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I just finished an exam, it has the following question: Where is the point on the plane $3x + 5y + z = 18$ has the shortest distance to $(0,0,0)$?

I found this question similar: Find the point on the plane $2x - y + 2z = 20$ nearest the origin

But I am not following the steps:

  1. Obtain the normal vector of the plane <3, 5, 1>
  2. Find the Unit normal vector by dividing $\sqrt{3^2 + 5^2 + 1^2}$
  3. Then how do I go about minimizing the problem? (Set up the constrain)?
  4. Do I use lagrange multipliers? (We were learning about this topic).
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    $\begingroup$ The Lagrange multiplier method is to minimize $x^2+y^2+z^2$ subject to the constraint $3x+5y+z=18.$ A different approach is run a normal line to the plane from (0,0,0) and find the point of intersection. Such a line would be of the form $f(t) = (0,0,0)+(3t,5t,1t)$ and would intersect the plane when $t=18/35.$ $\endgroup$ Commented May 7, 2015 at 3:16
  • $\begingroup$ @mattbiesecker How and why do we use $x^2 + y^2 + z^2$, this is a ... equation of a sphere... Sounds logical. Whenever we want to minimize the distance, we use a sphere? $\endgroup$
    – George
    Commented May 7, 2015 at 3:21
  • $\begingroup$ The distance from an arbitrary point $(x,y,z)$ to the the origin is $\sqrt{x^2+y^2+z^2}.$ So you need to examine all points $(x,y,z)$ on the plane that have the smallest distance. However, it is simple to minimize "distance-squared" $\endgroup$ Commented May 7, 2015 at 3:24
  • $\begingroup$ @mattbiesecker I was trying to work out the minimize, it seems really messy... Could you show the steps for the normal line and find the point of intersection method? $\endgroup$
    – George
    Commented May 7, 2015 at 3:34

2 Answers 2

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If your class was about calculus and you were learning about Lagrange multipliers you're supposed tu use it. So:

Distance from one point $(x,y,z)$ to $(0,0,0)$ is

$$ \lVert (x,y,z)-(0,0,0) \rVert = \sqrt{x^2+y^2+z^2} $$

You want to minimize it subject to $3x+5y+z=18$. But this is the same as minimizing $$ f(x,y,z) = x^2+y^2+z^2 $$ subject to $3x+5y+z=18$. Then we say $g(x,y,z)=3x+5y+8z$.

$$\nabla f(x,y,z) = (2x,2y,2z) \quad\text{and}\quad \nabla g(x,y,z)=(3,5,8)$$

Then we have to solve $\nabla f(x,y,z) = \lambda\nabla g(x,y,z)$ with the same subject. That's the system of equations: $$\left. \begin{array}{c} 2x = 3\lambda \\ 2y = 5\lambda \\ 2z = 8\lambda \\ 3x+5y+8z = 18 \\ \end{array} \right\} $$

After this, you have to prove if the point you find is actually a minimum.

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  • $\begingroup$ Thanks for the Lagrange Solution Danowsky! $\endgroup$
    – George
    Commented May 7, 2015 at 4:10
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The normal line through $(0,0,0)$ is $f(t)=(3t,5t,1t).$ Substitute this into the equation for the plane and you $3(3t) + 5(5t) + 1(1t)=18,$ which implies that $t=18/35.$ Therefore the point where the line intersects the plane is $\left(\frac{54}{35},\frac{90}{35}, \frac{18}{35}\right).$ The distance from this point to the origin is your desired answer.

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  • $\begingroup$ Thank you for showing me another method, although I think the exam was asking for the Lagrange solution. Yours is very simple and neat. Thanks! $\endgroup$
    – George
    Commented May 7, 2015 at 4:09

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