Question

Distance between two parrallel lines,
$\overline{r}={\overline{a}}_1+\lambda \overline{b}$ and $\overline{r}={\overline{a}}_2+\mu \overline{b}$, is given by

• A. $d=|\frac{({\overline{a}}_2-{\overline{a}}_1)}{\hat{b}}|$
• B. $d=|({\overline{a}}_2-{\overline{a}}_1)\times \hat{b}|$
• C. $d=|({\overline{a}}_2+{\overline{a}}_1)\times \hat{b}|$
• D. $d=|({\overline{a}}_2-{\overline{a}}_1)|$

Answer 1

The correct option is B $d=|({\overline{a}}_2-{\overline{a}}_1)\times \hat{b}|$

given lines

$\overrightarrow{r}=\overrightarrow{a_1}+\lambda \overrightarrow{b}$

$\overrightarrow{r}=\overrightarrow{a_2}+\mu \overrightarrow{b}$

we can find out the shortest distance between will be perpendicular distance

let a line intersect both parallel line with angle $\theta$

then

$SD=|\overrightarrow{a_2}-\overrightarrow{a_1}|\sin \theta$

$SD=\frac{\left|\right(\overrightarrow{a_2}-\overrightarrow{a_1})\times \overrightarrow{b}|}{\left|\overrightarrow{b}\right|}$

$d=\left|\right(\overrightarrow{a_2}-\overrightarrow{a_1})\times \hat{b}|$