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

Explanation:

Given lines in parametric form

line
`L_1`

`(x)/(3)=(y-1)/(-2) =(z-2)/(-3)`

direction vector of
`L_1 v_1=<3,-2,-3 >`

Line
`L_2`

direction vector of
`L_2 v_2=<-9,6,9 >`

therefore

`v_2=-3v_1`

thus lines are parallel.

(ii)distance between two lines is

`L_2` is given by

`(x-3)/(-9)=(y-1)/(6) =(z-4)/(9)`=s

`(x-3)/(-3)=(y-1)/(2) =(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}}{√(\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