{"id":1106,"date":"2019-10-19T05:54:14","date_gmt":"2019-10-19T05:54:14","guid":{"rendered":"https:\/\/muthu.co\/?p=1106"},"modified":"2021-05-24T02:42:08","modified_gmt":"2021-05-24T02:42:08","slug":"using-the-law-of-cosines-and-vector-dot-product-formula-to-find-the-angle-between-three-points","status":"publish","type":"post","link":"http:\/\/write.muthu.co\/using-the-law-of-cosines-and-vector-dot-product-formula-to-find-the-angle-between-three-points\/","title":{"rendered":"Using the law of cosines and vector dot product formula to find the angle between three points"},"content":{"rendered":"\n

For any 3 points A, B, and C on a cartesian plane. If we have to find the angle between these points, there are many ways we can do that. In this article I will talk about the two frequently used methods:<\/p>\n\n\n\n

  1. The Law of Cosines formula<\/li>
  2. Vector Dot product formula<\/li><\/ol>\n\n\n\n

    Law of Cosines<\/h2>\n\n\n\n

    For any given triangle ABC with sides AB, BC and AC, the angle formed by the lines AB and BC is given by the formula:<\/p>\n\n\n\n

    \"theta<\/a><\/figure><\/div>\n\n\n\n

    Here is how we can derive that formula:<\/p>\n\n\n\n

    \"triangle\"<\/a><\/figure><\/div>\n\n\n\n
    \"law<\/a><\/figure><\/div>\n\n\n\n

    The same can be extended for other angles as well.<\/p>\n\n\n\n

    \"angles<\/a><\/figure><\/div>\n\n\n\n

    Now, if we were given 3 points on a cartesian plane, we can find the distance between any two points using the Euclidean distance formula:<\/p>\n\n\n\n

    \"Euclidean<\/a><\/figure><\/div>\n\n\n\n

    So, to find the angle between three points A(x1,y1),<\/em> B(x2,y2)<\/em> and C(x3,y3), <\/em>our formula becomes:<\/p>\n\n\n\n

    \"angle<\/a><\/figure><\/div>\n\n\n\n

    In Python we can represent the above formula using the code:<\/p>\n\n\n\n

    import numpy as np\nimport math\n\ndef angle_between_three_points(pointA, pointB, pointC):\n    \n    x1x2s = math.pow((pointA[0] - pointB[0]),2)\n    x1x3s = math.pow((pointA[0] - pointC[0]),2)\n    x2x3s = math.pow((pointB[0] - pointC[0]),2)\n    \n    y1y2s = math.pow((pointA[1] - pointB[1]),2)\n    y1y3s = math.pow((pointA[1] - pointC[1]),2)\n    y2y3s = math.pow((pointB[1] - pointC[1]),2)\n\n    cosine_angle = np.arccos((x1x2s + y1y2s + x2x3s + y2y3s - x1x3s - y1y3s)\/(2*math.sqrt(x1x2s + y1y2s)*math.sqrt(x2x3s + y2y3s)))\n\n    return np.degrees(cosine_angle)\n\nA = np.array([2,4])\nB = np.array([8,7])\nC = np.array([9,1])\n\nprint(\"Angle between points:\", angle_between_three_points(A,B,C))\n# gives 72.89\n\nA = np.array([0,0])\nB = np.array([0,4])\nC = np.array([4,0])\n\n\nprint(\"Angle between points:\", angle_between_three_points(A,B,C))\n# gives 45.0<\/code><\/pre>\n\n\n\n
    Vector Dot Product<\/h2>\n\n\n\n
    The three points can also be represented as a vector as shown below.<\/p>\n\n\n\n
    \"vectors<\/a><\/figure><\/div>\n\n\n\n
    where,<\/p>\n\n\n\n
    \"vectors\"<\/a><\/figure><\/div>\n\n\n\n
    The scalar product or the dot product of two vectors is represented by the formula:<\/p>\n\n\n\n
    \"vector<\/a><\/figure><\/div>\n\n\n\n
    where || * || is the magnitude of the vector and \u03b8 is the angle made by the two vectors.<\/p>\n\n\n\n
    From the above formula we can represent the angle using the formula:<\/p>\n\n\n\n
    \"Cosines<\/a><\/figure><\/div>\n\n\n\n
    In Python we can represent the above formula using the code:<\/p>\n\n\n\n
    def angle_between_three_points(pointA, pointB, pointC):\n    BA = pointA - pointB\n    BC = pointC - pointB\n\n    try:\n        cosine_angle = np.dot(BA, BC) \/ (np.linalg.norm(BA) * np.linalg.norm(BC))\n        angle = np.arccos(cosine_angle)\n    except:\n        print(\"exc\")\n        raise Exception('invalid cosine')\n\n    return np.degrees(angle)<\/code><\/pre>\n\n\n\n
    The above is code from my post<\/a> where I talked about finding the collinearity of three points on a plane using the angle formed by the points.<\/p>\n\n\n\n
    References:<\/p>\n\n\n\n