Answer:
[tex]\left\{\begin{array}{l}x=t\\ \\y=0\\ \\z=8\end{array}\right.[/tex]
or
[tex]\overrightarrow{(t,0,8)}[/tex]
Step-by-step explanation:
yz-plane has the equation [tex]x=a,[/tex] where [tex]a[/tex] is a real constant. The normal vector of this plane (vector perpendicular to the plane) is
[tex]\overrightarrow {n}=(1,0,0)[/tex]
If the line is perpendicular to this plane, then it is parallel to the normal vector, so the equation of this line is
[tex]\dfrac{x-x_0}{1}=\dfrac{y-y_0}{0}=\dfrac{z-z_0}{0},[/tex]
where [tex](x_0,y_0,z_0)[/tex] 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
[tex]\dfrac{x-0}{1}=\dfrac{y-0}{0}=\dfrac{z-8}{0}[/tex]
Write a vector parametrization for this line
[tex]\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.[/tex]
