Questions: Write the parametric equations of the line through the point (-1,0,3) and perpendicular to the plane -5x-8y-4z=2

Transcript text: Write the parametric equations of the line through the point $(-1,0,3)$ and perpendicular to the plane $-5 x-8 y-4 z=2$

Solution

Solution Steps

To find the parametric equations of a line through a given point and perpendicular to a plane, we need to:

Identify the normal vector of the plane, which will be the direction vector of the line.
Use the given point as the initial point of the line.
Write the parametric equations using the direction vector and the initial point.

The normal vector of the plane $ -5x - 8y - 4z = 2 $ is $ \langle -5, -8, -4 \rangle $. The given point is $ (-1, 0, 3) $.

Step 1: Identify the Given Point and Normal Vector

The line passes through the point $ P(-1, 0, 3) $ and is perpendicular to the plane defined by the equation $ -5x - 8y - 4z = 2 $. The normal vector of the plane, which serves as the direction vector of the line, is given by $ \mathbf{n} = \langle -5, -8, -4 \rangle $.

Step 2: Write the Parametric Equations

Using the point $ P $ and the direction vector $ \mathbf{n} $, the parametric equations of the line can be expressed as follows: \[ \begin{align_} x(t) & = -1 - 5t, \\ y(t) & = 0 - 8t, \\ z(t) & = 3 - 4t. \end{align_} \]

Step 3: Evaluate the Parametric Equations at $ t = 1 $

Substituting $ t = 1 $ into the parametric equations: \[ \begin{align_} x(1) & = -1 - 5(1) = -6, \\ y(1) & = 0 - 8(1) = -8, \\ z(1) & = 3 - 4(1) = -1. \end{align_} \]