# Where is the point that has the shortest distance to the orgin?

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).
• 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 $t=18/35.$ Commented May 7, 2015 at 3:16
• @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? Commented May 7, 2015 at 3:21
• 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" Commented May 7, 2015 at 3:24
• @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? Commented May 7, 2015 at 3:34

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.

• Thanks for the Lagrange Solution Danowsky! Commented May 7, 2015 at 4:10

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.

• 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! Commented May 7, 2015 at 4:09