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Question

# Find the shortest distance between the following pairs of parallel lines whose equations are: (i) $\stackrel{\to }{r}=\left(\stackrel{^}{i}+2\stackrel{^}{j}+3\stackrel{^}{k}\right)+\lambda \left(\stackrel{^}{i}-\stackrel{^}{j}+\stackrel{^}{k}\right)\mathrm{and}\stackrel{\to }{r}=\left(2\stackrel{^}{i}-\stackrel{^}{j}-\stackrel{^}{k}\right)+\mu \left(-\stackrel{^}{i}+\stackrel{^}{j}-\stackrel{^}{k}\right)$ (ii) $\stackrel{\to }{r}=\left(\stackrel{^}{i}+\stackrel{^}{j}\right)+\lambda \left(2\stackrel{^}{i}-\stackrel{^}{j}+\stackrel{^}{k}\right)\mathrm{and}\stackrel{\to }{r}=\left(2\stackrel{^}{i}+\stackrel{^}{j}-\stackrel{^}{k}\right)+\mu \left(4\stackrel{^}{i}-2\stackrel{^}{j}+2\stackrel{^}{k}\right)$

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Solution

## (i) The vector equations of the given lines are $\stackrel{\to }{r}=\left(\stackrel{^}{i}+2\stackrel{^}{j}+3\stackrel{^}{k}\right)+\lambda \left(\stackrel{^}{i}-\stackrel{^}{j}+\stackrel{^}{k}\right)...\left(1\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\stackrel{\to }{r}=\left(2\stackrel{^}{i}-\stackrel{^}{j}-\stackrel{^}{k}\right)+\mu \left(-\stackrel{^}{i}+\stackrel{^}{j}-\stackrel{^}{k}\right)\phantom{\rule{0ex}{0ex}}=\left(2\stackrel{^}{i}-\stackrel{^}{j}-\stackrel{^}{k}\right)-\mu \left(\stackrel{^}{i}-\stackrel{^}{j}+\stackrel{^}{k}\right)...\left(2\right)$ These two lines pass through the points having position vectors $\stackrel{\to }{{a}_{1}}=\stackrel{^}{i}+2\stackrel{^}{j}+3\stackrel{^}{k}\mathrm{and}\stackrel{\to }{{a}_{2}}=2\stackrel{^}{i}-\stackrel{^}{j}-\stackrel{^}{k}$ and are parallel to the vector $\stackrel{\to }{b}=\stackrel{^}{i}-\stackrel{^}{j}+\stackrel{^}{k}$. Now, $\stackrel{\to }{{a}_{2}}-\stackrel{\to }{{a}_{1}}=\stackrel{^}{i}-3\stackrel{^}{j}-4\stackrel{^}{k}$ and $\left(\stackrel{\to }{{a}_{2}}-\stackrel{\to }{{a}_{1}}\right)×\stackrel{\to }{b}=\left(\stackrel{^}{i}-3\stackrel{^}{j}-4\stackrel{^}{k}\right)×\left(\stackrel{^}{i}-\stackrel{^}{j}+\stackrel{^}{k}\right)\phantom{\rule{0ex}{0ex}}=\left|\begin{array}{ccc}\stackrel{^}{i}& \stackrel{^}{j}& \stackrel{^}{k}\\ 1& -3& -4\\ 1& -1& 1\end{array}\right|\phantom{\rule{0ex}{0ex}}=-7\stackrel{^}{i}-5\stackrel{^}{j}+2\stackrel{^}{k}\phantom{\rule{0ex}{0ex}}⇒\left|\left(\stackrel{\to }{{a}_{2}}-\stackrel{\to }{{a}_{1}}\right)×\stackrel{\to }{b}\right|=\sqrt{{\left(-7\right)}^{2}+{\left(-5\right)}^{2}+{2}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{49+25+4}\phantom{\rule{0ex}{0ex}}=\sqrt{78}\phantom{\rule{0ex}{0ex}}$ The shortest distance between the two lines is given by $\frac{\left|\left(\stackrel{\to }{{a}_{2}}-\stackrel{\to }{{a}_{1}}\right)×\stackrel{\to }{b}\right|}{\left|\stackrel{\to }{b}\right|}=\frac{\sqrt{78}}{\sqrt{3}}=\sqrt{26}$ (ii) $\stackrel{\to }{r}=\left(\stackrel{^}{i}+\stackrel{^}{j}\right)+\lambda \left(2\stackrel{^}{i}-\stackrel{^}{j}+\stackrel{^}{k}\right)\mathrm{and}\stackrel{\to }{r}=\left(2\stackrel{^}{i}+\stackrel{^}{j}-\stackrel{^}{k}\right)+\mu \left(4\stackrel{^}{i}-2\stackrel{^}{j}+2\stackrel{^}{k}\right)\mathrm{or}\stackrel{\to }{r}=\left(2\stackrel{^}{i}+\stackrel{^}{j}-\stackrel{^}{k}\right)+2\mu \left(2\stackrel{^}{i}-\stackrel{^}{j}+\stackrel{^}{k}\right)$ These two lines pass through the points having position vectors $\stackrel{\to }{{a}_{1}}=\stackrel{^}{i}+\stackrel{^}{j}\mathrm{and}\stackrel{\to }{{a}_{2}}=2\stackrel{^}{i}+\stackrel{^}{j}-\stackrel{^}{k}$ and are parallel to the vector $\stackrel{\to }{b}=2\stackrel{^}{i}-\stackrel{^}{j}+\stackrel{^}{k}$. Now, $\stackrel{\to }{{a}_{2}}-\stackrel{\to }{{a}_{1}}=\stackrel{^}{i}-\stackrel{^}{k}$ and $\left(\stackrel{\to }{{a}_{2}}-\stackrel{\to }{{a}_{1}}\right)×\stackrel{\to }{b}=\left(\stackrel{^}{i}-\stackrel{^}{k}\right)×\left(2\stackrel{^}{i}-\stackrel{^}{j}+\stackrel{^}{k}\right)\phantom{\rule{0ex}{0ex}}=\left|\begin{array}{ccc}\stackrel{^}{i}& \stackrel{^}{j}& \stackrel{^}{k}\\ 1& 0& -1\\ 2& -1& 1\end{array}\right|\phantom{\rule{0ex}{0ex}}=-\stackrel{^}{i}-3\stackrel{^}{j}-\stackrel{^}{k}\phantom{\rule{0ex}{0ex}}⇒\left|\left(\stackrel{\to }{{a}_{2}}-\stackrel{\to }{{a}_{1}}\right)×\stackrel{\to }{b}\right|=\sqrt{{\left(-1\right)}^{2}+{\left(-3\right)}^{2}+{\left(-1\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{1+9+1}\phantom{\rule{0ex}{0ex}}=\sqrt{11}\phantom{\rule{0ex}{0ex}}$ The shortest distance between the two lines is given by $\frac{\left|\left(\stackrel{\to }{{a}_{2}}-\stackrel{\to }{{a}_{1}}\right)×\stackrel{\to }{b}\right|}{\left|\stackrel{\to }{b}\right|}=\frac{\sqrt{11}}{\sqrt{6}}$

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