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 xi yj zk ; x i y j a i b j c k Substituting these values in ( ) and equating the coefficients of ˆ , i j and k ˆ , we get x = x + λ a ; y = y + λ b ; z = z + λ c ... ( ) These are parametric equations of the line.
Eliminating the parameter λ from ( ), we get x – x y – y z – z ... ( ) This is the Cartesian equation of the line. A Note If l , m , n are the direction cosines of the line, the equation of the line is x – x y – y z – z Example Find the vector and the Cartesian equations of the line through the point ( , , – ) and which is parallel to the vector Solution We have Fig . Therefore, the vector equation of the line is ( Now, is the position vector of any point P( x , y , z ) on the line.
Therefore, xi y j z k ( ) ( ) ( ) + − −λ Eliminating λ , we get x − = − which is the equation of the line in Cartesian form. . 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 . . Let θ be the acute angle between OP