Finding the length from a point to a line in 3D??

Question

Here is the question:

What is the distance from the point $(4,1,-2)$ to the line given by : $$x=2+t$$

$$y=3+3t$$

$$z=4-t$$

Help would be greatly appreciate, as i do not even understand where to start. I am having a hard time understanding this work.Thanks

Answer 1

You are given $x$, $y$, and $z$ in terms of $t$. This gives you a vector $\mathbb{x}(t)$ See http://en.wikipedia.org/wiki/Distance_from_a_point_to_a_line#Vector_formulation where $\mathbb{p}$ is the point you are given.

Answer 2

Let $Q=(2,3,4), P=(4,1,-2)$, $\;\;\vec{a}=\langle1,3,-1\rangle, \;\;\;\vec{b}=\vec{QP}=\langle2,-2,-6\rangle$.

By the Pythagorean Theorem,

$\;\;\;d^2=\displaystyle|\vec{b}|^2-(\text{comp}_{\vec{a}}\vec{b})^2=\vec{b}\cdot\vec{b}-\left(\frac{\vec{a}\cdot\vec{b}}{|\vec{a}|}\right)^2=\vec{b}\cdot\vec{b}-\frac{(\vec{a}\cdot\vec{b})^2}{\vec{a}\cdot\vec{a}}$.

---

Alternatively, you can use $\displaystyle d=\frac{|\vec{a}\times\vec{b}|}{|\vec{a}|}$.

Answer 3

The distance between the given point and any point on the line is given by $$d^2(t)=(2+t-4)^2+(3+3t-1)^2+(4-t+2)^2=11t^2-4t+44.$$ This expression achieves a minimum when the first derivative vanishes, i.e. $22t-4=0$.

When $t=\frac{2}{11}$, $d(t)=\sqrt{\frac{480}{11}}$.

Answer 4

Let $\overrightarrow{r}=(1,3,-1)$, $P(4,1,-2)$ and $Q(2+t,3+3t,4-t)$, an arbitary point on the line.

$$\begin{pmatrix}1\\3\\-1\end{pmatrix}\perp\underbrace{\begin{pmatrix}t-2\\3t+2\\-t+6\end{pmatrix}}_{\overrightarrow{PQ}} \implies \begin{pmatrix}1\\3\\-1\end{pmatrix}\cdot\begin{pmatrix}t-2\\3t+2\\-t+6\end{pmatrix}=0 \implies t=\frac{2}{11}$$

The length is

$$\left|\begin{pmatrix}\frac{2}{11}-2\\3\cdot\frac{2}{11}+2\\-\frac{2}{11}+6\end{pmatrix}\right|=\left|\begin{pmatrix}-1\frac{9}{11}\\2\frac{6}{11}\\5\frac{9}{11}\end{pmatrix}\right|=\sqrt{(-1\frac{9}{11})^2+(2\frac{6}{11})^2+(5\frac{9}{11})^2}\approx6.61$$

Answer 5

The distance from a point $(x,y,z)$ on the line to the point $(4,1,-2)$ is given by:

$$d = \sqrt{(x - 4)^2 + (y - 1)^2 + (z + 2)^2} = \sqrt{(2 + t - 4)^2 + (3 + 3t - 1)^2 + (4 - t + 2)^2}$$

You want to find the minimum value of $d$. It is easier to find the minimum value of $d^2$, and then take the square root of the answer:

$$d^2 = (t - 2)^2 + (3t + 2)^2 + (6 - t)^2 = 11t^2 - 4t + 44$$

Now the parabola $at^2 + bt + c$ achieves a minimum value when $t = -b/2a$ (assuming $a > 0$). So one has in this case $t = -(-4)/(22) = 2/11$. Then:

$$d^2 = 11\left(\frac{2}{11}\right)^2 - 4\frac{2}{11} + 44 = \frac{4}{11} - \frac{8}{11} + \frac{484}{11} = \frac{480}{11}$$