1 / 5

BASIS FOR VECTORS IN 3D

BASIS FOR VECTORS IN 3D. Suppose that. and k = ( ). i = ( ). j = ( ). 0. 1. 0. 0. 0. 1. 1. 0. 0. then every vector in 3D can be expressed in terms of i , j and k. So we say that i , j and k “form a basis for vectors in 3D”. ( ).

preston
Download Presentation

BASIS FOR VECTORS IN 3D

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. BASIS FOR VECTORS IN 3D Suppose that and k = ( ) i = ( ) j = ( ) 0 1 0 0 0 1 1 0 0 then every vector in 3D can be expressed in terms of i , j and k. So we say that i , j and k “form a basis for vectors in 3D”. ( ) = ( ) + ( ) + ( ) = ( ) + 2( ) + 3( ) 1 1 0 0 1 0 0 eg 2 0 2 0 0 1 0 3 0 0 3 0 0 1 = i + 2j + 3k conversely 5( ) - 2( ) + 4( ) = ( ) - ( ) + ( ) = ( ) 1 0 0 5 0 0 5 5i - 2j + 4k = 0 1 0 0 2 0 -2 0 0 1 0 0 4 4

  2. No need for middle steps. Simply…. ( ) = ( ) = 8 0 8i – j + 3k 4j - 2k -1 4 3 -2 while ( ) ( ) 6 -1 6i + 5j – k = 3k – i = 5 0 -1 3

  3. RULES IN 3D The rules in 3D are exactly the same as in 2D but we have one extra component. Ex (+/-) Suppose that a = 2i + 3j + yk , b = 3i + xj + 4k and a + b = 5i – j + 9k then find x & y. ********* a + b = 5i – j + 9k so we have ( ) + ( ) = ( ) 2 3 5 3 x -1 x + 3 = -1 or x = -4 y 4 9 ( ) = ( ) 5 5 y + 4 = 9 or y = 5 x+3 -1 y+4 9

  4. Parallel Lines Ex S is (5,1,-3), T is (9,4,7) , U is (-3,7,-4) and V is (1,10,6). Prove that STVU is a parallelogram. NB: order of letters. ************* T V Since ST = UV and TV = SU S U ( ) - ( ) = ( ) 9 5 4 ST = t – s = 4 1 3 7 -3 10 then the opposite sides are equal and parallel so STVU must be a parallelogram. = ( ) ( ) - ( ) 4 1 -3 3 UV = v – u = 10 7 10 6 -4 ( ) - ( ) = ( ) 1 9 -8 TV = v – t = 10 4 6 6 7 -1 = ( ) ( ) - ( ) -8 -3 5 SU = u – s = 6 7 1 -1 -4 -3

  5. Ex JKLM is a parallelogram in which J is (3,5,-2) , K is (7,-1,8) and M is (5,5,-8). Find the coordinates of L. ************* J K Opposite sides are equal and parallel so JK = ML k - j = l - m M L l = k - j + m = ( ) - ( ) + ( ) 7 3 5 -1 5 5 8 -2 -8 So L is (9,-1,2) = ( ) 9 -1 2

More Related