Problem 14 - Essential Calculus Early Transcendentals

Question

(a) Find parametric equations for the line through \((2,4,6)\) that is perpendicular to the plane \(x-y+3 z=7\) (b) In what points does this line intersect the coordinate planes?

Step-by-Step Solution

Answer

Parametric equations: \(x = 2 + t\), \(y = 4 - t\), \(z = 6 + 3t\). Intersections: XY-plane \((0, 6, 0)\), XZ-plane \((6, 0, 18)\), YZ-plane \((0, 6, 0)\).

Step 1: Finding the Normal vector of the Plane

To find the line that is perpendicular to the plane, we first need to identify the normal vector of the plane, as this vector will be parallel to our line. The equation of the plane is given by \(x-y+3z=7\), and its normal vector is derived from the coefficients of \(x\), \(y\), and \(z\). Therefore, the normal vector is \(\mathbf{n} = \langle 1, -1, 3 \rangle\).

Step 2: Writing the Parametric Equations

The line passing through the point \((2, 4, 6)\) and parallel to the vector \(\mathbf{n} = \langle 1, -1, 3 \rangle\) can be described by the parametric equations: \[x = 2 + t,\] \[y = 4 - t,\] \[z = 6 + 3t\]. Here, \(t\) is the parameter.

Step 3: Finding the Intersection with the XY-plane

The line intersects the XY-plane when \(z = 0\). Using the parametric equation for \(z\), set \(6 + 3t = 0\) and solve for \(t\). This gives \(t = -2\). Substitute \(t = -2\) into the parametric equations for \(x\) and \(y\) to find the point of intersection: \(x = 0\), \(y = 6\). So, the line intersects the XY-plane at the point \((0, 6, 0)\).

Step 4: Finding the Intersection with the XZ-plane

The line intersects the XZ-plane when \(y = 0\). Using the parametric equation for \(y\), set \(4 - t = 0\) and solve for \(t\). This gives \(t = 4\). Substitute \(t = 4\) into the parametric equations for \(x\) and \(z\) to find the point \((6, 0, 18)\). Thus, the line intersects the XZ-plane at \((6, 0, 18)\).

Step 5: Finding the Intersection with the YZ-plane

The line intersects the YZ-plane when \(x = 0\). Using the parametric equation for \(x\), set \(2 + t = 0\) and solve for \(t\). This gives \(t = -2\). Substitute \(t = -2\) into the parametric equations for \(y\) and \(z\) to find the point \((0, 6, 0)\). Hence, the line intersects the YZ-plane at \((0, 6, 0)\).

Key Concepts

Vector calculus

Vector calculus is a branch of mathematics that deals with vector fields, which are vector quantities that vary with position in space. In the context of this exercise, vector calculus helps us define the directionality and orientation of lines and planes.

When we are given a plane equation like \(x-y+3z=7\), the normal vector plays a crucial role. The normal vector is derived straight from the coefficients of the variables in the plane equation. In this case, the normal vector is \mathbf{n} = \langle 1, -1, 3 \rangle\.

This vector is perpendicular to the plane, meaning any line parallel to this vector will also be perpendicular to the plane.

• Normal vectors are key in vector calculus as they define perpendicularity.
• Recognizing normal vectors from plane equations is a fundamental skill.

Understanding these concepts is essential because they allow us to solve for various properties of the line relative to the plane like intersections or parallelism.

Coordinate planes

Coordinate planes are flat, two-dimensional surfaces defined within three-dimensional space. In three-dimensional space, we commonly refer to the XY-plane, XZ-plane, and YZ-plane. Each of these planes corresponds to having a zero value in one of the coordinates.

For example, the XY-plane has coordinates of the form \(x, y, 0\), indicating that any point on this plane has a zero \(z\) coordinate. Similarly, the XZ-plane consists of points like \(x, 0, z\), and the YZ-plane consists of points like \(0, y, z\).

When determining where a line intersects these planes, we set the appropriate coordinate equal to zero.

• To find where a line intersects the XY-plane, set \(z=0\).
• To find where it intersects the XZ-plane, set \(y=0\).
• To find where it intersects the YZ-plane, set \(x=0\).

Doing so allows us to find the intersection points by solving the parametric equations of the line. This is a powerful technique in geometry and spatial analysis.

Intersection points

Finding intersection points is a central part of solving geometrical problems involving lines and planes in three-dimensional space. Intersection points show where geometric entities meet or cross each other, and are useful for understanding spatial relationships.

When a line intersects a plane, we determine these points by solving the parametric equations of the line for a specific plane equation.

For example, consider a line with parametric equations \x = 2 + t, y = 4 - t, z = 6 + 3t\. To find the intersection with the XY-plane, we set \z=0\. Solving \6 + 3t = 0\ gives us \t=-2\. We substitute this value back into the \x\ and \y\ equations to find the intersection point \ (0, 6, 0)\.

• Intersection points provide valuable insight about spatial alignment.
• Solving for \(t\) in parametric equations highlights where intersections occur.

By methodically checking the conditions where each coordinate (x, y, z) is zero with respect to their respective plane, we uncover the geometric relationships between different elements in space.