Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have 2 points with bearing's coming from them, I need to calculate a 3rd point(intersection of bearing's from point 1,2) but am unsure of the maths required to do this. Could someone give me an example of how to do this.

Edit - The bearing's I am referring to are degree's from north. So I have point A + Point B and an imaginary line coming from from each point on the a particular bearing. I wish to know at what point the imaginary lines would cross.

diagram

To explain further -

I have point X,Y on a map and a bearing to an object(point 3) from point 1, I also have point 2 on the map and a bearing from point 2 to the same object (point 3) as point 1. what I need to do is to calculate the X,Y for point 3 using points 1,2. If it helps I would imagine the max distances between point 3 and points 1,2 would be a mile or so.

Maths was never my strongest point so if someone could explain how to do this in basic steps that would be great

Thanks

Colin

share|improve this question
1  
What do you mean by "bearing's"? Rays? Spheres? –  El'endia Starman May 9 '11 at 22:05
    
On a sphere/globe or on a plane? –  Isaac May 9 '11 at 23:00
1  
"bearing" is usually, I think, direction given as an angle clockwise from North. Doing the problem, however, requires knowing whether we're working on a sphere/globe or on a plane. –  Isaac May 9 '11 at 23:21
2  
Points do not intersect. Rays from points or lines through points intersect. –  Ross Millikan May 10 '11 at 0:32
1  
Colin, but you could upload a pic to e.g. imgur.com and provide a link. Someone else can then include the pic into your post for you. –  t.b. May 10 '11 at 12:17

1 Answer 1

In the plane, if you are given two points $(x_1,y_1), (x_2,y_2)$ and the angles between the vertical and the vector to a third point $(x_3,y_3)$ as $\theta_1, \theta_2$ we have the slope of the line through $(x_1,y_1)$ and $(x_3,y_3)$ is $m_1=\tan(\theta_1+\frac{\pi}{2})$ and the slope of the line through $(x_2,y_2)$ and $(x_3,y_3)$ is $m_2=\tan(\theta_2+\frac{\pi}{2})$. Then $y_3-y_1=m_1(x_3-x_1)$ and $y_3-y_2=m_2(x_3-x_2)$. This gives two equations in two unknowns.

Added in response to comment: I used the point-slope form for the two lines. Some further discussion is at PurpleMath and at Mathwords. The slopes are given by your bearings. Normally the slope of a line is the tangent of the angle measured from the horizontal, but I assumed that your bearings are measured from the vertical (as they are usually taken from North). That accounts for the addition of $\frac{\pi}{2}=90^{\circ}$. Given two lines, the intersection is found by finding a point $(x_3,y_3)$ that lies on both. This gives two simultaneous equations to solve for the two coordinates.

share|improve this answer
    
Could you explain in terms an idiot could understand thanks colin –  meee May 10 '11 at 9:18
1  
@meee Colin, may I suggest making a drawing of what you want to do; that may vastly help us to explain the mathematics you need. –  J. M. May 10 '11 at 9:49
    
@J.M. I'm pretty sure Ross Millikan interpreted it right. You have two points and two slopes given as angles from the vertical ($\pi/2$) and want the intersection between the two lines thus determined. –  Ben Alpert May 10 '11 at 14:52
    
@Ben: I know Ross did it correctly; my point was that Ross's solution might have been more transparent to Colin had Colin sketched out his situation first... –  J. M. May 10 '11 at 15:09
    
@Ross: Given that bearings are angles clockwise from North, if the $\theta_i$ are bearings, then would we want $m_i=\tan(\frac{\pi}{2}-\theta_i)$? –  Isaac May 10 '11 at 16:35

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.