Answer:
x=-5+2t, y=-6+3t, z=-6+9t
Step-by-step explanation:
We need to find parametric equations for the line through the point ( - 5, - 6, - 6) and perpendicular to the plane 2x + 3y + 9z = 17 .
If a line passes through a point [tex](x_1,y_1,z_1)[/tex] and perpendicular to the plane [tex]ax+by+cz=d[/tex], then the Cartesian form of line is
[tex]\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}[/tex]
and parametric equations are
[tex]x=x_1+at,y=y_1+bt,z=z_1+ct[/tex]
For the given information, [tex]x_1=-5,y_1=-6,z_1=-6,a=2,b=3,c=9[/tex]. So, the Cartesian form of line is
[tex]\frac{x+5}{2}=\frac{y+6}{3}=\frac{z+6}{9}[/tex]
Parametric equations of the line are
[tex]x=-5+2t[/tex]
[tex]y=-6+3t[/tex]
[tex]z=-6+9t[/tex]
Therefore the parametric equation for the line are x=-5+2t, y=-6+3t, z=-6+9t.
