logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  3-D Geometry
0 votes

If the lines $\overrightarrow r=(\hat i-\hat j+\hat k)+\lambda(2\hat i+3\hat j+4\hat k)\:\:and\:\:\overrightarrow r=(3\hat i+k\hat j)+\mu(\hat i+2\hat j+\hat k)$ intersect then $k=?$

$\begin{array}{1 1} -1 \\ \frac{2}{9} \\\frac{9}{2} \\ 0 \end{array} $

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • If two non parallel lines intersect then the shortest distance (S.D) between the lines is $0$
  • $s.d.$ between thelines = $\bigg|\large\frac{(\overrightarrow {b_1}\times\overrightarrow {b_2}).(\overrightarrow a_1-\overrightarrow a_2)}{|\overrightarrow b_1\times\overrightarrow b_2|}\bigg|$
Given lines are $\overrightarrow r=(\hat i-\hat j+\hat k)+\lambda(2\hat i+3\hat j+4\hat k)$... ..(i) and
$\overrightarrow r=(3\hat i+k\hat j)+\mu(\hat i+2\hat j+\hat k)$.
If these lines intersect then the $S.D.$ between them =$0$
$\Rightarrow\:(\overrightarrow {b_1}\times\overrightarrow {b_2}).(\overrightarrow { a_1}-\overrightarrow {a_2})=0$
From (i) and (ii) $\overrightarrow a_1=(1,-1,1),\:\:\overrightarrow a_2=(3,k,0)$
$\overrightarrow a_1-\overrightarrow a_2=(-2,\:-1-k,\:1)$
$\overrightarrow {b_1}=(2,3,4)\:\:and\:\:\overrightarrow b_2=(1,2,1)$
$\overrightarrow b_1 \times\overrightarrow b_2=\left |\begin {array}{ccc} \hat i & \hat j &\hat k \\ 2 & 3 & 4 \\ 1 & 2 & 1\end {array}\right |=(-5,2,1)$
$(\overrightarrow b_1\times\overrightarrow b_2).(\overrightarrow a_1-\overrightarrow a_2)=(-5,2,1).(-2,(-1-k),1)=0$
$\Rightarrow\:10-2-2k+1=0$
$\Rightarrow\:k=\large\frac{9}{2}$
answered Dec 26, 2013 by rvidyagovindarajan_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
...