   Chapter 14.7, Problem 42E

Chapter
Section
Textbook Problem

Find the point on the plane x − 2y + 3z = 6 that is closest to the point (0, 1, 1).

To determine

To find: The point on the plane x2y+3z=6 .which is closest to the point (0,1,1)

Explanation

The given plane equation can be expressed as,

3z=6x+2yz=6x+2y3z=213x+23y

The distance d between (0,1,1) and any point (x,y,z) is,

d=(x0)2+(y1)2+(z1)2d2=x2+(y1)2+(z1)2

Substitute the value of z in the above distance equation,

d2=x2+(y1)2+[(213x+23y)1]2=x2+(y1)2+(113x+23y)2

Thus, f(x,y)=x2+(y1)2+(113x+23y)2 .

Take the partial derivative of f(x,y) with respect x and obtain fx .

fx=x(x2+(y1)2+(113x+23y)2)=x(x2)+x(y1)2+x(113x+23y)2=2x+0+2(113x+23y)(13)=2x23(113x+23y)

Simplify further as follows.

fx=2x23(113x+23y)=2x23+29x49y=209x49y23

Hence, fx=209x49y23 (1)

Take the partial derivative of f(x,y) with respect y and obtain fy

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