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The shortest distance between the lines \frac{x-3}{3}=\frac{y-8}{-1}=\frac{z-3}{1} and \frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-6}{4} is :
 
Option: 1 2\sqrt{30}
Option: 2 \frac{7}{2}\sqrt{30}
Option: 3 3
Option: 4 3\sqrt{30}
 

Answers (1)

best_answer

 

 

Shortest Distance between Two Lines -

Distance between two skew lines

If L1 and L2  are two skew lines, then there is one and only one line perpendicular to each of lines L1 and L2 which is known as the line of shortest distance.

Vector form

\\\mathrm{\;\;\;L_1:\;\;\;\;}\vec{\mathbf r} &=\vec{\mathbf r}_{0}+\lambda \vec{\mathbf b} \\\\\mathrm{\;\;\;L_2:\;\;\;\;}\vec{\mathbf r} &=\vec{\mathbf r'}_{0}+\mu \vec{\mathbf b'}

If \overrightarrow{PQ} is the shortest distance vector between L1 and L2, then it being perpendicular to both \vec{\mathbf b} and \vec{\mathbf b'}, therefore, the unit vector \hat{\mathbf n} along \overrightarrow{PQ} would be 

\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\hat{\mathbf n}=\frac{\vec {\mathbf b}\times \vec{\mathbf b'}}{\left | \vec{\mathbf b}\times \vec{\mathbf b'} \right |}\\\mathrm{Then,\;\;\;\;\;\;\;\;\;\;\;\;}\overrightarrow{PQ}=\mathit{d}\hat{\mathbf n}

where "d" is the magnitude of the shortest distance vector. Let θ be the angle between \overrightarrow{ST} and \overrightarrow{PQ}.

Then  

\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;PQ=ST\left |\cos\theta \right |}\\\\\text{but,}\;\;\;\;\;\;\;\;\;\;\;\;\;\cos \theta=\left|\frac{\overrightarrow{\mathrm{PQ}} \cdot \overrightarrow{\mathrm{ST}}}{|\overrightarrow{\mathrm{PQ}}||\overrightarrow{\mathrm{ST}}|}\right|\\\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}=\left|\frac{d \hat{n} \cdot\left(\vec{r'}_{0}-\vec{r}_{0}\right)}{d \;\mathrm{ST}}\right| \quad\left(\text { since } \overrightarrow{\mathrm{\;ST}}=\vec{r'}_{0}-\vec{r}_{0}\right)\\\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}=\left|\frac{\left(\vec{b} \times \vec{b'}\right) \cdot\left(\vec{r'}_{0}-\vec{r}_{0}\right)}{\mathrm{ST}\left|\vec{b} \times \vec{b'}\right|}\right|

Hence, the required shortest distance is

\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}d=\mathrm{PQ}=\mathrm{ST}|\cos \theta|\\\\\text{or}\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\mathbf d=\left|\frac{\left(\vec{\mathbf b} \times \vec{\mathbf b'}\right) \cdot\left(\vec{\mathbf r'}_{0} - \vec{\mathbf r}_{0}\right)}{\left|\vec{\mathbf b} \times \vec{\mathbf b'}\right|}\right|

-

 

 

\\\overrightarrow{\mathrm{AB}}=6 \hat{i}+15 \hat{j}+3 \hat{k}\\\vec p=3\hat i-\hat j+\hat k\\ \vec q=-3\hat i+2\hat j+4\hat k\\\vec p\times \vec q=\begin{vmatrix} \hat i &\hat j &\hat k \\ 3 &-1 &1 \\ -3 & 2 &4 \end{vmatrix}=\hat i(-4-2)-\hat j(12+3)+\hat k(6-3)\\\Rightarrow -6\hat i-15\hat j+3\hat k\\\text{shortest dist}=\frac{| \overrightarrow{\mathrm{AB}} \cdot(\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}})}{|\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}}|}

=3\sqrt{30}

Correct Option (4)

Posted by

Kuldeep Maurya

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