Question

a.)

Find the shortest distance from the point (0,1,2) to any point on the plane x - 2y +z = 2 by finding the function to optimize, finding its critical points and test for extreme values using the second derivative test.

b.)

Write the point on the plane whose distance to the point (0,1,2) is the shortest distance found in part a) above. All the work necessary to identify this point would be in part a). You just need to write the coordinates of the point here.

Answer 1

a)

general point on the plane x-2y+z=2 is (x,y,z)

distance from the point (0,1,2) to any point on the plane x - 2y +z = 2 is  [(x-0)²+(y-1)²+(z-2)²]^1/2

let f(x,y,z)=(x-0)²+(y-1)²+(z-2)²
=>f(x,y,z)=x²+(y-1)²+(z-2)²

x-2y+z=2
=> z-2=-x+2y

f(x,y)=x²+(y-1)²+(-x+2y)²
=>f(x,y)=x²+y²-2y+1+x²-4xy+4y²
=>f(x,y)=2x²+5y²-4xy-2y+1

f_x=4x+0-4y-0+0,f_y=0+10y-4x-2+0
=>f_x=4x-4y,f_y=10y-4x-2

for critical points f_x=0,f_y=0
=>4x-4y=0,10y-4x-2=0
=>4x=4y,10y-4x-2=0
=>10y-4y-2=0
=>6y=2
=>y=1_/3

4x-4y=0,y=1_/3
=>4x-4(1_/3)=0
=>4x=4(1_/3)
=>x=1_/3

(x,y)=(1_/3,1_/3) is the critical point

f_x=4x-4y,f_y=10y-4x-2
=>f_xx=4,f_yy=10,f_xy=-4

D=(f_xx)(f_yy)-(f_xy)²
=>D=(4)(10)-(-4)²
=>D=40-16
=>D=24

D>0,f_xx>0

=>f(x,y)=x²+(y-1)²+(-x+2y)² has minimum value at the point (1_/3,1_/3)

z-2=-x+2y,x=1_/3,y=1_/3
=> z-2=-(1_/3)+2(1_/3)
=> z-2=1_/3
=> z=2+(1_/3)
=> z=7_/3

shortest distance =[(1_/3-0)²+(1_/3-1)²+(7_/3-2)²]^1/2
=>shortest distance =[(1_/3)²+(-2_/3)²+(1_/3)²]^1/2
=>shortest distance =(2_/3)^1/2

shortest distance from the point (0,1,2) to any point on the plane x - 2y +z = 2 is units
or units

b)

point on the plane whose distance to the point (0,1,2) is the shortest distance is