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

Equation of the first line is r = ( i +2 j +3 k )+ ( i -3 j +2 k ) Comparing this equation with r = a_1 + b_1 , a_1 = i +2 j +3 k and b_1 = i -3 j +2 k Again equation of second line r = (4 i +5 j +6 k )+ (2 i +3 j + k ) Comparing this equation with r = a_2 + b_2 , a_2 =4 i +5 j +6 k and b_2 =2 i +3 j + k Now shortest distance (d)= | ( a_2 - a_1 ). ( b_1 b_2 ) | | b_1 b_2 | ...(i) Here a_2 - a_1 = (4 i +5 j +6 k )- ( i +2 j +3 k ) =3 i +3 j +3 k b_1 b_2 = i & j & k 1&-3&2 2&3&1 =(-3-6) i -(1-4) j +(3+6) k =-9 i +3 j +9 k | b _1 b _2 |=(-9)^2+(3)^2+(9)^2=171=319 ( a_2 - a_1 ). ( b_1 b_2 )=3 (-9)+(3 3)+(3 9) =-27+9+27=9 Putting these values in eq.(i), Shortest distance (d)=|9| 319 =9 319 =3 19 .

Find the shortest distance between the lines whose vector equatio…

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Question

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

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Answer

Equation of the first line is $\vec{\text{r}}=\Big(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}\Big)+\lambda\Big(\hat{\text{i}}-3\hat{\text{j}}+2\hat{\text{k}}\Big)$
Comparing this equation with $\vec{\text{r}}=\vec{\text{a}_1}+\lambda\vec{\text{b}_1},$
$\vec{\text{a}_1}=\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}\ \text{and}\ \vec{\text{b}_1}=\hat{\text{i}}-3\hat{\text{j}}+2\hat{\text{k}}$
Again equation of second line $\vec{\text{r}}=\Big(4\hat{\text{i}}+5\hat{\text{j}}+6\hat{\text{k}}\Big)+\mu\Big(2\hat{\text{i}}+3\hat{\text{j}}+\hat{\text{k}}\Big)$
Comparing this equation with $\vec{\text{r}}=\vec{\text{a}_2}+\mu\vec{\text{b}_2},$
$\vec{\text{a}_2}=4\hat{\text{i}}+5\hat{\text{j}}+6\hat{\text{k}}\ \text{and}\ \vec{\text{b}_2}=2\hat{\text{i}}+3\hat{\text{j}}+\hat{\text{k}}$
Now shortest distance $(\text{d})=\frac{\Big|\Big(\vec{\text{a}_2}-\vec{\text{a}_1}\Big).\Big(\vec{\text{b}_1}\times\vec{\text{b}_2}\Big)\Big|}{\Big|\vec{\text{b}_1}\times\vec{\text{b}_2}\Big|}\ \ \ \ ...(\text{i})$
Here $\vec{\text{a}_2}-\vec{\text{a}_1}=\Big(4\hat{\text{i}}+5\hat{\text{j}}+6\hat{\text{k}}\Big)-\Big(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}\Big)$
$=3\hat{\text{i}}+3\hat{\text{j}}+3\hat{\text{k}}$
$\vec{\text{b}_1}\times\vec{\text{b}_2}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\1&-3&2\\2&3&1\end{vmatrix}$
$=(-3-6)\hat{\text{i}}-(1-4)\hat{\text{j}}+(3+6)\hat{\text{k}}=-9\hat{\text{i}}+3\hat{\text{j}}+9\hat{\text{k}}$
$\Big|\vec{\text{b}}_1\times\vec{\text{b}}_2\Big|=\sqrt{(-9)^2+(3)^2+(9)^2}=\sqrt{171}=3\sqrt{19}$
$\Big(\vec{\text{a}_2}-\vec{\text{a}_1}\Big).\Big(\vec{\text{b}_1}\times\vec{\text{b}_2}\Big)=3\times(-9)+(3\times3)+(3\times9)$
$=-27+9+27=9$
Putting these values in eq.(i),
Shortest distance $(\text{d})=\frac{|9|}{3\sqrt{19}}=\frac{9}{3\sqrt{19}}=\frac{3}{\sqrt{19}}.$