Explanation Given: The point is ( 0 , 0 , 0 ) , and the plane is: x − y + z = 3 . Formula used: The distance from any point ( x , y , z ) to the point ( a , b , c ) is: d = ( x − a ) 2 + ( y − b ) 2 + ( z − c ) 2 And second derivative test is: M = f x x ( x , y ) ⋅ f y y ( x , y ) − [ f x y ( x , y ) ] 2 Calculation: The distance from any point ( x , y , z ) to the point ( 0 , 0 , 0 ) is: d = ( x − 0 ) 2 + ( y − 0 ) 2 + ( z − 0 ) 2 d 2 = x 2 + y 2 + z 2 Consider the provided plane is: x − y + z = 3 z = 3 − x + y …… (1) The point ( x , y , z ) lies on the provided surface. Then, forms a function in such a way, f ( x , y ) = d 2 = x 2 + y 2 + z 2 Substitute the value of z in the above equation, f ( x , y ) = ( x ) 2 + ( y ) 2 + ( 3 − x + y ) 2 Then, the partial derivatives are: f x ( x , y ) = ∂ ∂ x [ ( x ) 2 + ( y ) 2 + ( 3 − x + y ) 2 ] = 2 x + 2 ( 3 − x + y ) ( − 1 ) = 2 ( x − 3 + x − y ) = 2 ( 2 x − y − 3 ) And, f y ( x , y ) = ∂ ∂ y [ ( x ) 2 + ( y ) 2 + ( 3 − x + y ) 2 ] = 2 ( y ) + 2 ( 3 − x + y ) = 2 ( y + 3 − x + y ) = 2 ( − x + 2 y + 3 ) Now, set f x ( x , y ) = 0 , and f y ( x , y ) = 0 to get the critical points

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ISBN: 9781285057095

Author: Ron Larson, Bruce H. Edwards

Publisher: Cengage Learning

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Chapter 13.9, Problem 1E

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To calculate: The minimum distance from the point, (0,0,0) to the plane x−y+z=3.

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Chapter 13.9, Problem 20E

Chapter 13.9, Problem 2E

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The minimum distance from the point, ( 0 , 0 , 0 ) to the plane x − y + z = 3 .

Solution Summary: The author calculates the minimum distance from the point to the plane x-y+z=3.

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To calculate: The minimum distance from the point, (0,0,0) to the plane x−y+z=3.

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