Question

Find a vector parametrization for the line with the given description. Perpendicular to the yz-plane, passes through (0, 0, 8)

Answer 1

Answer:

$\left\{\begin{array}{l}x=t\\ \\y=0\\ \\z=8\end{array}\right.$

or

$\overrightarrow{(t,0,8)}$

Step-by-step explanation:

yz-plane has the equation $x=a,$ where $a$ is a real constant. The normal vector of this plane (vector perpendicular to the plane) is

$\overrightarrow {n}=(1,0,0)$

If the line is perpendicular to this plane, then it is parallel to the normal vector, so the equation of this line is

$\dfrac{x-x_0}{1}=\dfrac{y-y_0}{0}=\dfrac{z-z_0}{0},$

where $(x_0,y_0,z_0)$ are coordinates of the point the line is passing through.

In your case, the line is passing through the point (0,0,8), so the canonical equation of the line is

$\dfrac{x-0}{1}=\dfrac{y-0}{0}=\dfrac{z-8}{0}$

Write a vector parametrization for this line

$\left\{\begin{array}{l}\dfrac{x-0}{1}=t\\ \\\dfrac{y-0}{0}=t\\ \\\dfrac{z-8}{0}=t\end{array}\right.\Rightarrow \left\{\begin{array}{l}x=t\\ \\y=0\\ \\z=8\end{array}\right.$