By computing the shortest distance determine whether the followin…

Question

By computing the shortest distance determine whether the following pairs of lines intersect or not:
$\frac{\text{x}-1}{2}=\frac{\text{y}+1}{3}=\text{z}$ and $\frac{\text{x}+1}{5}=\frac{\text{y}-2}{1};\text{z}=2$

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Answer

Given equations of lines are,
$\frac{\text{x}-1}{2}=\frac{\text{y}+1}{3}=\frac{\text{z}}{1}=\lambda$ (say)
$\Rightarrow\text{x}=2\lambda+1,\text{y}=3\lambda-1,\text{z}=\lambda$
$\Rightarrow\vec{\text{r}}=\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}}$
$=\big(2\lambda+1\big)\hat{\text{i}}+(3\lambda-1)\hat{\text{j}}+(\lambda)\hat{\text{k}}$
$\vec{\text{r}}=\big(\hat{\text{i}}-\hat{\text{j}}\big)+\lambda\big(2\hat{\text{i}}+3\hat{\text{j}}+\hat{\text{k}}\big)$
$\Rightarrow\vec{\text{a}}_1=\big(\hat{\text{i}}-\hat{\text{j}}\big),\vec{\text{b}}_1=\big(2\hat{\text{i}}+3\hat{\text{j}}+\hat{\text{k}}\big)$
and, $\frac{\text{x}+1}{5}=\frac{\text{y}-2}{1}=\mu\text{ (say) },\text{z}=2$
$\Rightarrow\text{x}=5\mu-1,\text{y}=\mu+2,\text{z}=2$
$\Rightarrow\vec{\text{r}}=\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}}$
$=(5\mu-1)\hat{\text{i}}+(\mu+2)\hat{\text{j}}+2\hat{\text{k}}$
$\vec{\text{r}}=\big(-\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}\big)+\mu\big(5\hat{\text{i}}+\hat{\text{j}}\big)$
$\Rightarrow\vec{\text{a}}_2=\big(-\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}\big),\vec{\text{}b}_2=\big(5\hat{\text{i}}+\hat{\text{j}}\big)$
we know that, the shortest distance between $\vec{\text{r}}=\vec{\text{a}}_1+\lambda\vec{\text{b}}_1$ and $\vec{\text{r}}=\vec{\text{a}}_2+\lambda\vec{\text{b}}_2$ is given by
$\text{S.D.}=\Bigg|\frac{\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big).\big(\vec{\text{b}}_1\times\vec{\text{b}}_2\big)}{\big|\vec{\text{b}}_1\times\vec{\text{b}}_2\big|}\Bigg|\dots(1)$
$\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big)=\big(-\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}\big)-\big(\hat{\text{i}}-\hat{\text{j}}\big)$
$=-\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}-\hat{\text{i}}+\hat{\text{j}}$
$=-2\hat{\text{i}}+3\hat{\text{j}}+2\hat{\text{k}}$
$\vec{\text{b}}_1\times\vec{\text{b}}_2=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\2&3&1\\5&1&0 \end{vmatrix}$
$=\hat{\text{i}}(0-1)-\hat{\text{j}}(0-5)+\hat{\text{k}}(2+15)$
$\vec{\text{b}}_1\times\vec{\text{b}}_2=-\hat{\text{i}}+5\hat{\text{j}}-13\hat{\text{k}}$
$\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big)\big(\vec{\text{b}}_1\times\vec{\text{b}}_2\big)=\big(-2\hat{\text{i}}+3\hat{\text{j}}+2\hat{\text{k}}\big)\big(-\hat{\text{i}}+5\hat{\text{j}}-13\hat{\text{k}}\big)$
$=(-2)(-1)+(3)(5)+(2)(-13)$
$=-9$
$\big|\vec{\text{b}}_1\times\vec{\text{b}}_2\big|=\sqrt{(-1)^2+(5)^2+(-13)^2}$
$=\sqrt{1+25+169}$
$=\sqrt{195}$
Substituting the value of $\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big).\big(\vec{\text{b}}_1\times\vec{\text{b}}_2\big)$ and $\big|\vec{\text{b}}_1\times\vec{\text{b}}_2\big|$ in equation (1) to get shortest distance between given lines, so
$\text{S.D.}=\Big|\frac{-9}{\sqrt{195}}\Big|$
$=\frac{9}{\sqrt{195}}\text{ units}$

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