´A´, ´B´, ´C´ seien Punkte im ´RR^2´ mit den Koordinaten ´A(12,9)´, ´B(9,40)´ und ´C(0,0)´ (Koordinatenursprung). Berechne (näherungsweise) die Länge der Strecke ´|AB|´, den Winkel ´/_ ACB´ und den Flächeninhalt des Dreiecks /_\ ´ABC´.

Approach

´vec a = vec(CA)´ ´vec b = vec(CB)´ ´vec(AB) = vec b - vec a = ((9),(40)) - ((12),(9)) = ((-3),(31))´

´|vec(AB)| = sqrt((-3)^2 + 31^2) = sqrt(970)´

´a = sqrt(12^2 + 9^2) = 15´ ´b = sqrt(9^2 + 40^2) = 41´

´cos(phi) = (vec a @ vec b)/(a * b) = (9 * 12 + 40 * 9)/(15 * 41) = 468/615 = 156/205´

´phi = 40.45˚´

´A = 1/2 * |12 * 40 - 9 * 9| = 399/2 = 199.5´


Solution

  • ´|AB| = sqrt(970)´

    ´phi = 40.45˚´

    ´A = 199.5´

  • URL:
  • Language:
  • Subjects: math
  • Type: Calculate
  • Duration: 25min
  • Credits: 3
  • Difficulty: 0.4
  • Tags: hpi vector
  • Note:
    HPI, 2014-06-30, Mathe 2, Aufgabe 52
  • Created By: adius
  • Created At:
    2014-07-26 21:24:36 UTC
  • Last Modified:
    2014-07-26 21:24:36 UTC