Take an arbitrary vector (<em>x</em>, <em>y</em>, <em>z</em>), which goes from the origin to some point (<em>x</em>, <em>y</em>, <em>z</em>) on the plane we want to find.
Subtract from this vector, the vector that points to 
, which is (-3, 3, 1). This translates the first vector so that it starts at the point and is directed at some point (<em>x</em>, <em>y</em>, <em>z</em>). We get a new translated vector, (<em>x</em> + 3, <em>y</em> - 3, <em>z</em> - 1), which lies in the plane.
The normal vector to the plane is orthogonal to every vector in the plane. So taking the dot product of any vector in the plane with the normal to the plane will always result in 0. We use this to find the plane's equation:
and so the answer is D.
Answer:
(3, 2), (2, 3)
Step-by-step explanation:
x + y = 5
xy = 6
Solve the first equation for x.
x = 5 - y
Substitute 5 - y for x in the second equation.
xy = 6
(5 - y)y = 6
5y - y² = 6
y² - 5y + 6 = 0
Factor.
(y - 2)(y - 3) = 0
y - 2 = 0 or y - 3 = 0
y = 2 or y = 3
Now substitute 2 for y in the first equation and solve for x.
x + y = 5
x + 2 = 5
x = 3
One solution is x = 3; y = 2, or (3, 2).
Now substitute 3 for y in the first equation and solve for x.
x + y = 5
x + 3 = 5
x = 2
Another solution is x = 2; y = 3, or (3, 2).
Answer: (3, 2), (2, 3)
