Given:
A plane is normal to the vector = -2i+5j+k
It contains the point (-10,7,5).
To find:
The component equation of the plane.
Solution:
The equation of plane is
[tex]a(x-x_0)+b(y-y_0)+c(z-z_0)=0[/tex]
Where, [tex](x_0,y_0,z_0)[/tex] is the point on the plane and [tex]\left< a,b,c\right>[/tex] is normal vector.
Normal vector is -2i+5j+k and plane passes through (-10,7,5). So, the equation of the plane is
[tex]-2(x-(-10))+5(y-7)+1(z-5)=0[/tex]
[tex]-2(x+10)+5y-35+z-5=0[/tex]
[tex]-2x-20+5y-35+z-5=0[/tex]
[tex]-2x+5y+z-60=0[/tex]
[tex]-2x+5y+z=60[/tex]
Therefore, the equation of the plane is [tex]-2x+5y+z=60[/tex].
