8
$\begingroup$

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.

$\endgroup$
13
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
5
$\begingroup$

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 $(a+6,b-3,c-3) \cdot (-2,-6,1)=0$.

| cite | improve this answer | |
$\endgroup$
2
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

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 $$Q=(\frac {6}{41},\frac {18}{41}, \frac {-3}{41})$$

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.