- r Determine the distance from the point (2~ 'J~ 6) to the straight line parallel to the vector (3- 2- _1) and passes through the point (1-6- 4) (ON-has). Answer: The distance is ? solution :- step-I : Let the point c is ( 2, 5, 6) The vector form of c is (2 2+ 53+6k ) and let a and b is ( 1 , 6 , 4 ) and ( 3, 2 , - 1 ) vector form of a = ( 1+6 j+ 4 k ) vector form of b = (31+2]-k ) Step - II " Let 7 is the distance of the point , from the line parallel to the vector b and passing through the point a 1 ( 2- 2 ) x 5 1 1bl step - III . ( a - a ) = ( 2 2 + 5 ] + 6 k ) - ( 2 +6 ]+4k ) = (2-J+2k) step IVin ( c'- a ) xb = (2- 3+2k) * (32+ 2]-K) J K 1 -1 2 = 1 (-3 ) - J( -7 )+K (2+3 ) 3 2 - 1 7 - 36+75+5k step v : . The distance 7 = _ ( - 3) + ( 7 ) 2 + (5 ) ( 3) 2 + ( 2 ) 2 + ( - 1)2 83 . . The distance is = 14