Question

- Find a point in the plane x +2y+3z = 13 nearest to the point (1,1,1) using the method of Langrange's multipliers.​

Answer 1

Step-by-step explanation:

Here’s another method. Parameterize the plane as follows. Choose two non-parallel vectors a⃗ and b⃗ in the plane (perpendicular to n⃗ =(3,4,1)), say (−1,0,3) and (0,−1,4), and a point P in the plane, say (0,0,1). An equation for the plane is then {P+sa⃗ +tb⃗ :s,t∈R}={(3s−t,4s,s+3t+1):s,t∈R}. With calculus, find s0,t0 that minimize the distance squared from (3s−t,4s,s+3t+1) to (1,0,1): d(s,t)=(3s−t−1)2+(4s)2+(s+3t+1)2. (This minimizes the distance, but is easier.) The closest point is then (3s0−t0,4s0,s0+3t0+1). – Steve Kass Mar 23 '14 at 23:08

Answer 2

Answer:

The point on the plane x+2y+3z=13 closest to the given point (1, 1, 1) is  =

(1.5, 2, 2.5) .

Explanation:

A "plane" in three-dimensional mathematics is a two-dimensional shape that can extend indefinitely in all directions.

The general equation of a plane is written as ax+ by+ cz = d, where < a, b, c> is the normal vector to the plane.

The method of Lagrange’s multipliers is an important technique applied to solve the local maxima and minima of a given function of the form f(x, y, z) subject to equality constraints of the form of g(x, y, z) = k or g(x, y, z) = 0.

The distance of an arbitrary point of (x, y, z) from the origin is

d=$\sqrt{x^{2}+ y^{2} + z^{2} }$

It is geometrically proven that there is an absolute minimum of this function for (x, y, z) lying on the given plane. To find it, we will minimize the function-

d² = f(x, y, z) = x² + y² + z²

From subject to the constraint, g(x, y, z)= 0 where g(x, y, z) = x + 2y + 3z − 13.

The principle gradients of these two functions are ∇f = (2x, 2y, 2z), and ∇g = (1, 2, 3).

Since ∇g ≠ 0 is counted, the absolute minimum of the distance function will occur at a point where ∇f = λg, g = 0.

Getting rid of the 2’s in ∇f (which all get absorbed into the constant λ) and setting components of the gradient equation, we obtain the system of equations-

x =1+ λ

y = 1+2λ

z = 1+3λ [ as to find the closest point to (1, 1, 1)]

x + 2y + 3z = 13

Solving the first three equations gives y = 2x, z = 3x. Plugging these into the equation of the plane gives 1+x+2+4x+3+9x = 14x + 6 = 13, and so the point we are looking for is x = 1.5, y = 2, z = 2.5.

Therefore,

The point on the plane x+2y+3z=13 closest to the given point (1, 1, 1) is  =

(1.5, 2, 2.5) .

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