Chapter 3: Problem 4
Find a point that lies on the plane \(x+z=1\) and a normal vector to the plane.
Short Answer
Expert verified
A point on the plane is \((0, 0, 1)\); a normal vector is \((1, 0, 1)\).
Step by step solution
01
Understand the equation of the plane
The equation of the plane is given by \(x + z = 1\). This is a linear equation with variables \(x\) and \(z\), which implies that this plane is parallel to the y-axis. This plane can be visualized in a coordinate system where every point on this plane satisfies this equation.
02
Determine a point on the plane
To find a point on the plane, choose arbitrary values for any two variables satisfying the equation. For example, if we set \(x = 0\), then \(z = 1\) because \(x + z = 1\). Thus, one such point on the plane is \((0, y, 1)\). We can choose \(y = 0\) for simplicity, so a point on the plane is \((0, 0, 1)\).
03
Identify the normal vector to the plane
The normal vector to a plane in the form \(ax + by + cz = d\) is \((a, b, c)\). In this case, with the equation \(x + z - 1 = 0\), the coefficients correspond to \((1, 0, 1)\). Therefore, a normal vector to the plane is \((1, 0, 1)\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Plane Equation
In multivariable calculus, the equation of a plane is a fundamental concept. A plane can be defined using a linear equation involving three variables, generally written in the form:\[ ax + by + cz = d \]In the provided exercise, the plane equation is given as \(x + z = 1\). This is a simplified form where the coefficient of \(y\) is implicitly zero (\(b = 0\)), indicating that the plane is parallel to the y-axis. The "d" in our equation equates to 1, effectively shifting the plane along the x and z dimensions, while the lack of a y-term confirms that y values do not affect its position.
- The plane will consist of all points \((x, y, z)\) that satisfy this relationship.
- This type of plane equation is called a linear equation.
- By setting various values for \(x\) and \(z\), we can find different points lying on the plane.
Normal Vector
A normal vector is a vector that is perpendicular to a plane. For any plane represented by the equation \(ax + by + cz = d\), a normal vector can be expressed as \((a, b, c)\). This vector provides critical information for describing the plane's orientation in space.
In the step-by-step solution to the exercise, a normal vector for the plane \(x + z = 1\) is derived. Here's how:
In the step-by-step solution to the exercise, a normal vector for the plane \(x + z = 1\) is derived. Here's how:
- Rewrite the plane equation as \(x + z - 1 = 0\) to fit the general plane formula.
- Identify the coefficients: \(a = 1\), \(b = 0\), \(c = 1\).
- Thus, the normal vector is \((1, 0, 1)\).
Coordinate System
A coordinate system is essential for representing planes geometrically. It allows us to understand positions and movements in space. In three-dimensional space, every point can be described using a set of three values: \((x, y, z)\). Here’s how the coordinate system works:
- X-axis: Represents horizontal positions left and right.
- Y-axis: Represents vertical positions up and down.
- Z-axis: Represents depth positions in and out of the plane.
Linear Equation
A linear equation is any mathematical equation that results in a straight line when graphed. These are equations of the first degree, meaning they involve only variables raised to the power of one. In multivariable calculus, a linear equation of the form \(ax + by + cz = d\) represents a plane.The exercise gives a linear plane equation \(x + z = 1\), which simplifies operations:
- The relationship between \(x\) and \(z\) defines a two-variable linear equation.
- Adding the concept of y, which does not appear in the equation, confirms the plane's parallelism to the y-axis.