Answer: Q = (2/3, 4/3, 7/3)
Explanation:
To find the point of intersection, we can solve these two equations simultaneously:
x + y - z = 1
x + y - z = -1
Subtracting the second equation from the first, we get:
2z = 2
z = 1
Substituting this into either equation, we get:
x + y = 2
There are infinitely many solutions to this equation, but we want the one that is closest to P = (1, 0, 0). This means we want to minimize the distance |PQ|, which is given by:
|PQ| = sqrt((x-1)^2 + y^2 + 1)
We can use Lagrange multipliers to find the minimum of this distance subject to the constraint x + y = 2:
L(x, y, λ) = (x-1)^2 + y^2 + 1 + λ(x + y - 2)
Taking the partial derivatives and setting them equal to zero, we get:
∂L/∂x = 2(x-1) + λ = 0
∂L/∂y = 2y + λ = 0
∂L/∂λ = x + y - 2 = 0
Solving these equations, we get:
x = 2/3
y = 4/3
Substituting these values into z = x + y + 1, we get:
z = 7/3
So the point Q on the plane closest to P = (1, 0, 0) is Q = (2/3, 4/3, 7/3).