The shortest distance between the lines x-51=y-22=z-4-3 and x+31=y+54=z-1-5 is:

Question

The shortest distance between the lines $\dfrac{x-5}{1}=\dfrac{y-2}{2}=\dfrac{z-4}{-3}$ and $\dfrac{x+3}{1}=\dfrac{y+5}{4}=\dfrac{z-1}{-5}$ is:

Accepted Answer

$$
\begin{aligned}
& \mathrm{L}_1: \frac{x-5}{1}=\frac{y-2}{2}=\frac{z-4}{-3} \\
& \overrightarrow{a_1}=5 \hat{i}+2 \hat{j}+4 \hat{k} \\
& \overrightarrow{r_1}=\hat{i}+2 \hat{j}-3 \hat{k} \\
& \mathrm{~L}_2: \frac{x+3}{1}=\frac{y+5}{4}=\frac{z-1}{-5} \\
& \overrightarrow{a_2}=-3 \hat{i}-5 \hat{j}+\hat{k} \\
& \overrightarrow{r_2}=\hat{i}+4 \hat{j}-5 \hat{k} \\
& \overrightarrow{r_1} \times \overrightarrow{r_2}=\left|\[\begin{array}{lll} \hat{i} & \hat{j} & \hat{k} \ 1 & 2 & -3 \ 1 & 4 & -5 \end{array}\end{aligned}$$\right| \\
\& =2 \hat{i}+2 \hat{j}+2 \hat{k}

\]

$$
\begin{aligned}
& \text { Shortest distance (d) }=\left|\frac{\left(\overrightarrow{r_1} \times \overrightarrow{r_2}\right) \cdot\left(\overrightarrow{a_1}-\overrightarrow{a_2}\right)}{\left|\overrightarrow{r_1} \times \overrightarrow{r_2}\right|}\right| \\
& =\frac{36}{2 \sqrt{3}}=6 \sqrt{3}
\end{aligned}
$$

Solutions · 1