Answer:
5x+8y+2z = 68
Explanation:
Given the point P0 = (2, 5, 9), to Find an equation for a plane through that point and normal to the vector n = 5i+8j+2k the following steps must be followed:
The equation for the plane passing through the point is expressed as;
a(x-x0)+b(y-y0)+c(z-z0) = 0 where;
(x0, y0, z0) is the point on the plane and (a,b,c) is the normal vector n.
Given the point (2, 5, 9) and normal vector n =(5, 8, 2)
x0 = 2, y0 = 5, z0 = 9, a = 5, b = 8 and c= 2.
Substituting this values into the equation of the plane above will give;
5(x-2)+8(y-5)+2(z-9) = 0
On expansion:
5x-10+8y-40+2z-18 = 0
5x+8y+2z-10-40-18 = 0
5x+8y+2z-68 = 0
The required equation of the plane is 5x+8y+2z = 68