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 \(\overrightarrow r=(4\hat i+5\hat j+6\hat k)+\mu(2\hat i+3\hat j+\hat k)\)
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 \(\overrightarrow r=(4\hat i+5\hat j+6\hat k)+\mu(2\hat i+3\hat j+\hat k)\)
Solution and Explanation
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.
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Concepts Used:
Shortest Distance Between Two Parallel Lines
Formula to find distance between two parallel line:
Consider two parallel lines are shown in the following form :
\(y = mx + c_1\) …(i)
\(y = mx + c_2\) ….(ii)
Here, m = slope of line
Then, the formula for shortest distance can be written as given below:
\(d= \frac{|c_2-c_1|}{\sqrt{1+m^2}}\)
If the equations of two parallel lines are demonstrated in the following way :
\(ax + by + d_1 = 0\)
\(ax + by + d_2 = 0\)
then there is a little change in the formula.
\(d= \frac{|d_2-d_1|}{\sqrt{a^2+b^2}}\)