# line projection on top of a plane

If I have a horizontal line (a 3d point and 3d vector with zero z component) and another plane (could be an oblique or a horizontal; i have normal vector of the plane); then how do we get the direction (3d) of the 3d line which lie on top of the plane.

For that, I wish to project the above horizontal line on to the given plane.

(I made more clear the original post.)

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## 1 Answer

I'm not sure what a 2d vector is. I'm assuming you're specifying the line by the span of some vector $v$ translated by the 3d point $p$: $L = p + vt.$

You can specify a plane by two vectors and a point, or by a point and a vector. For the first, call the two vectors $v_1$ and $v_2$, and the point $q$. The plane is $rv_1 + sv_2 + q$.

If $p + vt$ does not intersect the plane, the projection can be written as a translation. If it does intersect the plane, pick $p$ and $q$ so that they coincide with the intersection of the line and the plane. Change coordinates so that $p=q=0$. Now all you do is project the vector $v$ onto $v_1$ and $v_2$. The projection map is

$$tv\mapsto t\langle v,v_1\rangle v_1 + t\langle v,v_2\rangle v_2.$$

If you want to work with a point $q$ and a single vector $w$ which specifies the plane by $\{x\ |\ \langle x,w\rangle = 0\} + q$, again translate coordinates to the intersection point $p=q=0$. Then project onto the span of $w$, and subtract that new line from the old line:

$$vt\mapsto vt - t\langle v,w\rangle w.$$

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I think that, with regards to the 2D vector, the OP is using the fact that the line is horizontal (i.e. it's really a 3D vector but with the third coordinate zero). – Lopsy Feb 19 '12 at 22:26
@Neal: sorry i didn't get you clearly. i know normal vector of the plane and the horizontal line (a point and direction of line). wouldn't it be taken by getting cross products of some vectors as i cannot get your previous explanation. – lenin Feb 19 '12 at 23:42
Yes, the a normal vector to a plane can be gotten by taking the cross product of two spanning vectors. – Neal Feb 19 '12 at 23:51
@lopsy That makes sense. I think of vectors and 1-forms as "1D", planes and 2-forms as "2D", and so forth, so referring to a vector by the dimension of its ambient space confused me. – Neal Feb 19 '12 at 23:53
@ Neal: No i want to know how can i get my required line (which lies on the plane) direction by getting the cross product of normal vectors? is it possible? if so, please tell me – lenin Feb 19 '12 at 23:56
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