Chapter 9: Problem 65
Given a plane through \((0,-2,1)\) and perpendicular to \(\left[\begin{array}{r}-1 \\ 1 \\ -1\end{array}\right]\), find a line through \((5,-1,0)\) that is parallel to the plane.
Short Answer
Expert verified
The line is given by \[ \mathbf{r}(t) = (5 + t, -1, t) \].
Step by step solution
01
Understanding the Problem
We need to find a line parallel to a plane defined by a normal vector. The plane passes through the point \((0, -2, 1)\) and is perpendicular to the vector \([-1, 1, -1]\).
02
Define Plane's Normal Vector
The normal vector to the plane is \(\mathbf{n} = [-1, 1, -1]\). A line parallel to the plane must be perpendicular to this normal vector.
03
Determine Direction Vector for the Line
A line parallel to the plane must have a direction vector \( \mathbf{v} \) that is perpendicular to the normal vector of the plane. Thus, \( \mathbf{v} \cdot \mathbf{n} = 0 \).Choose a direction vector \( \mathbf{v} = [a, b, c] \), and solve the equation:\[-a + b - c = 0\]
04
Choose an Example Direction Vector
To satisfy \(-a + b - c = 0\), a simple choice is to set \(a = 1\), \(b = 0\), and \(-c = -1\). Hence, \(a - c = 0\) holds and so does \(-a + b - c = 0\). The direction vector can then be \([1, 0, 1]\).
05
Write the Line Equation
Use the point \((5, -1, 0)\) and the direction vector \([1, 0, 1]\) to write the equation of the line:\[ \mathbf{r} = (5, -1, 0) + t \cdot (1, 0, 1) = (5 + t, -1, t) \] where \( t \) is the parameter.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
plane equation
Understanding the concept of a plane equation is crucial in vector calculus. A plane can be defined using a point on the plane and a normal vector. The normal vector is perpendicular to every line lying in the plane. When you have a normal vector \( \mathbf{n} = [a, b, c] \) and a point \( (x_0, y_0, z_0) \) on the plane, the equation of the plane is given by:\[ a(x - x_0) + b(y - y_0) + c(z - z_0) = 0 \]This equation indicates that any point \( (x, y, z) \) satisfying it lies on the plane. The point acts as a reference from where you measure the direction and magnitude defined by the components of the normal vector.
- The normal vector \( \mathbf{n} \) directs perpendicularly from the plane.
- The reference point \( (x_0, y_0, z_0) \) is any point embedded on the plane surface.
line direction vector
The direction vector of a line is fundamental in determining its orientation in space. A line can be determined by a point through which it passes and a direction vector. The direction vector shows the line's direction. For a line passing through point \( (x_0, y_0, z_0) \) with direction vector \( \mathbf{v} = [d_1, d_2, d_3] \), the parametric equation of the line is given by:\[ \mathbf{r}(t) = (x_0, y_0, z_0) + t \cdot (d_1, d_2, d_3) \]Here, \( t \) is a parameter that allows the equation to define every point along the line.
- \( \mathbf{v} \) determines how the line stretches indefinitely in its direction.
- The parameter \( t \) adjusts how far along the direction vector each point is from the initial point.
normal vector
In vector calculus, a normal vector is strategic in determining planes and their interactions with lines. The normal vector to a plane is perpendicular to the plane's surface, making it instrumental in defining the plane's orientation. If a plane has a normal vector \( \mathbf{n} = [a, b, c] \), then this vector is:
- Essential in forming the plane equation.
- A basis for identifying perpendicularity to other vectors or planes.
parallel lines
Parallel lines maintain a specific relationship with each other or with a plane. Two lines or line and plane are parallel if they never intersect, no matter how far they extend. For a line to be parallel to a plane:
- The line's direction vector must be perpendicular to the plane's normal vector.
- There should be no component of the line's direction vector that aligns in the direction of the normal vector.