Vectors (8) •2D Vector Equations •point of intersect •3D Vector Equations •do they intersect In 2D : lines either ……. Are parallel intersect Or are the same Intersect of 2D lines in vector form - 1 For example [] [] [] x y = 1 +s 3 2 2 and [] [] [] x y 6 +t -2 = -1 4 If the lines intersect, there must be values of s and t that give the position vector of the point of intersection. x part: 1 + 2s = 6 - t y part: 3 + 2s = -2 + 4t Subtract x from y : 2 = -8 + 5t 5t = 10 t=2 Substitute: 1 + 2s = 6 - 2 2s = 3 s = 1.5 [] [] [] 1 3 = + [] [] [] [y ] = [ ] x y x y x = 1 +1.5 2 3 2 3 3 4 6 position vector of the point of intersection Intersect of 2D lines in vector form - 2 Point of intersection? r = (i + 2j) + (4i - 2j) s = (2i - 6j) + (-2i + j) i coefficients : 1 + 4 = 2 -2 j coefficients : 2 - 2 = -6 + x2 : 4 - 4 = -12 + 2 add 5 = -10 … doesn’t work Direction vectors: (4i - 2j) and (-2i + j) are parallel ….. lines are parallel Example The lines r and s have the equations ... 2 4 r 3 1 5 3 and 4 2 s 7 2 2 3 Show they intersect and find the point of intersection If the lines intersect, there must be values of and that give the position vector of the point of intersection. 2 4 4 2 3 1 7 2 5 3 2 3 x+y : 5 + 3 = 11 Substitute x : 2 + 4 = 4 +2 y : 3 - = 7 - 2 z : 5 + 3 = 2 +3 3 = 6 =2 x : 2 + 4x2 = 4 +2 =3 Check the values in the 3rd equation z : 5 + 3x2 = 2 +3x3 11 = 11 Satisfied! Hence a point exists common to both lines Example (continued) The lines r and s have the equations ... 2 4 r 3 1 5 3 and 4 2 s 7 2 2 3 Show they intersect and find the point of intersection If the lines intersect, there must be values of and that give the position vector of the point of intersection. =2 =3 Values satisfy all equations Hence a point exists common to both lines 2 4 2 8 10 Substitute r 3 2 1 3 2 1 5 3 5 6 11 The 2 lines intersect at (10, 1, 11)
© Copyright 2026 Paperzz