Question

Find the shortest distance between the lines
\(\overrightarrow r=(\hat i+2\hat j+\hat k)+\lambda(\hat i-\hat j+\hat k)\) and
\(\overrightarrow r=2\hat i-\hat j-\hat k+\mu\,(2\hat i+2\hat j+2\hat k)\)

Answer 1

The equations of the given lines are
\(\overrightarrow r=(\hat i+2\hat j+\hat k)+\lambda(\hat i-\hat j+\hat k)\)
\(\overrightarrow r=2\hat i-\hat j-\hat k+\mu\,(2\hat i+2\hat j+2\hat k)\)

It is known that the shortest distance between the lines,

\(\overrightarrow r= \overrightarrow a_1+\lambda\overrightarrow b_1\)and \(\overrightarrow r=\overrightarrow a_2+\mu\overrightarrow b_2\) is given by,
d=|(b1→×b2→).(a2→-a2→)/b1→×b2→|....(1)

Comparing the given equations, we obtain
\(\overrightarrow a_1\)=\(\hat i+2\hat j+\hat k\)
\(\overrightarrow b_1\)=\(\hat i-\hat j+\hat k\)

\(\overrightarrow a_2\)=\(2\hat i-\hat j-\hat k\)
\(\overrightarrow b_2\)=\(2\hat i+\hat j+2\hat k\)

\(\overrightarrow a_2\)-\(\overrightarrow a_1\)=(\(2\hat i-\hat j-\hat k\))-(\(\hat i+2\hat j+\hat k\))=\(\hat i-3\hat j-2\hat k\)

\(\overrightarrow b_1*\overrightarrow b_2= \begin{vmatrix}\hat i&\hat j&\hat k\\1&-1&1\\2&1&2\end{vmatrix}\)

\(\overrightarrow b_1*\overrightarrow b_2\)=\((-2-1)\hat i-(2-2)\hat j+(1+2)\hat k=3\hat i+3\hat k\)

\(\Rightarrow\) |\(\overrightarrow b_1*\overrightarrow b_2\)|

=\(\sqrt{(-3)^2+(3)^2}\)
=\(\sqrt{9+9}\)
=\(\sqrt{18}\)
=3\(\sqrt2\)

Substituting all the values in equation(1), we obtain
d=\(\begin{vmatrix}\frac{(-3\hat i+3\hat k).(\hat i-3\hat j-2\hat k)}{3\sqrt2}\end{vmatrix}\)

\(\Rightarrow  d=\begin{vmatrix}\frac{(-3.1+3(-2))}{3\sqrt2}\end{vmatrix}\)

\(\Rightarrow d=\begin{vmatrix}\frac{-9}{3\sqrt2}\end{vmatrix}\)

\(\Rightarrow d=\frac{3}{2}\)

=\(\frac{3\sqrt2}{2}\)

Therefore, the shortest distance between the two lines is \(\frac{3\sqrt2}{2}\) units.