# Thread: Vector location in 3D space

1. ## Vector location in 3D space

I'm getting brain fried trying to work out simple vector positions.

I have two points in 3D space. In order x,y,z let's say one them is at 5,0,5 and the other at 6,5,0

I need to know the coordinates of the point that lies 1m (from the first point) along the line that stretches between those two points. know the length of that line via pythagoras, and I know with enough pythagoras and SOH CAH TOA I could work out the coordinates, but it would involve dozens of steps and I'm likely to get confused.

So....since I know little about vectors, how do I derive the coordinates of a point that is X units (in my case 1) along a vector linking two 3D vertices?

I'm hoping it's easy. Thanks

2. Originally Posted by jt7747
I'm getting brain fried trying to work out simple vector positions.

I have two points in 3D space. In order x,y,z let's say one them is at 5,0,5 and the other at 6,5,0

I need to know the coordinates of the point that lies 1m (from the first point) along the line that stretches between those two points. know the length of that line via pythagoras, and I know with enough pythagoras and SOH CAH TOA I could work out the coordinates, but it would involve dozens of steps and I'm likely to get confused.

So....since I know little about vectors, how do I derive the coordinates of a point that is X units (in my case 1) along a vector linking two 3D vertices?

I'm hoping it's easy. Thanks
If you want the coordinates that are one unit away from $(5, 0, 5)$ in the direction of $(6, 5, 0)$, first work out the unit vector that goes in that direction...

The vector that goes in that direction is

$\mathbf{x} = (6 - 5)\mathbf{i} + (5 - 0)\mathbf{j} + (0 - 5)\mathbf{k}$

$= \mathbf{i} + 5\mathbf{j} - 5\mathbf{k}$.

Now find the unit vector in that direction - so divide it by its length...

The unit vector is

$\frac{\mathbf{x}}{|\mathbf{x}|} = \frac{\mathbf{i} + 5\mathbf{j} - 5\mathbf{k}}{\sqrt{1^2 + 5^2 + (-5)^2}}$

$= \frac{\mathbf{i} + 5\mathbf{j} - 5\mathbf{k}}{\sqrt{1 + 25 + 25}}$

$= \frac{\mathbf{i} + 5\mathbf{j} - 5\mathbf{k}}{\sqrt{51}}$

$= \frac{\sqrt{51}}{51}\mathbf{i} + \frac{5\sqrt{51}}{51}\mathbf{j} - \frac{5\sqrt{51}}{51}\mathbf{k}$.

So that means that to find the coordinates of the point exactly one unit away from $(5, 0, 5)$ in the direction of $(6, 5, 0)$, you need to add $\frac{\sqrt{51}}{51}$ in the $x$ direction, $\frac{5\sqrt{51}}{51}$ in the $y$ direction, and subtract $\frac{5\sqrt{51}}{51}$ in the $z$ direction - in other words, add this vector to the starting point.

If you were trying to find the coordinates of any point in this line segment, multiply the unit vector by this length first, then add this resultant vector to your point.

3. Hello jt7747

Welcome to Math Help Forum!
Originally Posted by jt7747
I'm getting brain fried trying to work out simple vector positions.

I have two points in 3D space. In order x,y,z let's say one them is at 5,0,5 and the other at 6,5,0

I need to know the coordinates of the point that lies 1m (from the first point) along the line that stretches between those two points. know the length of that line via pythagoras, and I know with enough pythagoras and SOH CAH TOA I could work out the coordinates, but it would involve dozens of steps and I'm likely to get confused.

So....since I know little about vectors, how do I derive the coordinates of a point that is X units (in my case 1) along a vector linking two 3D vertices?

I'm hoping it's easy. Thanks
You'll need to use the formula for the position vector of a point that divides the line joining the points with position vectors $\vec a$ and $\vec b$ in the ratio $\lambda:\mu$, which is
$\frac{\mu\vec a + \lambda\vec b}{\lambda + \mu}$
I assume that the units you're using are metres, so by 1 m away from the first point, you mean 1 unit.

The distance between the points is $\sqrt{51}$. So we need the point that divides the line joining $\begin{pmatrix}5\\0\\5\end{pmatrix}$ to $\begin{pmatrix}6\\5\\0\end{pmatrix}$ in the ratio $1:\sqrt{51}-1$, which is:
$\dfrac{(\sqrt{51}-1)\begin{pmatrix}5\\0\\5\end{pmatrix}+1\begin{pmat rix}6\\5\\0\end{pmatrix}}{\sqrt{51}}$
$=\dfrac{1}{\sqrt{51}}\begin{pmatrix}5\sqrt{51}+1\\ 5\\5\sqrt{51}-5\end{pmatrix}$

$=\begin{pmatrix}5+\frac{1}{\sqrt{51}}\\ \frac{5}{\sqrt{51}}\\5-\frac{5}{\sqrt{51}}\end{pmatrix}$