Problem 63 - Calculus Early Transcendentals

Question

What point on the plane \(x-y+z=2\) is closest to the point (1,1,1)\(?\)

Step-by-Step Solution

Answer

Answer: The closest point on the plane to the point (1,1,1) is any point with coordinates (x, x, 2), where x can be any value.

Step 1: Write down the distance formula in 3D-coordinate system

The distance between points (x, y, z) and (1, 1, 1) is given by: $$D=\sqrt{(x-1)^2+(y-1)^2+(z-1)^2}$$

Step 2: Utilize the constraint from the plane

The constraint from the plane \(x-y+z=2\) is given by the equation: $$z = x-y+2$$

Step 3: Substitute the constraint into the distance formula

Replace z in the distance formula with the expression from the constraint: $$D=\sqrt{(x-1)^2+(y-1)^2+[(x-y+2-1)^2]}$$

Step 4: Simplify the distance formula

Simplify the distance formula: $$D=\sqrt{(x-1)^2+(y-1)^2+(x-y+1)^2}$$

Step 5: Minimize the square of the distance

To minimize the distance, we can minimize the square of the distance (since the square root function is monotonically increasing). The square of the distance is: $$D^2=(x-1)^2+(y-1)^2+(x-y+1)^2$$

Step 6: Take partial derivatives

Take the partial derivatives with respect to x and y: $$\frac{\partial{D^2}}{\partial{x}}=2(x-1)+2(x-y+1)=4x-2y+2$$ $$\frac{\partial{D^2}}{\partial{y}}=2(y-1)-2(x-y+1)=-2x+4y-2$$

Step 7: Set partial derivatives to zero and solve

Set the partial derivatives to zero and solve for x and y: $$4x-2y+2=0$$ $$-2x+4y-2=0$$ Multiplying the first equation by 2, we have: $$8x-4y+4=0$$ Comparing this equation to the second one, we can conclude that \(x = y\).

Step 8: Use the constraint from the plane to solve for z

Since \(x=y\), the equation of the constraint is: $$z=x - x + 2$$ $$z=2$$

Step 9: Evaluate the closest point

The closest point on the plane to the point (1,1,1) is: $$(x, y, z)=(x,x,2)$$ Since this problem is theoretical, the actual value of \(x\) does not matter, as any value of \(x\) would give a point on the plane with z-coordinate equal to 2. The closest point lies directly above or below the given point along the z-axis.