logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  TN XII Math  >>  Vector Algebra
0 votes

Find the shortest distance between the skew lines $\large\frac{x-6}{3}=\frac{y-7}{-1}=\frac{z-4}{1}$ and $\large\frac{x}{-3}=\frac{y+9}{2}=\frac{z-2}{4}$

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • Distance between two lines
  • (i) Shortest = 0 if they are intersecting.
  • (ii) Parallel lines $ \overrightarrow r = \overrightarrow a_1+t\overrightarrow u_1, \: \overrightarrow s=\overrightarrow a_2+s\overrightarrow u$
  • (iii) Skew line $ \overrightarrow r = \overrightarrow a_1+t\overrightarrow u, \: \overrightarrow r = \overrightarrow a_2+s\overrightarrow v$
  • $ d = \large\frac{[(\overrightarrow a_2-\overrightarrow a_1)\overrightarrow u\: \overrightarrow v]}{|\overrightarrow u \times \overrightarrow v|}$
  • The lines $ \overrightarrow r=\overrightarrow a_1+t\overrightarrow u, \: \overrightarrow r=\overrightarrow a_2+s\overrightarrow u$ intersect if $ [(\overrightarrow a_2-\overrightarrow a_1)\overrightarrow u\: \overrightarrow v]=0$
Step 1
$ \large\frac{x-6}{3}=\large\frac{y-7}{-1}=\large\frac{z-4}{1}$ (i)
$ \large\frac{x}{-3} = \large\frac{y+9}{2} =\large\frac{z-2}{4}$ (ii)
(i) passes through the point with pv $ \overrightarrow a_1 = 6\overrightarrow i+7\overrightarrow j+4\overrightarrow k$ and is parallel to the vector $ \overrightarrow u=3\overrightarrow i-\overrightarrow j+\overrightarrow k$
(ii) Passes through the point with pv $ \overrightarrow a_2 =-9\overrightarrow j+2\overrightarrow k$ and is parallel to the vector $ \overrightarrow v=-3\overrightarrow i+2\overrightarrow j+4\overrightarrow k$
The lines are not parallel therefore they are either intersecting or skew lines.
$ \overrightarrow a_2-\overrightarrow a_1 = -6\overrightarrow i-16\overrightarrow j-2\overrightarrow k$
Step 2
$ [\overrightarrow a_2-\overrightarrow a_1\: \: \overrightarrow u\: \overrightarrow v] = \begin{vmatrix} -6 & -16 & -2 \\ 3 & -1 & 1 \\ -3 & 2 & 4 \end{vmatrix} = -6(-4-2)+16(12+3)-2(6-3)$
$ = 36+240-6=270 \neq 0$
$ \therefore $ the lines are skew.
$ \overrightarrow u \times \overrightarrow v = \begin{vmatrix} \overrightarrow i & \overrightarrow j & \overrightarrow k \\ 3 & -1 & 1 \\ -3 & 2 & 4 \end{vmatrix} = (-4-2)\overrightarrow i-(12+3)\overrightarrow j+(6-3)\overrightarrow k$
$ = -6\overrightarrow i-15\overrightarrow j+3\overrightarrow k$
$ |\overrightarrow u \times \overrightarrow v|=\sqrt{36+225+9}=\sqrt{270}$
Step 3
Shortest distance
$ d = \large\frac{|[\overrightarrow a_2-\overrightarrow a_1\: \overrightarrow u\: \overrightarrow v]|}{|\overrightarrow u \times \overrightarrow v|} = \large\frac{270}{\sqrt{270}} = \sqrt{270}=3\sqrt{30}$ units

 

answered Jun 12, 2013 by thanvigandhi_1
edited Jun 24, 2013 by thanvigandhi_1
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...