Problem 65 - Calculus for Biology and Medicine

Question

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.

Step-by-Step Solution

Answer

The line is given by \[ \mathbf{r}(t) = (5 + t, -1, t) \].

Step 1: 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 \([-1, 1, -1]\).

Step 2: 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.

Step 3: 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:\[-a + b - c = 0\]

Step 4: 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 \([1, 0, 1]\).

Step 5: 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.

Key Concepts

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.

Understanding these components makes analyzing and solving problems involving planes, such as finding parallel lines, more accessible.

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.

This vector is essential when determining parallelism to a plane, as it needs to be perpendicular to the plane’s normal vector.

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.

To find a line parallel to a plane, the line's direction vector must be orthogonal to this normal vector. This condition implies:\[ \mathbf{v} \cdot \mathbf{n} = 0 \]Where \( \mathbf{v} \) is the line's direction vector. This formula guarantees that the plane is parallel and will never intersect at an angle with the direction determined by \( \mathbf{v} \). This orthogonality is crucial to solving geometry problems involving planes and parallel lines.

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.

This concept forms the basis for the condition \( \mathbf{v} \cdot \mathbf{n} = 0 \), ensuring the two never meet. In problems like the one given, determining an appropriate direction vector for the line that satisfies this condition ensures the lines or line and plane remain parallel, representing a harmony without intersection in the geometry presented.