Question

The shortest distance between lines $\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}$ and $\frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2}$ is given by

• A. $\left|\frac{\left|\begin{array}{ccc}{x}_2+x_1& y_2+y_1& z_2+z_1\\ a_1& b_1& c_1\\ a_2& b_2& c_2\end{array}\right|}{\sqrt{(b_1{c}_2-c_1{b}_2{)}^2+(c_1{a}_2-a_1{c}_2{)}^2+(a_1{b}_2-b_1{a}_2{)}^2}}\right|$
• B. $\left|\frac{\left|\begin{array}{ccc}{x}_2-x_1& y_2-y_1& z_2-z_1\\ a_1& b_1& c_1\\ a_2& b_2& c_2\end{array}\right|}{\sqrt{(b_1{c}_2-c_1{b}_2{)}^2+(c_1{a}_2-a_1{c}_2{)}^2+(a_1{b}_2-b_1{a}_2{)}^2}}\right|$
• C. $\left|\frac{\left|\begin{array}{ccc}{x}_2-x_1& y_2-y_1& z_2-z_1\\ a_1& b_1& c_1\\ a_2& b_2& c_2\end{array}\right|}{\sqrt{(b_1{c}_2-c_1{b}_2)+(c_1{a}_2-a_1{c}_2)+(a_1{b}_2-b_1{a}_2)}}\right|$
• D. $\left|\frac{\left|\begin{array}{ccc}{x}_2-x_1& y_2-y_1& z_2-z_1\\ a_1& b_1& c_1\\ a_2& b_2& c_2\end{array}\right|}{\sqrt{(b_1{c}_2+c_1{b}_2{)}^2+(c_1{a}_2+a_1{c}_2{)}^2+(a_1{b}_2+b_1{a}_2{)}^2}}\right|$

Answer 1

The correct option is B $\left|\frac{\left|\begin{array}{ccc}{x}_2-x_1& y_2-y_1& z_2-z_1\\ a_1& b_1& c_1\\ a_2& b_2& c_2\end{array}\right|}{\sqrt{(b_1{c}_2-c_1{b}_2{)}^2+(c_1{a}_2-a_1{c}_2{)}^2+(a_1{b}_2-b_1{a}_2{)}^2}}\right|$

Given lines

$\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}$------(1)

$\frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2}$------(2)

shortest distance between two line

$SD=\left|\frac{\overrightarrow{AC}\cdot (\overrightarrow{n_1}\times \overrightarrow{n_2})}{|\overrightarrow{n_1}\times \overrightarrow{n_2}|}\right|$

$AC=\overrightarrow{c}-\overrightarrow{a}$

where c and a are position vector of line 2 and 1

$AC=x_2\hat{i}+y_2\hat{j}+z_2\hat{k}-x_1\hat{i}-y_1\hat{j}-z_1\hat{k}$

$AC=(x_2-x_1)\hat{i}+(y_2-y_1)\hat{j}+(z_2-z_1)\hat{k}$

$\overrightarrow{n_1}=a_1\hat{i}+b_1\hat{j}+c_1\hat{k}$

$\overrightarrow{n_2}=a_2\hat{i}+b_2\hat{j}+c_2\hat{k}$

$\overrightarrow{n_1}\times \overrightarrow{n_2}=(a_1\hat{i}+b_1\hat{j}+c_1\hat{k})\times (a_2\hat{i}+b_2\hat{j}+c_2\hat{k})$

$\overrightarrow{n_1}\times \overrightarrow{n_2}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ a_1& b_1& c_1\\ a_2& b_2& c_2\end{array}\right|$

$\overrightarrow{n_1\times \overrightarrow{n_2}}=\hat{i}(b_1{c}_2-c_1{b}_2)-\hat{j}(c_1{a}_2-a_1{c}_2)+\hat{k}(a_1{b}_2-b_1{a}_2)$

$\left|\overrightarrow{n_1\times \overrightarrow{n_2}}\right|=\sqrt{(b_1{c}_2-c_1{b}_2{)}^2+(c_1{a}_2-a_1{c}_2{)}^2+(a_1{b}_2-b_1{a}_2{)}^2}$

from formula

$SD=\left|\frac{\overrightarrow{AC}\cdot (\overrightarrow{n_1}\times \overrightarrow{n_2})}{|\overrightarrow{n_1}\times \overrightarrow{n_2}|}\right|$

$SD=\left|\frac{\left[\overrightarrow{AC}\overrightarrow{n_1}\overrightarrow{n_2}\right]}{|\overrightarrow{n_1}\times \overrightarrow{n_2}|}\right|$

$SD=\left|\frac{\left|\begin{array}{ccc}{x}_2-x_1& y_2-y_1& z_2-z_1\\ a_1& b_1& c_1\\ a_2& b_2& c_2\end{array}\right|}{\sqrt{(b_1{c}_2-c_1{b}_2{)}^2+(c_1{a}_2-a_1{c}_2{)}^2+(a_1{b}_2-b_1{a}_2{)}^2}}\right|$