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 $\displaystyle \vec a$ and $\displaystyle \vec b$ in the ratio $\displaystyle \lambda:\mu$, which is$\displaystyle \frac{\mu\vec a + \lambda\vec b}{\lambda + \mu}$

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

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

Grandad