Question

Find the equation of line of shortest distance between:
$\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}and\frac{x-2}{3}=\frac{y-4}{4}=\frac{z-5}{5}$.

Answer 1

$l_1:\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}:\overrightarrow{b_1}=2\hat{i}+3\hat{j}+4\hat{k}$

$l_2:\frac{x-2}{3}=\frac{y-4}{4}=\frac{z-5}{5}:\overrightarrow{b_2}=3\hat{i}+4\hat{j}+5\hat{k}$

direction of line of shortest distance is perpendicular to $l_1$ & $l_2(\overrightarrow{b_1}\times \overrightarrow{b_2})$

$\therefore \overrightarrow{b_1}\times \overrightarrow{b_2}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 3& 4\\ 3& 4& 5\end{array}\right|=\widehat{-i}+2\hat{j}+(-1)\hat{k}$

Let $\left(x_1{y}_1{z}_1\right)=(2r+1,3r+2,4r+3)$ be the point of intersection of $l_1$ and line of shortest distance.

Let $\left(x_2{y}_2{z}_2\right)=(3k+2,4k+4,5k+5)$ be the point of intersection of $l_2$ and line of shortest distance.

$\therefore l:\frac{x-(2r+1)}{-1}=\frac{y-(3r+2)}{2}=\frac{z-(4r+3)}{-1}$

$\left(x_2{y}_2{z}_2\right)$ must satisfy $l$.

$\frac{(3k+2)-(2r+1)}{-1}=\frac{(4k+4)-(3r+2)}{2}=\frac{(5k+5)-(4r+3)}{-1}$

$\therefore 3k+2-(2r+1)=5k+5-(4r+3)$

$2k-2r+1=0---\left(i\right)$

$\therefore (4k+4)-(3r+2)=-2(5k-4r+2)$

$14k-11r+5=0---\left(ii\right)$

Solving $\left(i\right)$ and $\left(ii\right)$

$k=+\frac{1}{6},r=+\frac{2}{3}$

$L:\frac{x+\frac{7}{3}}{-1}=\frac{y-0.4}{2}=\frac{z-\frac{17}{3}}{-1}$

Ans: equation of the shortest distance line is : $\frac{x+\frac{7}{3}}{-1}=\frac{y-0.4}{2}=\frac{z-\frac{17}{3}}{-1}$