Understanding the distance between a line and a point in 3D space

Question

I know that there are quicker ways to do what I am about to present. But I want to understand why my approach does not work.

Let the point $P = (-6, 3, 3)$ and the line $L=(-2t,-6t,t)$.

I am trying to find the shortest distance between the point and the line. From my observation, I believe the line passes through the origin because it can be written as $$L=\begin{bmatrix}0 \\ 0 \\ 0 \end{bmatrix} +t \begin{bmatrix}-2 \\ -6 \\ 1 \end{bmatrix} $$.

Let $Q$ denoted $(a, b, c)$ be a point on $L$ such that $\vec{QP}$ is the shortest distance between $L$ and $P$. Note that $\vec{QP}$ is normal to $L$.

Therefore, I need to find $\vec{QP}$ which is $\vec{P}-\vec{Q}$.

$\vec{QP} = (-6 - a, 3 - b, 3 -c)$

We know that $\vec{QP}$ and $L$ are perpendicular so the dot product is 0.

$$-2(-6 - a) - 6(3 - b) + (3 - c) = 0$$

Simplifying gives us $$2a + 6b - c -3 = 0$$

let $a=0$, $b=1$, then by solving we know that $c=3$.

From my understanding, we should have found $Q$ which intersects $L$ and $\vec{QP}$. Unfortunately, it seems $||\vec{QP}||$ is not the correct answer. I think that the way I managed to pull out the $a$, $b$ and $c$ is the culprit, however I just don't understand what I did wrong.

Answer 1

You have ignored completely the constraint $Q$ being a point on $L$. That imposes $b=3a$, for example, which your proposed $a=0,b=1$ does not satisfy.

Answer 2

You need to find the point $Q$ on $L$ minimizing the distance from $P$. Since $Q=(a,b,c)$ lies in $L$, $a,b,c$ must satisfy the relations $a=-2c$ and $b=-6c$. Further, $QP$ must be orthogonal to $L$, so $(a+6,b-3,c-3) \cdot (-2,-6,1)=0$.

Answer 3

Let $v = (-6,3,3), u = (-2,-6,1)$

$v - \frac {u\cdot v}{\|u\|^2} u$

Describes a vector from a the line defined by $(0,0,0) + ut$ to the point $v$ that is orthogonal to $u.$

$\|v - \frac {u\cdot v}{\|u\|^2} u\|$ will be your distance.

$\big(v - \frac {u\cdot v}{\|u\|^2} u\big)\cdot\big(v - \frac {u\cdot v}{\|u\|^2} u\big) = \|v\|^2 - \frac {(u\cdot v)^2}{\|u\|^2}$

Alternatively.

$d^2 = (-6+2t)^2 + (3+6t^2) + (3-t)^2\\ d^2 = 54 +6t +41t^2\\ d^2 = 41(t + \frac {3}{41})^2 - \frac {9}{41} + 54$

Distance is minimized when $t = -\frac {3}{41}$

and $d^2 = 54 - \frac 9{41}\\ d = \sqrt {54 - \frac 9{41}}$

It is worth noting that

$54 = (-6,3,3)\cdot(-6,3,3) = \|v\|^2\\ -3 = (-6,3,3)\cdot(-2,-6,1) = u\cdot v\\$

and $41 = (-2,-6,1)\cdot(-2,-6,1) = \|u\|^2$

Answer 4

You want to minimize the distance $D$, but it is easier to minimize $$ D^2 = (2t-6)^2+(6t+3)^2+(t-3)^2$$

Differentiate to get $$4(2t-6)+12(3+6t)+2(t-3)=0$$ Solve for $t$ to get $t= \frac {-3}{41}$

That gives you the point on line $$Q=(\frac {6}{41},\frac {18}{41}, \frac {-3}{41})$$