Find an equation of the plane through the point(1, 5,-1) and perpendicular to the vector (1, 5, 1). Do this problem in the standard way.

Step-by-step explanation:

Given a plane passing through the point(1, 5,-1) and perpendicular to the vector (1, 5, 1), the equation of the plane can be expressed generally 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 perpendicular to the plane i.e (1,5,1)

Given the point P (1, 5, -1) and the normal vector n(1, 5, 1)

x₀ = 1, y₀ =5, z₀ = -1, a = 1, b = 5 and c = 1

Substituting this point in the formula we will have;

a(x-x₀)+b(y-y₀)+c(z-z₀) = 0

1(x-1)+5(y-5)+1(z-(-1)) = 0

(x-1)+5(y-5)+(z+1) = 0

x-1+5y-25+z+1 = 0

x+5y+z-1-25+1 = 0

x+5y+z-25 = 0

x+5y+z = 25

The final expression gives the equation of the plane.

13

step-by-step explanation:

distance formula is d=√(x2-x1)^2   +   (y2-y1)^2