Problem 18 - Calculus: Early Transcendentals

Question

Exercise 12.5.18 Give a prose description for each of the following processes: (a) Given two distinct points, find the line that goes through them. (b) Given three points (not all on the same line), find the plane that goes through them. Why do we need the caveat that not all points be on the same line? (c) Given a line and a point not on the line, find the plane that contains them both. (d) Given a plane and a point not on the plane, find the line that is perpendicular to the plane through the given point.

Step-by-Step Solution

Answer

(a) Find line through two points using point-slope formula. (b) Find plane from three non-collinear points using cross product. (c) Use a point and line direction to define plane with cross product. (d) Line from point perpendicular to plane uses plane's normal vector.

Step 1: Finding the Line through Two Points

To find a line that passes through two distinct points in a plane, denote these points as \(A(x_1, y_1)\) and \(B(x_2, y_2)\). Use the point-slope form of the equation of a line, which is given by \(y - y_1 = m(x - x_1)\), where \(m\) is the slope of the line. The slope \(m\) is calculated as \(m = \frac{y_2 - y_1}{x_2 - x_1}\). Plug the slope and the coordinates of either point into the point-slope equation to obtain the line's equation.

Step 2: Forming a Plane from Three Points

To determine a plane from three points \(A, B, C\), we require these points not to be collinear, meaning they should not all lie on the same line. Begin by expressing each point's position vector in terms of a common origin. Use the vectors \( \overrightarrow{AB} \) and \( \overrightarrow{AC} \) to find a normal vector \( \mathbf{n} \) to the plane by computing their cross product. With a normal vector \( \mathbf{n} = (a, b, c) \) and a point \( A(x_1, y_1, z_1)\) on the plane, the plane's equation becomes \(a(x - x_1) + b(y - y_1) + c(z - z_1) = 0\). The caveat ensures that the cross product \( \overrightarrow{AB} \times \overrightarrow{AC} \) results in a non-zero vector, confirming a well-defined plane exists.

Step 3: Plane through a Line and a Point

Given a line \(L\) and a point \(P\) not on \(L\), first determine a point \(Q\) on \(L\) and find the direction vector \( \mathbf{d}\) of the line. Use \(\overrightarrow{PQ}\) and \( \mathbf{d}\) to generate two non-parallel vectors that define the plane. Compute the cross product \( \overrightarrow{PQ} \times \mathbf{d} \) to obtain the plane's normal vector. Use this normal vector and point \(P\) or any point on line \(L\) to write the plane's equation.

Step 4: Perpendicular Line to Plane from External Point

Given a plane defined by \(ax + by + cz + d = 0\), and a point \(P(x_0, y_0, z_0)\) not on the plane, the line perpendicular to this plane through \(P\) is characterized by the plane's normal vector \(\mathbf{n} = (a, b, c)\). The parametric equations for the line can be established as \(x = x_0 + at, y = y_0 + bt, z = z_0 + ct\), where \(t\) is a parameter. The line \( extbf{r}(t) = extbf{r}_0 + t extbf{n} \) passes through \(P\) and aligns with the normal to the plane.

Key Concepts

Equation of a Line

To find the equation of a line that passes through two distinct points, we start by identifying our points, say \(A(x_1, y_1)\) and \(B(x_2, y_2)\). The key to forming a line equation is the slope \(m\), which tells us how steep the line is. The formula for the slope is simple: \(m = \frac{y_2 - y_1}{x_2 - x_1}\). Once we have the slope, we can use the point-slope form of the equation of a line:

• \(y - y_1 = m(x - x_1)\)

This equation is very handy because you can choose either of the two points to plug in for \((x_1, y_1)\). This form is flexible and can be easily adjusted to find the line's equation in the common slope-intercept form \(y = mx + b\).
Finding the equation of a line is essential in analytical geometry, as it allows you to understand relationships between points and helps solve geometric problems with precision.

Equation of a Plane

Determining a plane in three-dimensional space requires at least three non-collinear points, meaning these points should not lie on the same line. Suppose our points are \(A(x_1, y_1, z_1)\), \(B(x_2, y_2, z_2)\), and \(C(x_3, y_3, z_3)\). With these points, we first form vectors \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\). These vectors help define the plane's orientation.
To ensure the plane is well defined, we compute the cross product of these vectors, yielding a normal vector \(\mathbf{n} = \overrightarrow{AB} \times \overrightarrow{AC}\). This normal vector is perpendicular to the surface of the plane.

• The equation of the plane is: \(a(x - x_1) + b(y - y_1) + c(z - z_1) = 0\), where \(\mathbf{n} = (a, b, c)\).

The inclusion of "non-collinear" ensures that the cross product is non-zero, thus defining a unique plane. This concept is fundamental when exploring intersections of planes and lines in space.

Point-Slope Form

The point-slope form of the equation of a line is a valuable tool in geometry. It is especially useful when you know two points on the line or the slope of the line and a point through which it passes. The generic form is:

• \(y - y_1 = m(x - x_1)\)

Here, \(m\) represents the slope, and \((x_1, y_1)\) is a known point on the line.
This form is called "point-slope" because it straightforwardly incorporates the slope of the line and the coordinates of a point. It is particularly helpful for deriving the equation of a line quickly and accurately.
In practical scenarios, once you have the point-slope form, converting it into other forms, like the slope-intercept \(y = mx + b\), is a matter of simple algebra, making it highly versatile for various types of mathematical problems.

Cross Product

The cross product of two vectors in three-dimensional space is an operation that results in a vector perpendicular to both original vectors. Suppose you have vectors \(\mathbf{u}\) and \(\mathbf{v}\), their cross product \(\mathbf{u} \times \mathbf{v}\) can be computed using determinants.

• \(\mathbf{u} \times \mathbf{v} = (u_2v_3 - u_3v_2, u_3v_1 - u_1v_3, u_1v_2 - u_2v_1)\)

This result is useful because it provides a normal vector to define the orientation of a plane that contains both original vectors.
The direction of the cross product is determined using the right-hand rule, increasing its utility in physics and engineering. The magnitude of the cross product relates to the area of the parallelogram spanned by the two vectors, interrelating geometry and calculus in a fascinating way.
In practical terms, especially in analytical geometry, the cross product helps us determine perpendicularity and find normals to planes, which is crucial in defining planes in space.