Question

Find the distance between two skew lines x-1/2 = y-3/2 = z and x-2/4 = y+1/1 = z-4/8 ​

Answer 1

The distance between the two skew lines (x-1) ÷ (2) =(y-3) ÷ (2)= z

and (x-2) ÷ (4)= (y+1) ÷ (1) = (z-4) ÷ (8)  is 2.4189

Step-by-step explanation:

 The lines in three dimensional form which are not parallel and do not cross are called Skew lines. Skew lines are in different planes.

Given Data

two skew lines are (x-1) ÷ (2) =(y-3) ÷ (2)= z

and (x-2) ÷ (4)= (y+1) ÷ (1) = (z-4) ÷ (8)  

The given line equation is in the form of

(x-$x_1$) ÷ ($a_1$ ) = (y-$y_1$) ÷ ($b_1$)= (z- $z_1$) ÷ ( $c_1$ )

and (x-$x_2$) ÷ ($a_2$) = (y-$y_2$) ÷ ($b_2$ ) = (z-$z_2$) ÷ ($c_2$)

Where $x_1$=1,  $x_2$ =2, $y_1$ = 3, $y_2$ =-1, $z_1$ = 0, $z_2$ = 4, $a_1$ = 2, $a_2$  = 4,  $b_1$ = 2,  $b_2$ = 1, $c_1$ = 0,  $c_2$ = 8

Formula to find the distance between two lines is,

$=\left|\begin{array}{ccc}{x_{2}-x_{1}} & {y_{2}-y_{1}} & {z_{2}-z_{1}} \\{a_{1}} & {b_{1}} & {c_{1}} \\{a_{2}} & {b_{2}} & {c_{2}}\end{array}\right|$

÷ √ (($b_1$ $c_1$ -$b_2$ $c_2$)² + ($c_1$ $a_2$ - $c_2$$a_1$)² + ( $a_1$$b_2$ - $a_2$ $b_2$)²)

Substitute the values of x,y,z,a,b and c in their respective places

 $=\left|\begin{array}{ccc}{x_{2}-x_{1}} & {y_{2}-y_{1}} & {z_{2}-z_{1}} \\{a_{1}} & {b_{1}} & {c_{1}} \\{a_{2}} & {b_{2}} & {c_{2}}\end{array}\right| \div \sqrt{16-0)^{2}+(16-0)^{2}+(2-8)^{2}}$

$\begin{aligned}&=\left|\begin{array}{ccc}{2-1} & {-1-3} & {4-0} \\{2} & {2} & {0} \\{4} & {1} & {8}\end{array}\right| \div \sqrt{(256+(256)+(-6)^{2}\right})\\&=\left|\begin{array}{ccc}{1} & {-4} & {4} \\{2} & {2} & {0} \\{4} & {1} & {8}\end{array}\right| \div \sqrt{(256+256+24)}\end{aligned}$

=  (1(16-0)+4(16-0)+4(2-8)) ÷√((16-0)² +(16 - 0)² + (2-8)²)

=  (16+ 64-24) ÷ √(256 + (256) + (-6)²)

= ( 56) ÷ √(256 + 256 + 24)

=  (56) ÷√536

= (56) ÷ 23.151

= 2.4189

2.4189 is the distance between the two skew lines (x-1) ÷ (2) =(y-3) ÷ (2)= z

and (x-2) ÷ (4)= (y+1) ÷ (1) = (z-4) ÷ (8)  

To learn more ...

1. https://brainly.in/question/33165