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

r =6 i +2 j +2 k + ( i -2 j +2 k ) and r =-4 i - k + (3 i -2 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=6 i +2 j +2 k a _2=-4 i - k b _1= i -2 j +2 k b _2=3 i -2 j -2 k a _2- a _1=-10 i -2 j -3 k and b _1 b _2= i & j & k 1&-2&2 3&-2&-2 =8 i +8 j +4 k | b _1 b _2 |=8^2+8^2+4^2 =64+64+16 =144 =12 and ( a _2- a _1 ). ( b _1 b _2 ) = (-10 i -2 j -3 k ). (8 i +8 j +4 k ) =-80-16-12 =-108 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 | | = |-108 12 | =9

Find the shortest distance between the lines r =6 i +2 j +2 k + (…

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Question

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

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Answer

$\vec{\text{r}}=6\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}+\lambda\big(\hat{\text{i}}-2\hat{\text{j}}+2\hat{\text{k}}\big)$ and $\vec{\text{r}}=-4\hat{\text{i}}-\hat{\text{k}}+\mu\big(3\hat{\text{i}}-2\hat{\text{j}}-2\hat{\text{k}}\big)$
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
$\vec{\text{a}}_1=6\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}$
$\vec{\text{a}}_2=-4\hat{\text{i}}-\hat{\text{k}}$
$\vec{\text{b}}_1=\hat{\text{i}}-2\hat{\text{j}}+2\hat{\text{k}}$
$\vec{\text{b}}_2=3\hat{\text{i}}-2\hat{\text{j}}-2\hat{\text{k}}$
$\therefore\vec{\text{a}}_2-\vec{\text{a}}_1=-10\hat{\text{i}}-2\hat{\text{j}}-3\hat{\text{k}}$
and $\vec{\text{b}}_1\times\vec{\text{b}}_2=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\1&-2&2\\3&-2&-2\end{vmatrix}$
$=8\hat{\text{i}}+8\hat{\text{j}}+4\hat{\text{k}}$
$\Rightarrow\big|\vec{\text{b}}_1\times\vec{\text{b}}_2\big|=\sqrt{8^2+8^2+4^2}$
$=\sqrt{64+64+16}$
$=\sqrt{144}$
$=12$
and $\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big).\big(\vec{\text{b}}_1\times\vec{\text{b}}_2\big)$
$=\big(-10\hat{\text{i}}-2\hat{\text{j}}-3\hat{\text{k}}\big).\big(8\hat{\text{i}}+8\hat{\text{j}}+4\hat{\text{k}}\big)$
$=-80-16-12$
$=-108$
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
$\text{d}=\Bigg|\frac{\big(\vec{\text{a}}_2.\vec{\text{a}}_1\big).\big(\vec{\text{b}}_1\times\vec{\text{b}}_2\big)}{\big|\vec{\text{b}}_1\times\vec{\text{b}}_2\big|}\Bigg|$
$\Rightarrow\text{d}=\Big|\frac{-108}{12}\Big|$
$=9$