(Choices below) What's an equation for a plane with normal vector <1,1,-1> containing the point (1,1,-1)?

Question

possible choices.....

a. x + y + z = 3

b. x - y + z = 3

c. x + y - z = 3

d. -x + y + z = 3

e. x - y + z = 1

f. x - y - z = 1

there is no such plane (only if a candidate exists but is not listed below)
or
There is such a plane but it is none of these

Answer 1

The answer is C, normal vector is <1,1,-1> and 1 + 1 -(-1) = 3

Answer 2

What's an equation for a plane with normal vector <1,1,-1> containing the point (1,1,-1)?

This is just the information you need to write the equation of the plane.

1(x - 1) + 1(y - 1) - 1(z + 1) = 0
x - 1 + y - 1 - z - 1 = 0
x + y - z - 3 = 0

The answer is c.

Answer 3

C is right ans. the point (1,1,-1) is contined by both C and F; but the coefficients of the equation of the plane have to be 1,1,-1 for x,y,z respectively.

Answer 4

c. x + y - z = 3

r.n = a.n
r.(1 1 -1) = (1 1 -1).(1 1 -1)
r.(1 1 -1) = 3