Find the shortest distance between the lines whose vector equations are: r =(1-t) i +(t-2) j +(3-2t) k and r =(s+1) i +(2s-1) j -(2s+1) k

Equation of first line is r =(1-t) i +(t-2) j +(3-2t) k i -t i +t j -2 j +3 k -2t k = ( i -2 j +3 k )+t (- i + j -2 k ) Comparing this equation with a_1 +t b_1 , a_1 = i -2 j +3 k , b_1 =- i + j -2 k Equation of second line is r =(s+1) i +(2s-1) j +(2s+1) k s i + i +2s j - j -2s k - k = ( i - j - k )+s ( i +2 j -2 k ) Comparing this equation with a_2 +s b_2 , a_2 = i - j - k , b_2 = i +2 j -2 k Now shortest distance (d)= | ( a_2 - a_1 ). ( b_1 b_2 ) | | b_1 b_2 | ...(i) a_2 - a_1 = ( i - j - k )- ( i -2 j +3 k )= j -4 k b_1 b_2 = i & j & k -1&1&-2 1&2&-2 =(-2+4) i -(2+2) j +(-2-1) k =2 i -4 j -3 k | b_1 b_2 |=(2)^2+(-4)^2+(-3)^2=29 ( a_2 - a_1 ). ( b_1 b_2 )= ( j -4 k ). (2 i -4 j -3 k ) =0 × 2 + 1 × (-4) + (-4)(-3) = 8 Putting these values in eq.(i), Shortest distance (d)=|8| 29 =8 29 .

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:
$\vec{\text{r}}=(1-\text{t})\hat{\text{i}}+(\text{t}-2)\hat{\text{j}}+(3-2\text{t})\hat{\text{k}}\ \text{and}$
$\vec{\text{r}}=(\text{s}+1)\hat{\text{i}}+(2\text{s}-1)\hat{\text{j}}-(2\text{s}+1)\hat{\text{k}}$

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Answer

Equation of first line is $\vec{\text{r}}=(1-\text{t})\hat{\text{i}}+(\text{t}-2)\hat{\text{j}}+(3-2\text{t})\hat{\text{k}}$
$\hat{\text{i}}-\text{t}\hat{\text{i}}+\text{t}\hat{\text{j}}-2\hat{\text{j}}+3\hat{\text{k}}-2\text{t}\hat{\text{k}}$
$=\Big(\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}\Big)+\text{t}\Big(-\hat{\text{i}}+\hat{\text{j}}-2\hat{\text{k}}\Big)$
Comparing this equation with $\vec{\text{a}_1}+\text{t}\vec{\text{b}_1},$
$\vec{\text{a}_1}=\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}},\ \ \ \vec{\text{b}_1}=-\hat{\text{i}}+\hat{\text{j}}-2\hat{\text{k}}$
Equation of second line is $\vec{\text{r}}=(\text{s}+1)\hat{\text{i}}+(2\text{s}-1)\hat{\text{j}}+(2\text{s}+1)\hat{\text{k}}$
$\text{s}\hat{\text{i}}+\hat{\text{i}}+2\text{s}\hat{\text{j}}-\hat{\text{j}}-2\text{s}\hat{\text{k}}-\hat{\text{k}}$
$=\Big(\hat{\text{i}}-\hat{\text{j}}-\hat{\text{k}}\Big)+\text{s}\Big(\hat{\text{i}}+2\hat{\text{j}}-2\hat{\text{k}}\Big)$
Comparing this equation with $\vec{\text{a}_2}+\text{s}\vec{\text{b}_2},$
$\vec{\text{a}_2}=\hat{\text{i}}-\hat{\text{j}}-\hat{\text{k}},\ \ \ \vec{\text{b}_2}=\hat{\text{i}}+2\hat{\text{j}}-2\hat{\text{k}}$
Now shortest distance $(\text{d})=\frac{\Big|\Big(\vec{\text{a}_2}-\vec{\text{a}_1}\Big).\Big(\vec{\text{b}_1}\times\vec{\text{b}_2}\Big)\Big|}{\Big|\vec{\text{b}_1}\times\vec{\text{b}_2}\Big|}\ \ \ \ ...(\text{i})$
$\vec{\text{a}_2}-\vec{\text{a}_1}=\Big(\hat{\text{i}}-\hat{\text{j}}-\hat{\text{k}}\Big)-\Big(\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}\Big)=\hat{\text{j}}-4\hat{\text{k}}$
$\vec{\text{b}_1}\times\vec{\text{b}_2}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\-1&1&-2\\1&2&-2\end{vmatrix}$
$=(-2+4)\hat{\text{i}}-(2+2)\hat{\text{j}}+(-2-1)\hat{\text{k}}=2\hat{\text{i}}-4\hat{\text{j}}-3\hat{\text{k}}$
$\Big|\vec{\text{b}_1}\times\vec{\text{b}_2}\Big|=\sqrt{(2)^2+(-4)^2+(-3)^2}=\sqrt{29}$
$\Big(\vec{\text{a}_2}-\vec{\text{a}_1}\Big).\Big(\vec{\text{b}_1}\times\vec{\text{b}_2}\Big)=\Big(\hat{\text{j}}-4\hat{\text{k}}\Big).\Big(2\hat{\text{i}}-4\hat{\text{j}}-3\hat{\text{k}}\Big)$
=0 × 2 + 1 × (-4) + (-4)(-3) = 8
Putting these values in eq.(i),
Shortest distance $(\text{d})=\frac{|8|}{\sqrt{29}}=\frac{8}{\sqrt{29}}.$