Vector-points-angle problem

**hobbyist** · October 8th 2009, 12:00 PM

Hi all,

I have a 2 points - 1 vector - 1 angle problem in 3D:

- There are two points, S and N.
- The x,y,z-coordinates of S are known.
- Only the z-coordinate of N is known.
- The distance between N and S is known!
- there is a vector U which is the direction of a line going through S
- the vector V is the direction vector (N-S) of the points mentioned
- the angle between vector U and V is known!

Problem: calculate the other coordinates (x,y) of point N. If anybody could help me solve it, it would be greatly appreciated!!

**Grandad** · October 9th 2009, 05:43 AM

Hello hobbyist

Welcome to Math Help Forum!

Sorry, but your question doesn't make sense. What does this mean?

Originally Posted by hobbyist

- there is a vector U which is the direction of a line going through S

Grandad

**hobbyist** · October 10th 2009, 02:51 AM

point S is on a line. The direction vector of the line is U. So the line's equation would be $S + U$ . Is it clear ? There's another line which is the line going through point S and point N. The director vector of that line is V. That line's equation is $S+V$ . So there are two lines which are at an angle with each other. The angle is known. THe only thing that is not known is the X and Y coordinaties of point S , or (the same) the x + y vector components of direction vector V. If you have Vx and Vy, then you can calculate N (which is the goal), because you know the distance between S and N.

**Grandad** · October 10th 2009, 04:47 AM

Hello hobbyist

Originally Posted by hobbyist

point S is on a line. The direction vector of the line is U. So the line's equation would be $S + U$ . Is it clear ? There's another line which is the line going through point S and point N. The director vector of that line is V. That line's equation is $S+V$ . So there are two lines which are at an angle with each other. The angle is known. THe only thing that is not known is the X and Y coordinaties of point S , or (the same) the x + y vector components of direction vector V. If you have Vx and Vy, then you can calculate N (which is the goal), because you know the distance between S and N.

Sorry, but it still doesn't make sense. The information supplied about U being the direction of a line through S tells us nothing about the point S itself - its direction will be the same no matter where it is, and so will the angle between U and V. How does this supply any more information about the point S?

(Incidentally, the vector equation of the line with direction given by the vector $\vec{u}$ that passes through the point with position vector $\vec{s}$ is $\vec{r}=\vec{s} + \lambda\vec{u}$ , not $\vec{r} = \vec{s} + \vec{u}$ .)

Grandad

**hobbyist** · October 10th 2009, 05:38 AM

Hey Grandad, thanks for your reply.

The coordinates of points S are known! U is also known, and you're right, the equation is $ \vec{r} = \vec{s} + \lambda\vec{u}$

. What else do you want to know about S? I think the information about S and U are complete ?
The goal is to calculate point N. Only the z-coordinate of N is known.

To recapulate, there are two lines:
$ \vec{r} = \vec{s} + \lambda\vec{u}$

and

$\vec{v} = \vec{s} + \mu(\vec{n}-\vec{s})$

Assumptions:

line r is known
the angle between r and v is known.
the length of vector v is known
Only the x and y components of n are not known.

Goal: calculate x and y components of vector $\vec{v}$ .

I was thinking about using two equations to solve for Nx and Ny:

1) the formula for the angle between vectors $\vec{r}$ and $\vec{v}$ :

$\alpha = \arccos (\vec{r} \cdot \vec{v} ) \div ( \mid \vec{r} \mid \ast \mid \vec{r} \mid )$

2) and the formula for the length of the vector v :

$ \mid \vec{v} \mid = Vx^2 + Vy^2 + Vz^2$

by re-arranging (2) you get the expression for Vy, which you can fill in in the formula of (1) to solve for Vx. Once you have Vx, ofcourse you can use equation (2) to find Vy. What do you think?

**hobbyist** · October 10th 2009, 06:00 AM

there's a small mistake iny my previous reply, formula 1 should be:

$ \alpha = \arccos ( (\vec{r} \cdot \vec{v}) \div (\mid \vec{r} \mid \ast \mid \vec{v} \mid)) $

you can also take the normalized versions of r and v to make it simpler:
$ \alpha = \arccos (\vec{rnorm} \cdot \vec{vnorm}) $

but then you also have to change the 2nd formula to

$ 1= vnorm_{x}^2 + vnorm_{y}^2 + vnorm_z^2 $

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