Find the minimum distance from the point (2, -1, 1) to the plane x + y - z = 2.

Solution

VerifiedLet $A(2,-1,1)$ and $B(x,y,z)$, then we get the following:

$d(AB)=\sqrt{(x-2)^2+(y+1)^2+(z-1)^2}=\sqrt{(x-2)^2+(y+1)^2+(z-1)^2}$

So, we will minimize $d(AB)=\sqrt{(x-2)^2+(y+1)^2+(z-1)^2}$ if we minimize $D(AB)=(x-2)^2+(y+1)^2+(z-1)^2$. We are given that the plane is $x+y-z=2 \Rightarrow z=x+y-2$. Now, we get the following:

$D(x,y)=(x-2)^2+(y+1)^2+(x+y-3)^2 \Rightarrow D_x=2(x-2)+2(x+y-3)=0, D_y=2(y+1)+2(x+y-3)=0$

$\Rightarrow x=\dfrac{8}{3}, y=-\dfrac{1}{3}, z=\dfrac{1}{3}$

So, we got that the critical point is $\left(\dfrac{8}{3}, -\dfrac{1}{3} \right)$, then we get

$D_{xx}=4>0, D_{yy}=4, D_{xy}=2 \Rightarrow D_{xx}D_{yy}-(D_{yy})^2=12>0$

The conclusion is that the point $\left(\dfrac{8}{3}, -\dfrac{1}{3}, \dfrac{1}{3} \right)$ is the point of a local minimum of $D$, which means that is required point.

## Create a free account to view solutions

## Create a free account to view solutions

## Recommended textbook solutions

#### Thomas' Calculus

14th Edition•ISBN: 9780134438986 (11 more)Christopher E Heil, Joel R. Hass, Maurice D. Weir#### Calculus: Early Transcendentals

8th Edition•ISBN: 9781285741550 (6 more)James Stewart#### Calculus: Early Transcendentals

9th Edition•ISBN: 9781337613927 (1 more)Daniel K. Clegg, James Stewart, Saleem Watson#### Calculus

9th Edition•ISBN: 9781337624183 (1 more)Daniel K. Clegg, James Stewart, Saleem Watson## More related questions

1/4

1/7