Find the shortest distance between the lines r = ( i +2 j + k )+ ( i - j + k ) and, r =2 i - j - k + (2 i + j +2 k )

r = ( i +2 j + k )+ ( i - j + k ) and, r =2 i - j - k + (2 i + j +2 k ) Comparing the given equations with the equations r = a _1+ b _1 and r = a _2+ b _2, we get a _1= i +2 j + k a _2=2 i - j + k b _1= i - j + k b _2=2 i + j +2 k a _2- a _1= i -3 j -2 k and b _1 b _2= i & j & k 1&-1&1 2&1&2 =-3 i +3 k | b _1 b _2 |=(-3)^2+3^2 =9+9 =32 and ( a _2- a _1 ). ( b _1 b _2 )= ( i -3 j -2 k ). (-3 i +3 k ) =-3-6 =-9 The shortest distance between the lines r = a _1+ b _1 and r = a _2+ b _2 is given by d= | ( a _2- a _1 ). ( b _1 b _2 ) | b _1 b _2 | | = |-9 32 | =3 2

Find the shortest distance between the lines r = ( i +2 j + k )+ …

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Question

Find the shortest distance between the lines
$\vec{\text{r}}=\big(\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}}\big)+\lambda\big(\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}\big)$ and, $\vec{\text{r}}=2\hat{\text{i}}-\hat{\text{j}}-\hat{\text{k}}+\mu\big(2\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}\big)$

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Answer

$\vec{\text{r}}=\big(\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}}\big)+\lambda\big(\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}\big)$ and, $\vec{\text{r}}=2\hat{\text{i}}-\hat{\text{j}}-\hat{\text{k}}+\mu\big(2\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}\big)$
Comparing the given equations with the equations $\vec{\text{r}}=\vec{\text{a}}_1+\lambda\vec{\text{b}}_1$ and $\vec{\text{r}}=\vec{\text{a}}_2+\mu\vec{\text{b}}_2,$ we get
$\vec{\text{a}}_1=\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}}$
$\vec{\text{a}}_2=2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$
$\vec{\text{b}}_1=\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$
$\vec{\text{b}}_2=2\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}$
$\therefore\vec{\text{a}}_2-\vec{\text{a}}_1=\hat{\text{i}}-3\hat{\text{j}}-2\hat{\text{k}}$
and $\vec{\text{b}}_1\times\vec{\text{b}}_2=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\1&-1&1\\2&1&2\end{vmatrix}$
$=-3\hat{\text{i}}+3\hat{\text{k}}$
$\Rightarrow\big|\vec{\text{b}}_1\times\vec{\text{b}}_2\big|=\sqrt{(-3)^2+3^2}$
$=\sqrt{9+9}$
$=3\sqrt{2}$
and $\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big).\big(\vec{\text{b}}_1\times\vec{\text{b}}_2\big)=\big(\hat{\text{i}}-3\hat{\text{j}}-2\hat{\text{k}}\big).\big(-3\hat{\text{i}}+3\hat{\text{k}}\big)$
$=-3-6$
$=-9$
The shortest distance between the lines $\vec{\text{r}}=\vec{\text{a}}_1+\lambda\vec{\text{b}}_1$ and $\vec{\text{r}}=\vec{\text{a}}_2+\mu\vec{\text{b}}_2$ is given by
$\text{d}=\Bigg|\frac{\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big).\big(\vec{\text{b}}_1\times\vec{\text{b}}_2\big)}{\big|\vec{\text{b}}_1\times\vec{\text{b}}_2\big|}\Bigg|$
$\Rightarrow\text{d}=\Big|\frac{-9}{3\sqrt{2}}\Big|$
$=\frac{3}{\sqrt{2}}$

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