Answer:
[tex]9x-y+8z-44=0[/tex]
Step-by-step explanation:
It is given that the plane is perpendicular to the line x = 9t, y = 7 − t, z = 3 + 8t.
Coefficients of t are used to find the normal vector.
[tex]\overrightarrow {n}=<9,-1,8>[/tex]
If a plane passes through a point [tex](x_0,y_0,z_0)[/tex] with a normal vector <a,b,c>, then the equation of plane is
[tex]a(x-x_0)+b(y-y_0)+c(z-z_0)=0[/tex]
The plane passes through a point (4,0,1) with a normal vector <9,-1,8>, so the equation of plane is
[tex]9(x-4)+(-1)(y-0)+8(z-1)=0[/tex]
[tex]9x-36-y+8z-8=0[/tex]
[tex]9x-y+8z-44=0[/tex]
Therefore, the equation of plane is [tex]9x-y+8z-44=0[/tex].
