Problem 13 - Essential Calculus Early Transcendentals

Question

(a) Find symmetric equations for the line that passes through the point \((1,-5,6)\) and is parallel to the vector \(\langle- 1,2,-3\rangle .\) (b) Find the points in which the required line in part (a) intersects the coordinate planes.

Step-by-Step Solution

Answer

Symmetric equations: \( \frac{x-1}{-1} = \frac{y+5}{2} = \frac{z-6}{-3} \). Intersections: XY-plane at \((-1,-1,0)\), YZ-plane at \((0,-3,3)\), ZX-plane at \((-\frac{3}{2},0,-\frac{3}{2})\).

Step 1: Understanding the Problem

The problem requires finding the symmetric equations of a line given a point it passes through and a vector it is parallel to. Additionally, we need to find where this line intersects the XY, YZ, and ZX planes.

Step 2: Express the Line in Parametric Form

To express the line in parametric form, we use the given point \((1, -5, 6)\) as the starting point and the vector \(\langle -1, 2, -3 \rangle\) as the direction. The parametric equations of the line are: \[ x = 1 - t, \ y = -5 + 2t, \ z = 6 - 3t \] where \(t\) is the parameter.

Step 3: Convert to Symmetric Equations

To convert the parametric equations to symmetric form, we solve each equation for \(t\) and set them equal. This gives:\[ \frac{x - 1}{-1} = \frac{y + 5}{2} = \frac{z - 6}{-3} \] These are the symmetric equations of the line.

Step 4: Find Intersection with XY-Plane

On the XY-plane, \(z = 0\). Substitute \(z = 0\) into the parametric equation for \(z\):\[ 6 - 3t = 0 \Rightarrow t = 2 \]Substitute \(t = 2\) into the equations for \(x\) and \(y\):\[ x = 1 - 2 = -1, \quad y = -5 + 4 = -1 \]Thus, the line intersects the XY-plane at \((-1, -1, 0)\).

Step 5: Find Intersection with YZ-Plane

On the YZ-plane, \(x = 0\). Substitute \(x = 0\) into the parametric equation for \(x\):\[ 1 - t = 0 \Rightarrow t = 1 \]Substitute \(t = 1\) into the equations for \(y\) and \(z\):\[ y = -5 + 2 = -3, \quad z = 6 - 3 = 3 \]Thus, the line intersects the YZ-plane at \((0, -3, 3)\).

Step 6: Find Intersection with ZX-Plane

On the ZX-plane, \(y = 0\). Substitute \(y = 0\) into the parametric equation for \(y\):\[ -5 + 2t = 0 \Rightarrow t = \frac{5}{2} \]Substitute \(t = \frac{5}{2}\) into the equations for \(x\) and \(z\):\[ x = 1 - \frac{5}{2} = -\frac{3}{2}, \quad z = 6 - \frac{15}{2} = -\frac{3}{2} \]Thus, the line intersects the ZX-plane at \((-\frac{3}{2}, 0, -\frac{3}{2})\).

Key Concepts

Symmetric Equations

Symmetric equations are a way of representing lines in three-dimensional space. They are particularly useful because they show the relationship between all three coordinates (x, y, and z) directly. To form symmetric equations from parametric equations, the key step is to solve each parametric equation for the common parameter (usually denoted as \(t\)), and then set these expressions equal to each other.

In the given problem, we start with the parametric equations \(x = 1 - t\), \(y = -5 + 2t\), and \(z = 6 - 3t\). Solving each for \(t\) gives us:

• \(t = 1 - x\)
• \(t = \frac{y + 5}{2}\)
• \(t = \frac{z - 6}{-3}\)

By setting these expressions equal, we form the symmetric equation:
\[ \frac{x - 1}{-1} = \frac{y + 5}{2} = \frac{z - 6}{-3} \]
These equations concisely describe the line by showing how x, y, and z depend on each other through a single parameter.

Parametric Equations

Parametric equations describe a line by expressing each of its coordinates as functions of a parameter, often \(t\). This method provides a straightforward way of describing a path in space. It's particularly handy when dealing with problems involving direction and points, like ours with the vector \(\langle -1, 2, -3 \rangle\) and the point \((1, -5, 6)\).

Here's how parametric equations were formed in the example:

• Start with a given point on the line, \((1, -5, 6)\).
• Consider a direction vector, \(\langle -1, 2, -3 \rangle\), which indicates how the point moves through space.

Each coordinate of the line can be described parametrically as:

• \(x = 1 - t\)
• \(y = -5 + 2t\)
• \(z = 6 - 3t\)

These equations give a comprehensive description of every point on the line as the parameter \(t\) varies. These equations also translate into symmetric equations, providing more math tools to analyze the situation.

Coordinate Plane Intersections

Finding the intersection of a line with coordinate planes (XY, YZ, and ZX) can reveal significant insights. This involves setting one of the coordinates to zero as per the specific plane and solving for the other coordinates. In the exercise, determining intersection points involved substituting values that set each coordinate suitable for each plane.

### Intersection with the XY-Plane- Here, the condition is \(z = 0\).- Substitute into the z-parametric equation to find \(t\).- Solve for \(t = 2\), and substitute in the x and y equations, resulting in point \((-1, -1, 0)\).

### Intersection with the YZ-Plane- Set \(x = 0\) and solve the corresponding equation, \(1 - t = 0\).- Get \(t = 1\) and use it in the equations for y and z. The intersection point is \((0, -3, 3)\).

### Intersection with the ZX-Plane- Let \(y = 0\), substituting in the corresponding parametric equation.- Solve for \(t = \frac{5}{2}\), then find x and z coordinates. This gives the intersection point \((-\frac{3}{2}, 0, -\frac{3}{2})\).
Understanding these intersections helps in visualizing and verifying the geometry of lines in three-dimensional spaces.