Problem 65 - Calculus for Biology and Medicine

Question

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.

Step-by-Step Solution

Answer

The line is given by \(x = 5 + t, y = -1, z = t.\)

Step 1: 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 \(-x + y - z = -3\).

Step 2: 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.

Step 3: 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 \(-1(1) + 1(0) - 1(1) = -1 + 0 - 1 = 0.\)

Step 4: 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.

Key Concepts

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

• 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.

• 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 \[ \mathbf{n} \cdot \mathbf{d} = 0 \]

where in \[\mathbf{d}\] is – the direction vector of the line. Thisgives usthe equation\[-a + b - c = 0\],providing specific constraints to selectcertain vectors.
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\(\mathbf{d} = \left[1, 0, 1 \right]\).

We express the line using parametric equations:

• \[x = 5 + t,\]
• \[y = -1,\]
• \[z = t.\]

Here, \( t \)can be varied, allowing us to pinpoint any specific location along the line,

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 \[ \mathbf{n} = \left[ -1, 1, -1 \right] \],

we choose a direction vector like

• \( \mathbf{d} = \left[ 1, 0, 1 \right] \),

satisfying:\[-1\cdot 1 + 1\cdot 0 - 1 \cdot 1 = 0.\]

• This selection guarantees that the line remains orthogonal and parallel to the plane's normal vector, ensuring trueparallelism.

Using this vector provides a backbone for further geometric scenarios or for translating restrictions into practical solutions.