Question

Find the shortest distance between the lines whose vector equations are

\(\overrightarrow r=(\hat i+2\hat j+3\hat k)+\lambda(\hat i-3\hat j+2\hat k)\)

and \(\overrightarrow r=(4\hat i+5\hat j+6\hat k)+\mu(2\hat i+3\hat j+\hat k)\)

Answer 1

The given lines are \(\overrightarrow r=(\hat i+2\hat j+3\hat k)+\lambda(\hat i-3\hat j+2\hat k)\)and \(\overrightarrow r=(4\hat i+5\hat j+6\hat k)+\mu(2\hat i+3\hat j+\hat k)\)

It is known that the shortest distance between the lines, \(\overrightarrow r=\overrightarrow a_1+\lambda b_1\) and \(\overrightarrow r=\overrightarrow a_2+\mu b_2\), is given by,
d=|(b1×b2).(a2-a2)/|b1×b2||...(1)

Comparing the given equations with \(\overrightarrow r=\overrightarrow a_1+\lambda b_1\) and \(\overrightarrow r=\overrightarrow a_2+\mu b_2\), we obtain

\(a_1=\hat i+2\hat j+3\hat k\)
\(b_1=\hat i-3\hat j+2\hat k\)
\(a_2=4\hat i+5\hat j+6\hat k\)
\(b_2=2\hat i+3\hat j+\hat k\)

\(\overrightarrow a_2-\overrightarrow a_1\)

=\((4\hat i+5\hat j+6\hat k)\)-\((\hat i+2\hat j+3\hat k)\)

=\(3\hat i+3\hat j+3\hat k\)

\(\overrightarrow b_1*\overrightarrow b_2\)

=\(\begin{vmatrix}\hat i&\hat j&\hat k\\1&-3&2\\2&3&1\end{vmatrix}\)

=(-3-6)\(\hat i\)-(1-4)\(\hat j\)+(3+6)\(\hat k\)

=-9\(\hat i\)+3\(\hat j\)+9\(\hat k\)

\(\Rightarrow \mid \overrightarrow b_1*\overrightarrow b_2\mid\)
=\(\sqrt{(-9)^2+(3)^2+(9)^2}\)
=\(\sqrt{81+9+81}\)
=\(\sqrt{171}\)
=\(3\sqrt{19}\)

\((\overrightarrow b_1*\overrightarrow b_2)\).\((\overrightarrow a_2-\overrightarrow a_1)\)
=(-9\(\hat i\)+3\(\hat j\)+9\(\hat k\)).(3\(\hat i\)+3\(\hat j\)+3\(\hat k\))
=-9×3+3×3+9×3
=9

Substituting all the values in equation (1), we obtain
d=|\(\frac{9}{3\sqrt{19}}\)|
=\(\frac{3}{\sqrt {19}}\)

Therefore, the shortest distance between the two given lines is \(\frac{3}{\sqrt {19}}\) units.