Question
Prove The Theorem : The distance between parallel lines $\bar{r}=\bar{a}_1+\lambda \bar{b}$ and $\bar{r}=\bar{a}_2+\lambda \bar{b}$ is $\left|\left(\bar{a}_2-\bar{a}_1\right) \times \hat{b}\right|$
✓
Answer
Let lines represented by $\bar{r}=\bar{a}_1+\lambda \bar{b}$ and $\bar{r}=\bar{a}_2+\lambda \bar{b}$ be $l_1$ and $l_2$
Lines $\mathrm{L}_1$ and $\mathrm{L}_2$ pass through $\mathrm{A}\left(\bar{a}_1\right)$ and $\mathrm{B}\left(\overline{a_2}\right)$ respectively.
Let $\mathrm{BM}$ be perpendicular to $\mathrm{L}_1$. To find $\mathrm{BM}$
$\triangle A M B$ is a right angle triangle. Let $\mathrm{m} \angle B A M=\theta$
$
\begin{aligned}
& \sin \theta=\frac{B M}{A B} \\
& \therefore \quad \mathrm{BM}=\mathrm{AB} \sin \theta=\mathrm{AB} \cdot 1 \sin \theta=\mathrm{AB} \cdot|\hat{b}| \cdot \sin \theta \\
& |\overline{A B} \times \hat{b}|=\left|\left(\bar{a}_2-\bar{a}_1\right) \times \hat{b}\right|
\end{aligned}
$
$\therefore \quad$ The distance between parallel lines $\bar{r}=\bar{a}_1+\lambda \bar{b}$ and $\bar{r}=\bar{a}_2+\lambda_1 \bar{b}$ is given by
$
d=B M=\left|\left(\bar{a}_2-\bar{a}_1\right) \times \hat{b}\right|
$
Lines $\mathrm{L}_1$ and $\mathrm{L}_2$ pass through $\mathrm{A}\left(\bar{a}_1\right)$ and $\mathrm{B}\left(\overline{a_2}\right)$ respectively.
Let $\mathrm{BM}$ be perpendicular to $\mathrm{L}_1$. To find $\mathrm{BM}$
$\triangle A M B$ is a right angle triangle. Let $\mathrm{m} \angle B A M=\theta$
$
\begin{aligned}
& \sin \theta=\frac{B M}{A B} \\
& \therefore \quad \mathrm{BM}=\mathrm{AB} \sin \theta=\mathrm{AB} \cdot 1 \sin \theta=\mathrm{AB} \cdot|\hat{b}| \cdot \sin \theta \\
& |\overline{A B} \times \hat{b}|=\left|\left(\bar{a}_2-\bar{a}_1\right) \times \hat{b}\right|
\end{aligned}
$
$\therefore \quad$ The distance between parallel lines $\bar{r}=\bar{a}_1+\lambda \bar{b}$ and $\bar{r}=\bar{a}_2+\lambda_1 \bar{b}$ is given by
$
d=B M=\left|\left(\bar{a}_2-\bar{a}_1\right) \times \hat{b}\right|
$
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