Question

Find the symmetric equations of the line through \((4,0,6)\) and perpendicular to the plane \(x-5 y+2 z=10\).

Short answer

Symmetric equations: \(x-4 = \frac{y}{-5} = \frac{z-6}{2}\).

Step-by-step solution

Identify the normal vector of the plane

The normal vector of the plane is derived from its equation by taking the coefficients of \(x\), \(y\), and \(z\). For the plane \(x-5y+2z=10\), the normal vector \(\mathbf{n}\) is \(\langle 1, -5, 2 \rangle\).

Recognize line's direction vector

The line is perpendicular to the plane, meaning its direction vector is the same as the normal vector of the plane. Therefore, the direction vector \(\mathbf{d}\) of the line is also \(\langle 1, -5, 2 \rangle\).

Use the point-direction form of a line

The symmetric equations of a line passing through a point \((x_0, y_0, z_0)\) with a direction vector \(\mathbf{d} = \langle a, b, c \rangle\) are \(\frac{x-x_0}{a} = \frac{y-y_0}{b} = \frac{z-z_0}{c}\).

Substitute the known point and direction vector

Using the point \((4, 0, 6)\) and the direction vector \(\langle 1, -5, 2 \rangle\), substitute into the equation: \(\frac{x-4}{1} = \frac{y-0}{-5} = \frac{z-6}{2}\).

Write the symmetric equations

The symmetric equations of the line are: \(x-4 = \frac{y}{-5} = \frac{z-6}{2}\).

Key concepts

Understanding Normal Vectors

A normal vector is crucial when working with planes in three-dimensional space. It is a vector that is perpendicular to the plane. In simple terms, it points straight out from the plane, like a pin sticking out from a board. To find the normal vector, look at the coefficients of the equation of the plane. For the plane equation \(x - 5y + 2z = 10\), the coefficients of \(x\), \(y\), and \(z\) are 1, -5, and 2 respectively.

• Normal vector, \(\mathbf{n}\), for this plane is \(\langle 1, -5, 2 \rangle\).

Understanding the normal vector helps in determining directions related to the plane, such as when finding lines that are perpendicular to it.

Direction Vectors and Their Role

A direction vector gives you the orientation of a line in space. It tells you which way the line points. When a line is perpendicular to a plane, its direction is the same as the normal vector of the plane.
Thus, if you know a plane's normal vector, you automatically know the direction vector of any line perpendicular to that plane. From our example, the plane's normal vector \(\langle 1, -5, 2 \rangle\) serves as the direction vector for the line.

• Direction vector, \(\mathbf{d}\), of the line: \(\langle 1, -5, 2 \rangle\)

This vector shows us how the line spans across the 3D space and is essential for writing equations of the line.

Line-Plane Perpendicularity Explained

The concept of line-plane perpendicularity is straightforward once you grasp normal and direction vectors. A line and a plane are perpendicular if the line's direction vector is the normal vector of the plane. Think of it like this: if the line's path follows the direction the normal vector points, it never touches the plane but rather crosses it perpendicularly.

• Perpendicular line shares the plane's normal vector
• Contact point doesn't affect perpendicularity

In our exercise, this property allowed us to quickly derive the line's direction vector from the plane's normal vector without additional calculations.

Point-Direction Form of a Line

To describe a line in 3D, one efficient way is using point-direction form. This form gives a set of symmetric equations that relate a point the line passes through and its direction.
The formula is: \(\frac{x-x_0}{a} = \frac{y-y_0}{b} = \frac{z-z_0}{c}\), where \((x_0, y_0, z_0)\) is any point on the line and \(\langle a, b, c \rangle\) is the line's direction vector.
For our case:

• Point \((4, 0, 6)\) on the line
• Direction vector \(\langle 1, -5, 2 \rangle\)

Substituting these into the formula gives: \(\frac{x-4}{1} = \frac{y}{-5} = \frac{z-6}{2}\). These symmetric equations provide a full description of the line in space.