Find the shortest distance between the lines r = i +2 j +3 k + ( i -3 j +2 k ) and r =4 i +5 j +6 k + (2 i +3 j + k )

r = i +2 j +3 k + ( i -3 j +2 k ) and r =4 i +5 j +6 k + (2 i +3 j + k ) Comparing the given equation with the equations r = a _1+ b _1 and r = a _2+ b _2, we get a _1= i +2 j +3 k a _2=4 i +5 j +6 k b _1= i -3 j +2 k b _2=2 i +3 j + k a _2- a _1=3 i +3 j +3 k and b _1 b _2= i & j & k 1&-3&2 2&3&1 =-9 i +3 j +9 k | b _1 b _2 |=(-9)^2+3^2+9^2 =81+9+81 =171 and ( a _2- a _1 ). ( b _1 b _2 ) = (3 i +3 j +3 k ). (-9 i +3 j +9 k ) =-27+9+27 =9 The shortest distence 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 | | = |9 171 | =3 19

Find the shortest distance between the lines r = i +2 j +3 k + ( …

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Question

Find the shortest distance between the lines
$\vec{\text{r}}=\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}+\lambda\big(\hat{\text{i}}-3\hat{\text{j}}+2\hat{\text{k}}\big)$ and $\vec{\text{r}}=4\hat{\text{i}}+5\hat{\text{j}}+6\hat{\text{k}}+\mu\big(2\hat{\text{i}}+3\hat{\text{j}}+\hat{\text{k}}\big)$

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$\vec{\text{r}}=\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}+\lambda\big(\hat{\text{i}}-3\hat{\text{j}}+2\hat{\text{k}}\big)$ and $\vec{\text{r}}=4\hat{\text{i}}+5\hat{\text{j}}+6\hat{\text{k}}+\mu\big(2\hat{\text{i}}+3\hat{\text{j}}+\hat{\text{k}}\big)$
Comparing the given equation 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
$\vec{\text{a}}_1=\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}$
$\vec{\text{a}}_2=4\hat{\text{i}}+5\hat{\text{j}}+6\hat{\text{k}}$
$\vec{\text{b}}_1=\hat{\text{i}}-3\hat{\text{j}}+2\hat{\text{k}}$
$\vec{\text{b}}_2=2\hat{\text{i}}+3\hat{\text{j}}+\hat{\text{k}}$
$\therefore\vec{\text{a}}_2-\vec{\text{a}}_1=3\hat{\text{i}}+3\hat{\text{j}}+3\hat{\text{k}}$
and $\vec{\text{b}}_1\times\vec{\text{b}}_2=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\1&-3&2\\2&3&1\end{vmatrix}$
$=-9\hat{\text{i}}+3\hat{\text{j}}+9\hat{\text{k}}$
$\Rightarrow\big|\vec{\text{b}}_1\times\vec{\text{b}}_2\big|=\sqrt{(-9)^2+3^2+9^2}$
$=\sqrt{81+9+81}$
$=\sqrt{171}$
and $\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big).\big(\vec{\text{b}}_1\times\vec{\text{b}}_2\big)$
$=\big(3\hat{\text{i}}+3\hat{\text{j}}+3\hat{\text{k}}\big).\big(-9\hat{\text{i}}+3\hat{\text{j}}+9\hat{\text{k}}\big)$
$=-27+9+27$
$=9$
The shortest distence 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
$\text{d}=\Bigg|\frac{\big(\vec{\text{a}}_2.\vec{\text{a}}_1\big).\big(\vec{\text{b}}_1\times\vec{\text{b}}_2\big)}{\big|\vec{\text{b}}_1\times\vec{\text{b}}_2\big|}\Bigg|$
$\Rightarrow\text{d}=\Big|\frac{9}{\sqrt{171}}\Big|$
$=\frac{3}{\sqrt{19}}$

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