Chapter 14: Problem 19
Find an equation of the plane that satisfies the stated conditions. Through \(P(6,-7,4)\) and parallel to (a) the \(x y\) -plane \(\quad\) (b) the \(y z\) -plane \(\quad\) (c) the \(x z\) -plane
Short Answer
Expert verified
(a) \( z = 4 \); (b) \( x = 6 \); (c) \( y = -7 \).
Step by step solution
01
Understanding the Plane Equation
The general equation of a plane is given by \( ax + by + cz = d \), where \(a\), \(b\), and \(c\) are the coefficients of the normal vector to the plane, and \(d\) is a constant. To find a specific plane equation, we need a point on the plane and the direction of the normal vector.
02
Plane Parallel to the xy-plane
A plane parallel to the xy-plane has a normal vector of \( \langle 0, 0, 1 \rangle \), meaning it is perpendicular to the z-axis. The equation of the plane parallel to the xy-plane and passing through point \((6,-7,4)\) is \( z = 4 \).
03
Corresponding Plane Equation
Since the plane is parallel to the xy-plane, and its normal vector only affects the \(z\)-direction, the equation is defined simply by the z-coordinate of the point. Thus, the equation is \( z = 4 \).
04
Plane Parallel to the yz-plane
A plane parallel to the yz-plane has a normal vector of \( \langle 1, 0, 0 \rangle \), meaning it is perpendicular to the x-axis. The equation of this plane through point \((6,-7,4)\) is \( x = 6 \).
05
Constructing the Equation for yz-plane Parallelism
Here, the plane is expressed by the x-coordinate alone, indicating its parallelism to the yz-plane. The equation is \( x = 6 \).
06
Plane Parallel to the xz-plane
A plane parallel to the xz-plane has a normal vector of \( \langle 0, 1, 0 \rangle \), indicating it is perpendicular to the y-axis. The equation of this plane passing through point \((6,-7,4)\) is \( y = -7 \).
07
Equation for xz-plane Parallelism
Based on the y-coordinate of the given point, the complete plane equation is formulated as \( y = -7 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Normal Vector
A normal vector is key to understanding planes in three-dimensional space. It is essentially a vector that is perpendicular to the plane. In the plane equation \( ax + by + cz = d \), the coefficients \(a\), \(b\), and \(c\) form the components of the normal vector \( \langle a, b, c \rangle \).
- This vector serves as a directional indicator that is perpendicular to every line lying on the plane.
- It determines the plane's orientation in 3D space.
xy-plane
The xy-plane is one of the three principal planes considered in 3D space, where the x-axis and y-axis intersect, and the z-coordinate is zero. A plane that is parallel to the xy-plane does not change in the z-axis directions.
- The normal vector for such a plane is \( \langle 0, 0, 1 \rangle \), meaning it is perpendicular to the z-axis.
- In practical terms, if a plane is parallel to the xy-plane, its equation simplifies to something like \( z = k \), where \(k\) is a constant.
yz-plane
The yz-plane in three-dimensional space is where the y-axis and z-axis intersect, with the x-coordinate being zero. A plane parallel to the yz-plane shows no variation along the x-axis.
- Such a plane has a normal vector of \( \langle 1, 0, 0 \rangle \), which indicates its perpendicularity to the x-axis.
- The equation of a plane parallel to the yz-plane takes the form \( x = k \), with \(k\) being constant.
xz-plane
The xz-plane is composed of the x-axis and z-axis, where the y-coordinate remains zero. A plane parallel to the xz-plane maintains a steady y-coordinate.
- It is defined by a normal vector \( \langle 0, 1, 0 \rangle \), indicating it is perpendicular to the y-axis.
- An equation describing this plane can be expressed as \( y = k \), where \(k\) is a constant value.