Chapter 9: Problem 65
Given is a plane through \((0,-2,1)\) and perpendicular to \(\left[\begin{array}{r}-1 \\ 1 \\ -1\end{array}\right] .\) Find a line through \((5,-1,0)\) and that is parallel to the plane.
Short Answer
Expert verified
The line is given by \(x = 5 + t, y = -1, z = t.\)
Step by step solution
01
Understanding the Plane Equation
The equation of a plane through point \((x_0, y_0, z_0)\) and perpendicular to vector \(\mathbf{n} = \left[\begin{array}{r} a \ b \ c \end{array}\right]\) is \(a(x-x_0) + b(y-y_0) + c(z-z_0) = 0\). Here, the point is \((0, -2, 1)\) and \(\mathbf{n} = \left[\begin{array}{r} -1 \ 1 \ -1 \end{array}\right]\). So, the plane equation becomes \(-1(x - 0) + 1(y + 2) - 1(z - 1) = 0\). Simplifying, we have \(-x + y + 2 - z + 1 = 0\), which leads to the equation \(-x + y - z = -3\).
02
Identifying Parallel Directions
A line is parallel to a plane if its direction vector is orthogonal to the plane's normal vector. For our plane, the normal vector is \(\mathbf{n} = \left[\begin{array}{r} -1 \ 1 \ -1 \end{array}\right].\) If a line is parallel to the plane, its direction vector \(\mathbf{d}\) must satisfy \(\mathbf{n} \cdot \mathbf{d} = 0.\) Let \(\mathbf{d} = \left[\begin{array}{r} a \ b \ c \end{array}\right].\) Then: \(-1 \cdot a + 1 \cdot b - 1 \cdot c = 0.\) Thus, \(-a + b - c = 0.\) This gives us a constraint for the direction vector.
03
Choosing a Direction Vector
We can choose any vector \( \mathbf{d} \) that satisfies the constraint \(-a + b - c = 0.\) A simple choice is \(a = 1, b = 0, c = 1,\) which gives us the direction vector \(\mathbf{d} = \left[\begin{array}{r} 1 \ 0 \ 1 \end{array}\right].\) This vector satisfies \(-1(1) + 1(0) - 1(1) = -1 + 0 - 1 = 0.\)
04
Writing the Equation of the Line
The line passes through the point \((5, -1, 0)\) and follows the direction vector \(\mathbf{d} = \left[\begin{array}{r} 1 \ 0 \ 1 \end{array}\right].\) The parametric form of the line is: \[ x = 5 + t, \] \[ y = -1, \] \[ z = t, \] 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
A plane in three-dimensional space can be defined using a plane equation. This equation is derived from a point through which the plane passes and a normal vector that is perpendicular to the plane. Consider a point
and a normal vector
The plane equation is then expressed as:\[ a(x-x_0) + b(y-y_0) + c(z-z_0) = 0 \] . This represents all points egin{align*}(x, y, z),each located on the plane.
after expanding and simplifying, we get our plane equation as
- defined by coordinates egin{align*}(x_0, y_0, z_0),e.g.,(0, -2, 1)
and a normal vector
- defined bythe tripletegin{align*}(a, b, c),e.g.,(-1, 1, -1).
The plane equation is then expressed as:\[ a(x-x_0) + b(y-y_0) + c(z-z_0) = 0 \] . This represents all points egin{align*}(x, y, z),each located on the plane.
after expanding and simplifying, we get our plane equation as
- \[-x + y - z = -3.\]
Vector Orthogonality
Orthogonality in vectors is a crucial concept, especially when relating lines and planes in geometry. Two vectors are orthogonal, or perpendicular, if their dot product is zero.
If path of a line is to be parallel to a plane, its direction vector must be orthogonal to the plane's normal vector.
The normal vector of a plane can guide us in defining this relation.
This makes it straightforward to generate a direction vector for use indescribing a line parallelto the plane.
If path of a line is to be parallel to a plane, its direction vector must be orthogonal to the plane's normal vector.
The normal vector of a plane can guide us in defining this relation.
- Using this concept, the normal vector is \[\mathbf{n} = \left[ -1, 1, -1 \right]\],
- The direction vector for a line that is parallel to the plane should satisfy \[ \mathbf{n} \cdot \mathbf{d} = 0 \]
This makes it straightforward to generate a direction vector for use indescribing a line parallelto the plane.
Parametric Equations
Parametric equations offer a highly flexible way of describing the path of a line in space. Using a point and direction vector, the line can be expressed as a set of equations depending on a parameter, generally denoted as \( t \).
- Consider a point through which the line passes, e.g.,\((5, -1, 0)\),
- and a direction vector, such as\(\mathbf{d} = \left[1, 0, 1 \right]\).
- \[x = 5 + t,\]
- \[y = -1,\]
- \[z = t.\]
streamlining calculations and enhancing intuitive understanding of its geometry.
Direction Vector
The direction vector is critical when determining the orientation and path of a line. This vector indicates which direction the line extends within a vector space. It plays a pivotal role in constructing parametric equations.
- To find a direction vector parallel to a given plane, we should ensure it is orthogonal to the plane's normal vector.
- In an example where a normal vector is \[ \mathbf{n} = \left[ -1, 1, -1 \right] \],
- \( \mathbf{d} = \left[ 1, 0, 1 \right] \),
- This selection guarantees that the line remains orthogonal and parallel to the plane's normal vector, ensuring trueparallelism.