Question

29 रेखाओं r = (i^+2j^+k^)t(i^-j^+k^) और r = (2i^-j^-k^)+u(2i^+j^+2k^
)
के बीच की न्यूनतम दूरी ज्ञात कीजिए।​

Answer 1

SOLUTION

TO DETERMINE

The shortest distance between the lines

$\vec{r} = ( \hat{i} + 2\hat{j} + \hat{k}) + t( \hat{i} - \hat{j} + \hat{k})$

$\vec{r} = ( 2\hat{i} - \hat{j} - \hat{k}) + u(2 \hat{i} + \hat{j} +2 \hat{k})$

CONCEPT TO BE IMPLEMENTED

General equation of any line is

$\vec{r} = \vec{a} + t \vec{b}$

Then the shortest distance between two lines

$\displaystyle \: = \frac{ | \: ( \vec{a_2 } - \vec{a_1 }). (\vec{b_1 } \times \vec{b_2 }) \: | }{ | \: \vec{b_1 } \times \vec{b_2 } \: | }$

EVALUATION

Here the given equation of the lines are

$\vec{r} = ( \hat{i} + 2\hat{j} + \hat{k}) + t( \hat{i} - \hat{j} + \hat{k})$

$\vec{r} = ( 2\hat{i} - \hat{j} - \hat{k}) + u(2 \hat{i} + \hat{j} +2 \hat{k})$

Comparing with the general equation

$\vec{r} = \vec{a} + t \vec{b}$

we get

$\vec{a_1 } = ( \hat{i} + 2\hat{j} + \hat{k})$

$\vec{b_1 } = ( \hat{i} - \hat{j} + \hat{k})$

$\vec{a_2 } =( 2\hat{i} - \hat{j} - \hat{k})$

$\vec{b_2 } =(2 \hat{i} + \hat{j} +2 \hat{k})$

Now

$\vec{a_2 } - \vec{a_1 }$

$= ( 2\hat{i} - \hat{j} - \hat{k}) - ( \hat{i} + 2\hat{j} + \hat{k})$

$= ( \hat{i} - 3\hat{j} - 2 \hat{k})$

Again

$\vec{b_1 } \times \vec{b_2 }$

$= \displaystyle\begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 1 & - 1 & 1 \\ 2 & 1 & 2 \end{vmatrix}$

$= - 3 \hat{i} + 0\hat{j} + 3 \hat{k}$

Now

$| \vec{b_1 } \times \vec{b_2 }|$

$= \sqrt{ {( - 3)}^{2} + {(3)}^{2} }$

$= \sqrt{ 9 + 9}$

$= \sqrt{ 18}$

$= 3 \sqrt{2}$

Again

$( \vec{a_2 } - \vec{a_1 }). (\vec{b_1 } \times \vec{b_2 })$

$= ( \hat{i} - 3\hat{j} - 2 \hat{k}).( - 3 \hat{i} + 0\hat{j} + 3 \hat{k})$

$= - 3 - 6$

$= - 9$

Hence the required shortest distance

$\displaystyle \: = \frac{ | \: ( \vec{a_2 } - \vec{a_1 }). (\vec{b_1 } \times \vec{b_2 }) \: | }{ | \: \vec{b_1 } \times \vec{b_2 } \: | }$

$\displaystyle \sf{ = \frac{ | - 9| }{3 \sqrt{2} } \: \: \: unit}$

$\displaystyle \sf{ = \frac{ 9 }{3 \sqrt{2} } \: \: \: unit}$

$\displaystyle \sf{= \frac{3 }{ \sqrt{2} } \: \: unit}$

$\displaystyle \sf{ = \frac{3 \sqrt{2} }{2} \: \: \: \: unit }$

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