Problem 39 - Advanced Engineering Mathematics

Question

In Problems, find an equation of the plane that contains the given point and is perpendicular to the indicated vector. $$ (5,1,3) ; 2 \mathrm{i}-3 \mathrm{j}+4 \mathrm{k} $$

Step-by-Step Solution

Answer

Equation of the plane: \(2x - 3y + 4z = 19\).

Step 1: Understand the Equation of a Plane

The equation of a plane in 3D space can be written as \( ax + by + cz = d \), where \((a, b, c)\) is a normal vector to the plane, and \((x, y, z)\) is any point on the plane. \(d\) is a constant determined by a specific point on the plane.

Step 2: Identify the Perpendicular Vector

From the problem, the vector \(2\mathrm{i} - 3\mathrm{j} + 4\mathrm{k}\) is given as perpendicular to the plane. This vector, \((a, b, c) = (2, -3, 4)\), will be the coefficients in the plane equation.

Step 3: Substitute the Point into the Plane Equation

The plane passes through the point \((5, 1, 3)\). Using the equation form \(ax + by + cz = d\), substitute \(a = 2\), \(b = -3\), \(c = 4\), and the point \((x, y, z) = (5, 1, 3)\).

Step 4: Calculate the Plane Equation Constant \(d\)

Substitute into the plane equation to find \(d\):\[2(5) - 3(1) + 4(3) = d\]Calculate: \[10 - 3 + 12 = d\]Thus, \[d = 19.\]

Step 5: Write the Final Equation of the Plane

The equation of the plane is \[2x - 3y + 4z = 19.\] This equation uses the normal vector coefficients and the calculated \(d\).

Key Concepts

Normal Vector

In the context of a plane in 3D space, a normal vector is a critical component. It is a vector that is perpendicular to every point on the plane, essentially defining the orientation of the plane.
Consider a plane in space. It can be likened to a flat piece of paper floating. The normal vector is an arrow jutting out of this paper, ensuring it's always vertical to it.
In mathematical terms, if we have a normal vector represented as

• (a, b, c),

this means it directly influences the plane's equation, written as

• ax + by + cz = d.

This relationship shows how closely the normal vector is tied to the plane's equation, making it indispensable for characterizing the plane's overall orientation.

Plane in 3D Space

A plane in 3D space is an infinite flat surface. You can think of it like a giant, endless sheet of paper extending in all directions within a three-dimensional environment. Understanding a plane helps you comprehend how it interacts with other geometrical entities such as lines and points.
A plane is typically described using the equation:

• ax + by + cz = d.

In this equation:

• (a, b, c) are the coefficients from the normal vector, indicating the plane's tilt and rotation in space.
• (x, y, z) represents any point lying on this plane.
• d is a fixed value that shows where the plane sits relative to the origin.

So, when someone refers to a plane in 3D, they are talking about an infinite surface defined uniquely by these components, offering a foundational structure for 3D geometry.

Perpendicular Vector

A perpendicular vector in the context of a plane is particularly interesting and useful. Imagine a vector that sticks out perfectly at right angles to a plane, much like a pole standing perpendicular to a flat surface.
When determining the equation of a plane, this perpendicular vector becomes the normal vector. Its components are drawn directly into the equation of the plane as coefficients.

• For example, if the given perpendicular vector is
  • (2, -3, 4),
  it will shape the plane's equation as
  • 2x - 3y + 4z = d.

Because it's perpendicular to the plane, this vector helps determine how the plane slices through the 3D space, thereby playing a crucial role in defining the spatial orientation of the plane.