the Derivation of cartesian form from vector form Let the coordinates of the given point A be x y z and the direction ratios of the line be a b c Consider the coordinates of any point P be x y z Then 7 xi y zk a x i j z k and b al bj ck Substituting these values in 1 and equating the coefficients of i and k we get x x a y y b z z kc 2 These are parametric equations of the line Eliminating the parameter from 2 we get x X 2 4 b This is the Cartesian equation of the line Note If 1 m n are the direction cosines of the line the equation of the line is 2 4 11 m Example 6 Find the vector and the Cartesian equations of the line through the point 5 2 4 and which is parallel to the vector 3 2j 8k Solution We have a Si 2 4k and b 3i 2j 8k cos 8 Rationalised 2023 24 Therefore the vector equation of the line is F 57 23 4k 37 27 8k Now F is the position vector of any point P x y z on the line Therefore xi yj zk 5i 2j 4k 3 2j 8k Eliminating we get 5 3 2 2X j 4 8 k 5y 2 4 8 THREE DIMENSIONAL GEOMETRY 383 which is the equation of the line in Cartesian form 11 4 Angle between Two Lines Let L and L be two lines passing through the origin and with direction ratios a b c and a b c respectively Let P be a point on L and Q be a point on L Consider the directed lines OP and OQ as given in Fig 11 6 Let 8 be the acute angle between OP and OQ Now recall that the directed line segments OP and OQ are vectors with components a b c and a b c respectively Therefore angle 8 between them is given by aa bb cc a b c b c The angle between the lines in terms of sin 8 is given by sin 0 1 cos 0 3 a bb 56 Fig 11 4 resublished Q L