How to find a vector orthogonal to a line

[furqankaimkha]

Hi
I have two (two dimensional) linear equations (in the plane). and i am required to find out the angle between them. I have found the solution somewhere.
the solution uses two vectors orthogal to the two lines.
The problem now i have is: How to determine those orthogonal vectors?

e.g: two equations are: 3x + y = 5 and 2x - y =4

Now the solution goes like this:

n1 = 3i + j and n2 = 2i - j.
Then the angle between them is found out by using Dot product. This is the required angle.
How did they determine n1 and n2?

 

[vela]

Look at the coefficients of x and y in an equation and compare them to the corresponding normal vector.

 

[JonF]

I’m confused on your approach.

Do you know these facts about vectors, if so it should be fairly easy:

a dot b = a₁b₁ + a₂b₂ + ...

theta = cos^-1( [a dot b]/[|a||b|])

the cross product gives you a normal vector

A line can be represented as: <x,y> = <a,b> + tv where (a,b) is on your line and v is the same direction as the line. v is easy to find: If A = (x₁, y₁) and B = (x₂,y₂) are both on your linen the v = <x₁ - x₂, y₁ - y₂>

How i would do your problem, find the direction vector of each line. Then use that property of the dot product i posted.

 

[hunt_mat]

I think that you're thinking too hard, write the equations as y=mx+c, m is the tangent of the angle that the line makes with the x-axis. So to find the angle made by both of them, take the arctan of the gradient and then just subtract these two angles to get your answer.

 

[HallsofIvy]

As Hunt_Mat says, you can rewrite 3x + y = 5 and 2x - y =4 as y= -3x+ 5 and y= 2x- 4 seeing that the slopes are -3 and 2 respectively. I would NOT, however, take the arctan of each to find the angles! Remember the simplest introduction to slope: "rise/run". Slope -3 tells you "if x increases by 1 then y decreases by 3" which corresponds to the vector <1, -3>. Similarly, slope 2 tells you "if x increases by 1 then y increases by 2" which corresponds to the vector <1, 2>.

Those are NOT the vectors given because they are vectors in the direction of the lines, not orthogonal to them. But you can use the "dot product" as you suggest: if a vector orthogonal to <1, -3> is given by <a, b> then we must have <1, -3>.<a, b>= a- 3b= 0. From that, a= 3b so if we take b= 1, a= 3- an orthogonal vector is <3, 1>. Similarly, if <a, b> is orthogonal to <1, 2> we must have <1, 2>.<a, b>= a+2b= 0 or a= -2b. Taking b= 1, a= -2 so an orthogonal vector is <-2, 1>. Those are the two vectors you give.

However, if the original problem was to find the angle between the two lines, I cannot imagine why they have chosen to use orthogonal vectors! The simplest way to do it is to just use the angle in the direction of the lines, <1, -3> and <1, 2>. $u\cdot v= |u||v|cos(\theta)$ which here gives
$$1(1)+ (-3)(2)= 1- 6= -5= \sqrt{1+ 9}\sqrt{1+ 4}cos(\theta)$$
[tex]= \sqrt{50}cos(\theta)= 5\sqrt{2} cos(\theta)[/itex].<br /> <br /> That is, $cos(\theta)= \frac{-5}{5\sqrt{2}}= -\frac{1}{\sqrt{2}}$ and it should be easy to get the angle from that. Of course, if l and m are lines, l' is perpendicular to l, m' is perpendicular to m, then the angle between l' and m' is exactly the same as the angle between l and m. The two methods give the same answer- I just don't see why one should go to the trouble of finding the orthogonal vectors when it is not necessary.[/tex]

 

[hunt_mat]

Or what you could do, is calculate the point at which they intersect, call this point (x,y), choose a>x (you are free to choose a, it could be x+1 for example), and consider the point (a,2a-4). Construct the nornal to y=2x-4 at this point and exanine where this intersects the line y=-3x+5, call this point (c,d).

What you have done is created a right angled triangle from the two lines that you want to find the angle of intersection of. let l_1 be the distance from (x,y) to (a,2a-4) (you can compute this easily) and let l_2 be the distance from (a,2a-4) and (c,d), then is theta is the angle between the two lines, you know that:
$$\tan\theta =\frac{l_{1}}{l_{2}}$$
This method is somewhat long winded but it avoids the use of vectors and is quite clear geometrically. Food for thought.

 

[furqankaimkha]

I thank all of you for your precious guidance. This topic is now very much clear to me now. Love you all.