Problem 59 - Advanced Engineering Mathematics

Question

In Problems, find an equation of the plane that satisfies the given conditions. Contains \((2,4,8)\) and is perpendicular to the line \(x=10-3 t\), \(y=5+t, z=6-\frac{1}{2} t\)

Step-by-Step Solution

Answer

The plane equation is \(-3x + y - \frac{1}{2}z = 0\).

Step 1: Identify the Problem

We need to find an equation of a plane that passes through a point and is perpendicular to a given line. The line is represented parametrically.

Step 2: Extract Direction Vector

The line is given by the parametric equations: \( x = 10 - 3t \), \( y = 5 + t \), and \( z = 6 - \frac{1}{2}t \). The direction vector of this line can be found from the coefficients of \( t \): \( \langle -3, 1, -\frac{1}{2} \rangle \).

Step 3: Use the Normal Vector

Since the plane is perpendicular to the line, the direction vector of the line \(\langle -3, 1, -\frac{1}{2} \rangle\) is also the normal vector to the plane.

Step 4: Substitute Point in Plane Equation

A plane equation is given by \( ax + by + cz = d \). Using the normal vector as \( \langle a, b, c \rangle = \langle -3, 1, -\frac{1}{2} \rangle \) and the point \((2, 4, 8)\), substitute into the plane equation: \(-3(2) + 1(4) + (-\frac{1}{2})(8) = d\).

Step 5: Calculate the Constant

Compute \(-3(2) + 4 - 4 = 0\). This means \( d = 0 \).

Step 6: Write the Plane Equation

The plane equation is \(-3x + y - \frac{1}{2}z = 0\).

Key Concepts

Parametric Equations

To understand parametric equations, it's helpful to think of them as a way to express a line or curve using parameters, typically denoted with variables like \(t\). They allow us to describe a path or motion by specifying both the direction and position at specific parameter values. In our exercise, the parametric equations for the line are given as \(x = 10 - 3t\), \(y = 5 + t\), and \(z = 6 - \frac{1}{2}t\). Here:

• \(x\), \(y\), and \(z\) represent the coordinates of points on the line.
• \(t\) is the parameter that varies, leading to different points along the line as it changes.

From these equations, we obtain a direction vector that helps describe how the line moves through space, a crucial element in understanding the nature of lines and planes.

Normal Vector

The normal vector is a key component in defining a plane. It is a vector that is perpendicular to the plane itself. In terms of equations of planes, the normal vector helps determine the orientation of the plane in three-dimensional space. In our task, the normal vector is directly extracted from the direction vector of the parametric line since the plane is perpendicular to this line.

• The normal vector here is \(\langle -3, 1, -\frac{1}{2} \rangle\), taken from the coefficients in the parametric equations.
• This vector defines the orientation of the plane, which is vital for writing the plane's equation.

Recognizing this connection facilitates solving for the equation of the plane, as the normal vector components are used as coefficients \(a\), \(b\), \(c\) in the general plane equation \(ax + by + cz = d\). The plane can be considered a 'flat sheet' for which the normal vector points directly outwards.

Direction Vector

A direction vector communicates the direction in which a line runs, essentially demonstrating the line's path through space. It is inferred from parametric line equations by taking the coefficients of \(t\). In the given problem, our direction vector is \(\langle -3, 1, -\frac{1}{2} \rangle\).

• This vector indicates how each coordinate (\(x, y, z\)) changes as \(t\) increases.
• Negative or positive values show direction in respective axes.

Understanding the direction vector not only helps in drafting plane equations but also visualizes the line's orientation. For the exercise, since the plane is perpendicular to the given line, this same vector becomes the plane's normal vector, seamlessly bridging between line and plane properties.