Find the shortest distance between the following given lines l_1 … - Vidyadip

Question

Find the shortest distance between the following given lines $l_1$ and $l_2$
$
\begin{array}{l}
\vec{r}=\hat{i}+2 \hat{j}-4 \hat{k}+\lambda(2 \hat{i}+3 \hat{j}+6 \hat{k}) \\
\text { and } \vec{r}=3 \hat{i}+3 \hat{j}-5 \hat{k}+\lambda(2 \hat{i}+3 \hat{j}+6 \hat{k})
\end{array}
$

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Answer

Both the lines are parallel, because
$\overrightarrow{a_1}=\hat{i}+2 \hat{j}-4 \hat{k}, \quad \overrightarrow{a_2}=3 \hat{i}+3 \hat{j}-5 \hat{k}$ and
$\overrightarrow{b_1}=2 \hat{i}+3 \hat{j}+6 \hat{k}=\vec{b}_2$
Hence the distance between the parallel lines
$
\begin{array}{l}
d=\left|\frac{\vec{b} \times\left(\vec{a}_2-\vec{a}_1\right)}{|\vec{b}|}\right| \\
\vec{a}_2-\vec{a}_1=(3 \hat{i}+3 \hat{j}-5 \hat{k})-(\hat{i}+2 \hat{j}-4 \hat{k})=2 \hat{i}+\hat{j}-\hat{k} \\
\therefore \vec{b} \times\left(\vec{a}_2-\vec{a}_1\right)=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
2 & 3 & 6 \\
2 & 1 & -1
\end{array}\right|=(-3-6) \hat{i} \\
-(-2-12) \hat{j}+(2-6) \hat{k} \\
=-9 \hat{i}+14 \hat{j}-4 \hat{k} \\
|\vec{b}|=\sqrt{(2)^2+(3)^2+(6)^2} \\
=\sqrt{4+9+36}=\sqrt{49}=7 \\
\text { Hence }\left|\frac{\vec{b} \times\left(\vec{a}_2-\vec{a}_1\right)}{|\vec{b}|}\right|=\frac{\sqrt{(-9)^2+(14)^2+(-4)^2}}{7} \\
=\frac{\sqrt{81+196+16}}{7}=\frac{\sqrt{293}}{7} \text { }
\end{array}
$

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