Minimum Distance to Disk

The equation of the disk can be parametrized as: $\vec{r}(R,\theta)=\vec{C}+R\cos\theta \hat{e_1} + R\sin\theta \hat{e_2}$ where $0 \leq R \leq 3$ , $0 \leq \theta \leq 2\pi$ and $\{\hat{e_1},\hat{e_2}\}$ is an orthonormal basis for the plane in which the disk is contained. It's not hard to see that $(-2,0,1) \times (7,2,0) = \vec{N}$ , so apply Gram–Schmidt process to $\{(-2,0,1),(7,2,0)\}$ to obtain $\hat{e_1}=\dfrac{1}{\sqrt{5}}(-2,0,1)$ and $\hat{e_2}=\dfrac{1}{\sqrt{345}}(7,10,14)$ .

Let $f(R,\theta)=\lVert \vec{r} - (2,3,6) \rVert^2$ , then: $\begin{aligned} f(R,\theta) &= \lVert (1,-2,-1)+R\cos\theta \hat{e_1} + R\sin\theta \hat{e_2} \rVert^2 \\ &= \lVert (1,-2,1) \rVert^2 + R^2\cos^2\theta \lVert \hat{e_1} \rVert^2 + R^2\sin^2\theta \lVert \hat{e_2} \rVert^2 + 2R\cos\theta(1,-2,-1) \cdot \hat{e_1} + 2R\sin\theta(1,-2,-1) \cdot \hat{e_2} + R^2\cos\theta\sin\theta\hat{e_1} \cdot \hat{e_2} \\ &= 6 + R^2 - \dfrac{6}{\sqrt{5}}R\cos\theta - \dfrac{54}{\sqrt{345}}R\sin\theta \end{aligned}$ Now, to find the minimum distance, find $(R,\theta)$ such that $\nabla f = \vec{0}$ : $\begin{aligned} 0 &= \dfrac{\partial f}{\partial \theta} = \dfrac{6}{\sqrt{5}}R\sin\theta - \dfrac{54}{\sqrt{345}}R\cos\theta\\ 0 &= \dfrac{\partial f}{\partial R} = 2R - \dfrac{6}{\sqrt{5}}\cos\theta - \dfrac{54}{\sqrt{345}}\sin\theta \end{aligned}$ We obtain the solution $(R,\theta)=\left(\dfrac{3}{23}\sqrt{230},\arctan \left(\dfrac{9}{\sqrt{69}}\right)\right)$ . So: $\begin{aligned} f(R,\theta) &= 6 + \dfrac{90}{23} - \dfrac{6}{\sqrt{5}} \cdot \dfrac{3\sqrt{230}}{23} \cdot \dfrac{\sqrt{69}}{5\sqrt{6}} - \dfrac{54}{\sqrt{345}} \cdot \dfrac{3\sqrt{230}}{23} \cdot \dfrac{9}{5\sqrt{6}} \\ &= \dfrac{228}{23} - \dfrac{18}{5} - \dfrac{486}{115} \\ &= \dfrac{48}{23} \end{aligned}$ Finally, the minimum distance is $\sqrt{\dfrac{48}{23}} \approx \boxed{1.4446}$ , when $\vec{r}=(3,1,5) + \dfrac{3\sqrt{230}}{23} \cdot \dfrac{\sqrt{69}}{5\sqrt{6}} \cdot \dfrac{1}{\sqrt{5}}(-2,0,1) + \dfrac{3\sqrt{230}}{23} \cdot \dfrac{9}{5\sqrt{6}} \cdot \dfrac{1}{\sqrt{345}}(7,10,14) = (3,1,5)+\dfrac{3}{5}(-2,0,1)+\dfrac{9}{115}(7,10,14) = \dfrac{1}{23}(54,41,154)$ .

[This is not a rigorous approach]

Our disk (circle) lies on the plane $p$ with equation: $ax + by + cz = d$ . Notice that Surface Normal Vector is the vector normal to that plane and actually can be seen as its gradient, represented with elements equal to partial derivatives of a plane: $\vec{N} = \left ( \frac{\partial p}{\partial x}, \frac{\partial p}{\partial y}, \frac{\partial p}{\partial z}\right ) = \left ( -2, 7, -4 \right )$ Hold on, but we know what are these partial derivatives, they are just $a$ , $b$ and $c$ (look at the equation and derive them yourself if you don't believe me). Thus, now we have $a = -2$ , $b = 7$ and $c = -4$ . Point $C = \left ( 3, 1, 5 \right )$ , center of the disk, also lies on the plane and we can use its coordinates to calculate $d$ :

$\begin{aligned} d & = -2\times 3 + 7\times 1 + -4\times 5 \\ & = -19 .\end{aligned}$ .

Comparing the positions of point $C$ and point $\left (2, 3, 6\right)$ , I found that they are pretty close, so that minimum distance between point and disk is minimum distance between point and plane $p$ : $D = \frac{-2\times 2 + 7\times 3 - 4\times 6 -(-19)}{\sqrt{(-2)^{2} + 7^{2} + (-4)^{2}}} = \frac{12}{\sqrt{69}} = 1.4446302...$

Here I used the general formula for distance between point and plane which can be derived easily by those who know Linear Algebra.

So, my approach is pretty common. First, I found out the equation of plane in which the disc is contained using given normal and point(centre of the disc) contained. I got it as 2x-7y+4z=19. Now, there can only be 2 possibilities for the point (2,3,6) to lie in.

1). It can be inside/on the cylinder containing the disc as cross section or

2). It can lie outside the cylinder

Now, I found out the perpendicular distance of the point from the plane because it is the shortest distance from a plane. I got it as 12/√69 units, which is also the shortest distance from the disc surface, if the point satisfies Case 1. Now, I found out the distance between given point and the centre which is √6 units. So, this will be the hypotenuse of the triangle formed by the given point, centre of the disc and foot of perpendicular of (2,3,6) on the plane.

Now, I found out the distance between the foot of perpendicular of (2,3,6) on the plane containing the disc and the centre of the disc. I found out that this distance is less than the radius of the disc. So, the point should lie inside the cylinder and should satisfy Case 1. Since it satisfies case 1, the shortest distance is 12/√69 which is ≈1.44463....

If the point were to lie outside the cylinder, the shortest distance from the surface of the disc would have been the distance between the given point (2,3,6) and a point on the disc which lies on the circumference of the disc and also lies on the line joining foot of perpendicular and the centre.

Also, just by judging that √6 (hypotenuse) is less than the radius, we can conclude that (2,3,6) lies within the cylinder itself. So, 12/√69 should be the least distance