# Math Help - How to find the shortest distance from a plane to a point?

1. ## How to find the shortest distance from a plane to a point?

From the point (2, -1, 3) to the plane z - 2x = 3

2. Originally Posted by AlphaRock
From the point (2, -1, 3) to the plane z - 2x = 3
We have the distance from $(x,y,z)$ to $(2,-1,3)$ is given by $d=\sqrt{(x-2)^2+(y+1)^2+(z-3)^2}$

We know though that $z=2x+3$ and also the critical points of $d$ will be the same as the critical points of $d^2$. So let's minimize $d^2$ instead.

$d^2 = f(x,y) = (x-2)^2+(y+1)^2+4x^2$

Now we solve $\frac{\partial f}{\partial x} = 2(x-2)+8x=0$ and $\frac{\partial f}{\partial x} = 2(y+1)=0$.

Thus we get one critical point, namely $(x,y)=\left(\frac25,-1\right)$.

I'll leave it to you to verify it minimizes the distance. After that, plug $(x,y)$ into $d$ and you'll have your answer.

3. Originally Posted by AlphaRock
From the point (2, -1, 3) to the plane z - 2x = 3
$D = \hat{n} \cdot (\vec{p} - \vec{q})$ where $\hat{n}$ is a unit normal vector to the plane, $\vec{p}$ is the position vector of a known point in the plane and $\vec{q}$ is the position vector of the point from which you're calculating the distance from.

4. First, we need to identify a point on the plane.

Lets pick (0,0,3)

n=<-2,0,1> vector

Vector PQ=<2-0,-1-0,3-3>=<2,-1,0>

D=absolute value of the dot product of pq dot n and then divide by the norm of n

4/sqr(5)

5. Originally Posted by mr fantastic
$D = \hat{n} \cdot (\vec{p} - \vec{q})$ where $\hat{n}$ is a unit normal vector to the plane, $\vec{p}$ is the position vector of a known point in the plane and $\vec{q}$ is the position vector of the point from which you're calculating the distance from.
This solution seems simple. I'm interested in learning more about it. Can you elaborate more on this and perhaps show me how you would use this formula with this particular question? Because my professor never taught us vectors.