)

Step 1: Identify the Normal Vector of the Plane

The normal vector to the plane given by the equation \(-7x+2y+3z=1\) is formed by the coefficients of \x, y, \text{and } z. Use these coefficients to identify the normal vector \((-7, 2, 3)\). This vector will be the direction vector of the line since the line is perpendicular to the plane.

Step 2: Write Parametric Equations for the Line

With the direction vector \((-7, 2, 3)\) and a point on the line \((-4, 1, 7)\), use the following parametric form to find the equations: \(\begin{align*}x(t) &= x_0 + at\y(t) &= y_0 + bt\z(t) &= z_0 + ct\end{align*}\), where \((x_0, y_0, z_0)\) is a point on the line and \((a, b, c)\) is the direction vector. Substitute \((-4, 1, 7)\) and \((-7, 2, 3)\): \(x(t) = -4 - 7t\), \(y(t) = 1 + 2t\), \(z(t) = 7 + 3t\).

Key Concepts

Normal Vector

A normal vector is a vital concept in understanding the orientation of a plane in three-dimensional space. Simply put, it is a vector that is perpendicular, or "normal," to a surface at a given point.
In the context of a plane's equation like \(-7x+2y+3z=1\), the normal vector can be extracted directly from the coefficients of \(x, y, ext{and } z\). This means the normal vector here is \((-7, 2, 3)\).

• Direction: The normal vector shows the direction in which the plane "pushes out" of its surface.
• Calculation: Extract it from the coefficients in the plane's equation.

This concept is essential when dealing with perpendicular lines, as it often serves as a direction vector for such lines.

Direction Vector

A direction vector provides information about the direction of a line in space. It is a vector pointing along the line and is essential for forming parametric equations of the line.
In our example, since the line in question is perpendicular to the plane, the direction vector of the line is the same as the normal vector of the plane, \((-7, 2, 3)\).

• Purpose: It defines how the line extends through space, giving us direction.
• Representation: It consists of components \((a, b, c)\), which affect the line's slope and direction.

Having a direction vector is key to forming parametric equations.

Perpendicular Lines

Perpendicular lines form a right angle ( 90 degrees ) with each other in space. In the context of a plane and a line, if a line is perpendicular to a plane, it means it intersects the plane at a right angle.
When you have an equation of a plane, the line that is perpendicular to it can be found using the plane's normal vector as the direction vector.

• Intersection: A perpendicular line intersects the surface at a single point with a 90-degree angle.
• Normal to the surface: Using the normal vector of the plane, a perpendicular line can be defined.

Understanding perpendicularity is crucial for finding angles between vectors and defining relationships among geometric entities.

Parametric Form

Parametric form allows us to express the coordinates of each point on a line in a variable-driven equation, often in terms of a parameter \(t\).
This is particularly useful for representing lines in three-dimensional space, as it provides a complete set of every point on the line.
Here's the general formula for parametric equations:

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

In this formula, \((x_0, y_0, z_0)\) is a known point on the line, and \((a, b, c)\) is the direction vector.
In the exercise, substituting the point \((-4, 1, 7)\) and direction vector \((-7, 2, 3)\) gives us the parametric equations:

• \(x(t) = -4 - 7t\)
• \(y(t) = 1 + 2t\)
• \(z(t) = 7 + 3t\)

This provides a clear way of identifying every point along that line involving a parameter \(t\), which can be any real number.