Matlab – Angle between two vectors in Matlab

geometrymathMATLAB

I have two points, a and b. I need to calculate the angle between them, so I treat them like vectors. However vector a will always be defined as [0 0 0]. Reading over the MATLAB Newsreader, "Angle between two vectors", three solutions are provided:

x1 = 0;
y1 = 0;
z1 = 0;
x2 = 0;
y2 = 1;
z2 = 0;
a = [x1,y1,z1]; b= [x2,y2,z2];

theta = rad2deg(atan2(norm(cross(a,b)),dot(a,b)))
theta = rad2deg(acos(dot(a,b)))
theta = rad2deg(atan2(x1*y2-x2*y1,x1*x2+y1*y2))

However as acos has accuracy problems as theta nears zero, yet out of the three equations, only acos provides the correct solution.

Should I continue to use acos or is there a better solution?

Best Answer

A vector has magnitude and direction, while a and b are just coordinate points in space. When you treat a and b as vectors, you are implicitly defining [0 0 0] as the origin point for the two vectors. However, since point a is at [0 0 0], then it will be a vector with zero length.

If a vector has zero length, which direction does it point in? The answer is nowhere. It doesn't point in any direction, and thus you can't find the angle between it and another vector.

I think maybe you've defined your problem poorly. Does your coordinate system have an origin other than [0 0 0]? Are you actually trying to find the angle between the line formed by a and b and the x-y plane?

Related Solutions

Calculate distance between two latitude-longitude points? (Haversine formula)

This link might be helpful to you, as it details the use of the Haversine formula to calculate the distance.

Excerpt:

This script [in Javascript] calculates great-circle distances between the two points – that is, the shortest distance over the earth’s surface – using the ‘Haversine’ formula.

function getDistanceFromLatLonInKm(lat1,lon1,lat2,lon2) {
  var R = 6371; // Radius of the earth in km
  var dLat = deg2rad(lat2-lat1);  // deg2rad below
  var dLon = deg2rad(lon2-lon1); 
  var a = 
    Math.sin(dLat/2) * Math.sin(dLat/2) +
    Math.cos(deg2rad(lat1)) * Math.cos(deg2rad(lat2)) * 
    Math.sin(dLon/2) * Math.sin(dLon/2)
    ; 
  var c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1-a)); 
  var d = R * c; // Distance in km
  return d;
}

function deg2rad(deg) {
  return deg * (Math.PI/180)
}

Determine Whether Two Date Ranges Overlap

(StartA <= EndB) and (EndA >= StartB)

Proof:
Let ConditionA Mean that DateRange A Completely After DateRange B

_                        |---- DateRange A ------|
|---Date Range B -----|                          _

(True if StartA > EndB)

Let ConditionB Mean that DateRange A is Completely Before DateRange B

|---- DateRange A -----|                        _ 
_                          |---Date Range B ----|

(True if EndA < StartB)

Then Overlap exists if Neither A Nor B is true -
(If one range is neither completely after the other,
nor completely before the other, then they must overlap.)

Now one of De Morgan's laws says that:

Not (A Or B) <=> Not A And Not B

Which translates to: (StartA <= EndB) and (EndA >= StartB)

NOTE: This includes conditions where the edges overlap exactly. If you wish to exclude that,
change the >= operators to >, and <= to <

NOTE2. Thanks to @Baodad, see this blog, the actual overlap is least of:
{ endA-startA, endA - startB, endB-startA, endB - startB }

(StartA <= EndB) and (EndA >= StartB) (StartA <= EndB) and (StartB <= EndA)

NOTE3. Thanks to @tomosius, a shorter version reads:
DateRangesOverlap = max(start1, start2) < min(end1, end2)
This is actually a syntactical shortcut for what is a longer implementation, which includes extra checks to verify that the start dates are on or before the endDates. Deriving this from above:

If start and end dates can be out of order, i.e., if it is possible that startA > endA or startB > endB, then you also have to check that they are in order, so that means you have to add two additional validity rules:
(StartA <= EndB) and (StartB <= EndA) and (StartA <= EndA) and (StartB <= EndB) or:
(StartA <= EndB) and (StartA <= EndA) and (StartB <= EndA) and (StartB <= EndB) or,
(StartA <= Min(EndA, EndB) and (StartB <= Min(EndA, EndB)) or:
(Max(StartA, StartB) <= Min(EndA, EndB)

But to implement Min() and Max(), you have to code, (using C ternary for terseness),:
(StartA > StartB? Start A: StartB) <= (EndA < EndB? EndA: EndB)

Best Answer

Related Solutions

Calculate distance between two latitude-longitude points? (Haversine formula)

Determine Whether Two Date Ranges Overlap

Related Topic