Answer:
Step-by-step explanation:
Given lines in parametric form
line [tex]L_1[/tex]
[tex]\frac{x}{3}=\frac{y-1}{-2} =\frac{z-2}{-3}[/tex]
direction vector of [tex]L_1 v_1=<3,-2,-3 >[/tex]
Line [tex]L_2[/tex]
direction vector of [tex]L_2 v_2=<-9,6,9 >[/tex]
therefore
[tex]v_2=-3v_1[/tex]
thus lines are parallel.
(ii)distance between two lines is
[tex]L_2[/tex] is given by
[tex]\frac{x-3}{-9}=\frac{y-1}{6} =\frac{z-4}{9}[/tex]=s
[tex]\frac{x-3}{-3}=\frac{y-1}{2} =\frac{z-4}{3}[/tex]=3s
[tex]\left | \frac{\begin{vmatrix}x_2-x_1 &y_2-y_1 &z_2-z_1 \\ a_1&b_1&c_1 \\ a_2&b_2&c_2\end{vmatrix}}{\sqrt{\left ( a_1b_2-a_2b_1 \right )^2+\left ( b_1c_2-b_2c_1 \right )^2+\left ( c_1a_2-c_2a_1 \right )^2}}\right |[/tex]
where [tex]a_1[/tex]=3
[tex]b_1[/tex]=-2
[tex]c_1[/tex]=-3
[tex]a_2[/tex]=-9
[tex]b_2[/tex]=6
[tex]c_2[/tex]=9
distance(d)=0 units since value of the matrix
[tex]\begin{vmatrix}x_2-x_1 &y_2-y_1 &z_2-z_1\\ a_1&b_1&c_1 \\a_2&b_2 &c_2 \end{vmatrix}[/tex]
is zero
