Parametric equation of\({L_1}.\)
Substitute\(2{\rm{ for }}{x_0},3{\rm{ for }}{y_0},1{\rm{ for }}{z_0},1{\rm{ for }}a, - 2{\rm{ for }}b{\rm{,}}\)and\( - 3{\rm{ for }}c\) in equation (4),
\(\begin{array}{l}x = 2 + (1)t,y = 3 + ( - 2)t,z = 1 + ( - 3)t\\x = 2 + t,y = 3 - 2t,z = 1 - 3t\end{array}\)
The parametric equation of the line\({L_1}\)is\({\rm{x = 2 + t, y = 3 - 2 t, z = 1 - 3 t}}.\)
From equation (3), it is clear that the point and the direction vector\(\left( {{v_2}} \right)\) for line\({L_2}\)\((3, - 4,2)\)and\(\langle 1,3, - 7\rangle \)respectively.
Consider the constant term \(t\)in equation (4) as\(s\)for the calculation of parametric equations for the line\({L_2}.\)
Parametric equation of\({L_2}\)
Substitute\(3{\rm{ for }}{x_0}, - 4{\rm{ for }}{y_0},2{\rm{ for }}{z_0},1{\rm{ for }}a,3{\rm{ for }}b, - 7{\rm{ for }}c{\rm{,}}\) and\(s{\rm{ for }}t\) in equation (4),
\(\begin{array}{l}x = 3 + (1)s,y = - 4 + (3)s,z = 2 + ( - 7)s\\x = 3 + s,y = - 4 + 3s,z = 2 - 7s\end{array}\)
The parametric equation of the line\({L_2}\)is\({\rm{x = 3 + s, y = - 4 + 3 s, z = 2 - 7 s}}.\)
The parametric equations of the two lines are expressed for the lines to be intersected as follows.
\(2 + t = 3 + s\) …… (5)
\(3 - 2t = - 4 + 3s\) …… (6)
\(1 - 3t = 2 - 7s\) ...... (7)
Solve the equations (5) and (6) as follows.
Rearrange the equation (5).
\(\begin{array}{l}2 + t = 3 + s\\t = 3 + s - 2\\t = 1 + s\end{array}\)
Substitute\(1 + s\) for\(t\)in equation (6),
\(\begin{array}{l}3 - 2(1 + s) = - 4 + 3s\\3 - 2 - 2s = - 4 + 3s\\5s = 5\\s = 1\end{array}\)
In the expression\({\rm{t = 1 + s}},\)substitute\(t\) for\(s.\)
\(\begin{array}{l}t = 1 + (1)\\ = 2\end{array}\)
Substitute\(1\) for\(s\)and\(2\) for\(t\) in equation (7),
\(\begin{array}{l}1 - 3(2) = 2 - 7(1)\\1 - 6 = 2 - 7\\ - 5 = - 5\end{array}\)
From the calculations, it is clear that the two line are satisfied the condition of intersection of two lines.
Thus, the two lines are intersected lines.
As the condition of intersection is satisfied for the two line at\(s = 1\)and\(t = 2\),the point of intersection is determined either by substituting\({\rm{t}}\)value in the parametric equations of line\({L_1}\)$ or by substituting the\(s\)value in the parametric equations of line\({L_2}\).
Rewrite the parametric equations of line \({L_1}\)as follows.
\({\rm{x = 2 + t, y = 3 - 2 t, z = 1 - 3 t}}\)
In the parametric equation of line\({L_1}\), substitute\(2{\rm{ for }}t\)to find the\(x - ,y - \) and\(z{\rm{ - }}\)coordinates of intersection points of two lines.
\(\begin{array}{l}x = 2 + 2,y = 3 - 2(2),z = 1 - 3(2)\\x = 4,y = - 1,z = - 5\end{array}\)
Thus, the point of intersection of two lines\({L_1}\) and\({L_2}\)is\({\rm{(4, - 1, - 5)}}.\)
