Question

In Exercises 1 and \(2,\) the figure shows the graph of a line given by the parametric equations. (a) Draw an arrow on the line to indicate its orientation. To print an enlarged copy of the graph, go to the website \(w w w .\) mathgraphs.com. (b) Find the coordinates of two points, \(P\) and \(Q,\) on the line. Determine the vector \(\overrightarrow{P Q}\). What is the relationship between the components of the vector and the coefficients of \(t\) in the parametric equations? Why is this true? (c) Determine the coordinates of any points of intersection with the coordinate planes. If the line does not intersect a coordinate plane, explain why. $$ \begin{array}{l} x=1+3 t \\ y=2-t \\ z=2+5 t \end{array} $$

Short answer

The line defined by the given parametric equations is oriented in the direction of the vector (3, -1, 5), and the vector connecting any two points P and Q on the line is equal to this vector. The line does not intersect the x,y, or z planes at any point.

Step-by-step solution

Define the direction of the line

The line's direction corresponds to the vector given by the coefficients of \( t \), i.e. \( (3,-1,5) \). In the positive \( t \) direction, the gradient vector points. Therefore, the line is oriented in the direction of the arrowhead of this vector.

Identify two points P and Q on the line

To find points on the line, we just vary \( t \). Let's pick \( t = 0 \) and \( t = 1 \) for simplicity. This gives us point \( P(1, 2, 2) \) when \( t = 0 \) and point \( Q(4, 1, 7) \) when \( t = 1 \).

Determine the vector \( \overrightarrow{P Q} \)

The vector \( \overrightarrow{P Q} \) is the difference in coordinates of \( Q \) and \( P \). Therefore \( \overrightarrow{P Q} = (4, 1, 7) - (1, 2, 2) = (3, -1, 5) \). As we see, the vector is equivalent to the gradient vector and coefficients of \( t \). This equivalence exists because the parametric equation of a line defines how it moves along specific dimensions.

Determine intersection points with the coordinate planes

A line intersects a coordinate plane where the variable corresponding to that plane equals zero. For the x-plane (y=0, z=0), when \( y = 2 - t = 0 \) and \( z = 2 + 5t = 0 \), we get \( t = 2 \), and \( t = -2/5 \) respectively. This leads to two different values of \( t \) which means that there is no intersection with the x-plane. We do a similar analysis for the y-plane (x=0, z=0) and the z-plane (x=0, y=0), and we find that the line does not intersect with these planes either.

Key concepts

Coordinate Planes Intersection

Understanding where a line intersects coordinate planes is essential for visualizing geometric relationships in three-dimensional space. A coordinate plane is a two-dimensional surface defined by two of the three axes in 3D space - the XY, YZ, and ZX planes.

When discussing intersections of a line and a coordinate plane, it is crucial to grasp that an intersection happens where the line passes through the plane. For the line represented by the given parametric equations \(x=1+3t, y=2-t, z=2+5t\), we need to determine when any of the coordinates (x, y, z) becomes zero, since this signifies the line crossing that respective plane.

However, if a line does not intersect a plane, we may discover that no value of \(t\) brings a coordinate to zero, or that different values of \(t\) are required to zero out different coordinates. This discrepancy indicates the line runs parallel to or does not touch that particular coordinate plane.

For our line, setting \(y\) and \(z\) to zero does not yield a consistent value of \(t\), thus indicating no intersection with the x-plane. The same goes for the y-plane and z-plane, implying that our line never intersects any of the three coordinate planes.

Vector Direction

The direction of a vector is a fundamental concept that gives insight into the orientation and flow of quantities in space. Vectors are described by both magnitude and direction, and in the context of a parametrically defined line, the direction vector conveys which way the line progresses as \(t\) increases.

To establish the vector's direction from the parametric equations, we look at the coefficients of \(t\) - in this case, \(3, -1, 5\). These values form the direction vector of the line \(\overrightarrow{d} = (3, -1, 5)\), indicating how much the line moves in each of the x, y, and z directions for one unit increase in \(t\).

These coefficients reflect the line's slope in each dimension. For instance, a positive coefficient expresses a movement in the positive direction along the corresponding axis, and vice versa. The vector \(\overrightarrow{PQ}\), calculated between two points on the line, not only establishes the line’s slope but also represents the line's orientation. In this case, the vector points in the direction where x and z increase while y decreases.

Parametric Line Representation

A parametric line representation offers a dynamic way to describe a line in space by using one or more parameter(s), commonly \(t\). The parameters represent a set of coordinates that vary continuously, defining the line's position at every instance.

For the given exercise, the line is defined by three individual parametric equations: \(x=1+3t\), \(y=2-t\), and \(z=2+5t\). Each equation details how the respective coordinate changes as \(t\) changes. By setting \(t\) to any real number, we get a unique point on the line, and by varying \(t\), we can trace the entire line.

The advantage of parametric representation is its clear illustration of how each coordinate depends on \(t\). It allows for easy visualization of the line's direction and orientation. For example, if we were to choose two points, say when \(t=0\) and \(t=1\), like points \(P\) and \(Q\) in our step-by-step solution, we get a tangible sense of the line's trajectory through space.