Chapter 13: Problem 22
Find an equation of a plane that satisfies the given conditions. Through (2,-1,3) perpendicular to \(5 \vec{i}+4 \vec{j}-\vec{k}\)
Short Answer
Expert verified
The plane equation is \(5x + 4y - z - 3 = 0\).
Step by step solution
01
Understanding the Requirements
To find the equation of a plane, we need a point on the plane and a normal vector to the plane. We are given the point (2, -1, 3) and a vector perpendicular to the plane: \(\vec{n} = 5\vec{i} + 4\vec{j} -\vec{k}\). Thus, the vector \(\vec{n}\) is our normal vector.
02
Using the Point-Normal Form
The equation of a plane can be given by the point-normal form: \(a(x-x_0) + b(y-y_0) + c(z-z_0) = 0\), where \((a, b, c)\) is the normal vector and \((x_0, y_0, z_0)\) is the point on the plane. Here, \(a = 5, b = 4, c = -1\), and \((x_0, y_0, z_0) = (2, -1, 3)\).
03
Substituting Into the Equation
Substitute \(a = 5, b = 4, c = -1, x_0 = 2, y_0 = -1, z_0 = 3\) into the point-normal form: \(5(x - 2) + 4(y + 1) - 1(z - 3) = 0\).
04
Simplifying the Equation
Distribute and combine like terms in the equation: \(5x - 10 + 4y + 4 - z + 3 = 0\). Simplify it to get: \(5x + 4y - z - 3 = 0\).
05
Final Equation of the Plane
Thus, the equation of the plane is \(5x + 4y - z - 3 = 0\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Point-Normal Form
Understanding the point-normal form is essential when aiming to find the equation of a plane in three-dimensional space. The point-normal form relies on two key components: a point through which the plane passes and a normal vector that is perpendicular to the plane. The general formula for the point-normal form of a plane is given by:
- \[ a(x - x_0) + b(y - y_0) + c(z - z_0) = 0 \]
- Here, \((a, b, c)\) represents the components of the normal vector, and \((x_0, y_0, z_0)\) refers to the coordinates of the point on the plane.
Normal Vector
The concept of a normal vector is fundamental in understanding planes within geometry and vector algebra. A normal vector, by definition, is a vector that is perpendicular to a plane or a surface at a given point. It's like having a pole stick straight out from the surface.
- In three-dimensional geometry, a normal vector has three components: \(a, b, c\), which correspond to its projections along the x, y, and z-axes, respectively.
- For example, in our exercise, the given normal vector was \(\vec{n} = 5\vec{i} + 4\vec{j} - \vec{k}\), translating to components \((5, 4, -1)\).
- This vector tells us that the plane is tilted or oriented in space in a specific way.
Vector Algebra
Vector algebra is a branch of mathematics concerned with vector quantities. These quantities have both magnitude and direction, making them useful for representing physical entities like forces and velocities.
- In the context of plane equations, understanding vector algebra helps in computing and interpreting normal vectors and points in space effectively.
- Key operations in vector algebra include vector addition, scalar multiplication, and dot product computation.
- The dot product is especially relevant here as it is part of the standard form of a plane equation. It measures how much one vector projects onto another, which is vital for working with the normal vector.
Plane Through a Point
A plane through a point is a basic concept in geometry where the plane is described as extending infinitely but defined by a specific starting location. This concept is essential because:
- By knowing a point on the plane, such as \((2, -1, 3)\) in our exercise, together with a normal vector, one can uniquely determine the equation of that plane.
- This point serves as a reference from which the plane is oriented or projected into space.