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i have an object and a 3d direction vector and position for it . I would like to know how do i determine if a certain point X is in the space below the plan determined by my direction ?

Here is an image that i have drawn to make it more clear . In this image i've made the vector 2d

enter image description here

Please Help!

Regards, Aleandru Badescu

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draw a vector from x to the start of the arrow and ask what the cross product of the two vectors tells you. – anon Oct 25 '10 at 8:46
the cross product returns an vector that is perpendicular to both my first vectors. How could i use this information to solve my problem? – Alex Oct 25 '10 at 8:54
*to both vectors – Alex Oct 25 '10 at 9:00
I'm not sure I'm understanding your question: a "direction" (straight line) in the space doesn't determine a plane: you need two straight lines. – a.r. Oct 25 '10 at 9:02
i didn't quite explain right; let me tell you exactly what i want : i have a spaceship A that has a position and a direction, it can be rotated up/down, left/right. If another spaceship B is below Ai would like my spaceship A to rotate down ( so that i may move to it ) . The same think must be done if spaceship B is in A's left or right : if B on my left, rotate left, if B on right, rotate right . – Alex Oct 25 '10 at 9:12

1 Answer 1

up vote 1 down vote accepted

You have a point $x$, vector $v$, and also some vertical vector, pointing down. Lets call it $g$. If I have understood your problem correctly, your plane contains $(0,0,0)$ (or otherwise you can write $x$ in the coordinate system with origin on the plane), vector $v$, and a vector $w=v\times g$ --- horizontal vector, orthogonal to $v$. I want to interpret $x$ as a vector (from the origin to the point $x$). Then for some $\alpha\in\mathbb{R}$ vector $x-\alpha g$ is on the plane and $x$ is below the plane if and only if $\alpha>0$. We have $$(x-\alpha g, v\times w)=0,$$ $$(x-\alpha g, v\times[v\times g])=0,$$ $$(x-\alpha g, v(v,g)-g(v,v))=0,$$ $$(x, v)(v,g)-(x,g)(v,v)=\alpha\bigl((g,v)(g,v)-(g,g)(v,v)\bigr),$$ $$\alpha=\bigl((x, v)(v,g)-(x,g)(v,v)\bigr)/\bigl((g,v)(g,v)-(g,g)(v,v)\bigr).$$

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the vector v is pointing down relative to the objects rotation or parallel to the 0y axis ? – Alex Oct 25 '10 at 10:01
If Oz is the vertical axis then g=(0,0,-1) and $v$ is the given vector (in the problem you have asked you have a vector and a point). I haven't written anything about rotation. – Fiktor Oct 25 '10 at 10:05
ah yes, i got lost in my head. If i would want to find out if an object is in my right, what would i have to change? the g vector ? – Alex Oct 25 '10 at 10:11
You should find a vector $r$ pointing right and check if $(x,r)>0$. In my notation you can take $r=w=v\times g$. – Fiktor Oct 25 '10 at 10:14
Sorry, it should be $r=-w=-v\times g$ – Fiktor Oct 25 '10 at 12:19

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