Find the shortest distance between the lines whose vector equations are
\(\overrightarrow r=(1-t)\hat i+(t-2)\hat j+(3-2t)\hat k\) and
\(\overrightarrow r=(s+1)\hat i+(2s-1)\hat j-(2s+1)\hat k\)
Find the shortest distance between the lines whose vector equations are
\(\overrightarrow r=(1-t)\hat i+(t-2)\hat j+(3-2t)\hat k\) and
\(\overrightarrow r=(s+1)\hat i+(2s-1)\hat j-(2s+1)\hat k\)
Solution and Explanation
The given lines are
\(\overrightarrow r=(1-t)\hat i+(t-2)\hat j+(3-2t)\hat k\)
\(\Rightarrow\overrightarrow r=(\hat i-2\hat j+3\hat k)+t(-\hat i+\hat j-2\hat k)\)...(1)
\(\overrightarrow r=(s+1)\hat i+(2s-1)\hat j-(2s+1)\hat k\)
\(\Rightarrow\overrightarrow r=(\hat i-\hat j+\hat k)+s(\hat i+2\hat j-2\hat k)\)...(2)
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→||...(3)
For the given equations,
\(\overrightarrow a_1= \hat i-2\hat j+3\hat k\)
\(\overrightarrow b_1= -\hat i+\hat j-2\hat k\)
\(\overrightarrow a_2= \hat i-\hat j-\hat k\)
\(\overrightarrow b_2= \hat i+2\hat j-2\hat k\)
\(\overrightarrow a_2-\overrightarrow a_1\)
=\((\hat i-\hat j-\hat k)\)-\((\hat i-2\hat j+3\hat k)\)=\(\hat j-4\hat k\)
\(\overrightarrow b_1.\overrightarrow b_2\) = \(\begin{vmatrix}\hat i&\hat j&\hat k\\-1&1&-2\\1&2&-2\end{vmatrix}\)
=\((-2+4)\hat i-(2+2)\hat j+(-2-1)\hat k\)
=\(2\hat i-4\hat j-3\hat k\)
\(\Rightarrow \mid \overrightarrow b_1*\overrightarrow b_2\mid\)
=\(\sqrt{(2)^2+(-4)^2+(-3)^2}\)
=\(\sqrt{4+16+9}\)
=\(\sqrt{29}\)
∴(\( \overrightarrow b_1*\overrightarrow b_2\)).(\(\overrightarrow a_2-\overrightarrow a_1\))
=(\(2\hat i-4\hat j-3\hat k\)).(\(\hat j-4\hat k\)^)
=-4+12
=8
Substituting all the values in equation (3), we obtain
d=|\(\frac{8}{\sqrt{29}}\)|
=\(\frac{8}{\sqrt{29}}\)
Therefore, the shortest distance between the lines is \(\frac{8}{\sqrt{29}}\) 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.
\(d= \frac{|d_2-d_1|}{\sqrt{a^2+b^2}}\)