Find a point on the plane x + y z = 1 closest to the point (1, 6, -6). Also find the distance from this point to the plane. (a) We would like to minimize the distance of a point with coordinates (x, y, z) to the point (1,6, -6). In order to make it easier to take partial derivatives, let's minimize the squared distance instead. Let f(x, y, z) be the squared distance from (x, y, z) to (1,6, −6) in terms of x, y, z. f(x, y, z) = == f (b) Let g(x, y, z) = x + y − z − 1. Find the gradients ▼ƒ and Vg. Vf(x, y, z) = - , , Vg(x, y, z) = = , , (c) Find the point on the given plane closest to (1,6, —6). Enter answers as integers or fractions, no decimals. Point: (d) Find the distance from the point (1, 6, -6) to the plane. Enter an exact answer. Distance:
Find a point on the plane x + y z = 1 closest to the point (1, 6, -6). Also find the distance from this point to the plane. (a) We would like to minimize the distance of a point with coordinates (x, y, z) to the point (1,6, -6). In order to make it easier to take partial derivatives, let's minimize the squared distance instead. Let f(x, y, z) be the squared distance from (x, y, z) to (1,6, −6) in terms of x, y, z. f(x, y, z) = == f (b) Let g(x, y, z) = x + y − z − 1. Find the gradients ▼ƒ and Vg. Vf(x, y, z) = - , , Vg(x, y, z) = = , , (c) Find the point on the given plane closest to (1,6, —6). Enter answers as integers or fractions, no decimals. Point: (d) Find the distance from the point (1, 6, -6) to the plane. Enter an exact answer. Distance:
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Transcribed Image Text:Find a point on the plane x + y z = 1 closest to the
point (1, 6, -6). Also find the distance from this point to
the plane.
(a) We would like to minimize the distance of a point with
coordinates (x, y, z) to the point (1,6, -6). In order to
make it easier to take partial derivatives, let's minimize the
squared distance instead. Let f(x, y, z) be the squared
distance from (x, y, z) to (1,6, −6) in terms of x, y, z.
f(x, y, z) =
==
f
(b) Let g(x, y, z) = x + y − z − 1. Find the gradients ▼ƒ
and Vg.
Vf(x, y, z) =
-
,
,
Vg(x, y, z) =
=
,
,
(c) Find the point on the given plane closest to (1,6, —6).
Enter answers as integers or fractions, no decimals.
Point:
(d) Find the distance from the point (1, 6, -6) to the
plane. Enter an exact answer.
Distance:
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