Question

For the lines x=3t, y=1-2t, z=2-3t and x (3, 1, 4) +s(-9, 6, 9) (a) Show that the lines are parallel. (b) Calculate the distance between the paralle lines.

Answer 1

Answer:

Step-by-step explanation:

Given lines in parametric form

line $L_1$

$\frac{x}{3}=\frac{y-1}{-2} =\frac{z-2}{-3}$

direction vector of $L_1 v_1=$

Line $L_2$

direction vector of $L_2 v_2=$

therefore

$v_2=-3v_1$

thus lines are parallel.

(ii)distance between two lines is

$L_2$ is given by

$\frac{x-3}{-9}=\frac{y-1}{6} =\frac{z-4}{9}$=s

$\frac{x-3}{-3}=\frac{y-1}{2} =\frac{z-4}{3}$=3s

$\left | \frac{\begin{vmatrix}x_2-x_1 &y_2-y_1 &z_2-z_1 \\ a_1&b_1&c_1 \\ a_2&b_2&c_2\end{vmatrix}}{\sqrt{\left ( a_1b_2-a_2b_1 \right )^2+\left ( b_1c_2-b_2c_1 \right )^2+\left ( c_1a_2-c_2a_1 \right )^2}}\right |$

where $a_1$=3

$b_1$=-2

$c_1$=-3

$a_2$=-9

$b_2$=6

$c_2$=9

distance(d)=0 units since value of the matrix

$\begin{vmatrix}x_2-x_1 &y_2-y_1 &z_2-z_1\\ a_1&b_1&c_1 \\a_2&b_2 &c_2 \end{vmatrix}$

is zero