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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”. ( ).

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BASIS FOR VECTORS IN 3D

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  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 – f + 9k then find x & y. ********* a + b = 5i – f + 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 JKML 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

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