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  • Find the value of  for which the following lines are perpendicular to each other: <img alt="\frac{x-5}{5\lambda +2}= \frac{2-y}{5}= \frac{1-z}{-1};\: \:

Find the value of \lambda for which the following lines are perpendicular to each other:
\frac{x-5}{5\lambda +2}= \frac{2-y}{5}= \frac{1-z}{-1};\: \: \frac{x}{1}= \frac{y+\frac{1}{2}}{2\lambda }= \frac{z-1}{3}
Hence,find whether the lines intersect or not.

 

 

 

 
 
 
 
 

Answers (1)

\frac{x-5}{5\lambda +2}= \frac{y-2}{-5}= \frac{z-1}{1}----(1)\left [ given \right ]
& \frac{x}{1}= \frac{y+\frac{1}{2}}{2\lambda }= \frac{z-1}{3}---(2)\left [ given \right ]
a_{1}= 5\lambda +2,\: b_{1}= -5,\: c= 1 &
a_{2}= 1,\: b_{2}= 2\lambda ,\: c_{2}= 3
a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}= 0\: \: \left [ \because they\: are\: perpendicular \right ]\left ( 5\lambda +2 \right )-5\left ( 2\lambda \right )+1\left ( 3 \right )= 0
-5\lambda +5= 0,\, \lambda = 1

The shortest distance between two lines d = \left|\frac{\left|\begin{array}{ccc} {x_{2}-x_{1}} & {y_{2}-y_{1}} & {z_{2}-z_{1}} \\ {a_{1}} & {b_{1}} & {c_{1}} \\ {a_{2}} & {b_{2}} & {c_{2}} \end{array}\right|}{\sqrt{\left(b_{1} c_{2}-b_{2} c_{1}\right)^{2}+\left(c_{1} a_{2}-c_{2} a_{1}\right)^{2}+\left(a_{1} b_{2}-a_{2} b_{1}\right)^{2}}}\right|

= \left|\frac{\left|\begin{array}{ccc} {-5} & {\frac{5}{2}} & {0} \\ {7} & {-5} & {1} \\ {1} & {2} & {3} \end{array}\right|}{\sqrt{\left(b_{1} c_{2}-b_{2} c_{1}\right)^{2}+\left(c_{1} a_{2}-c_{2} a_{1}\right)^{2}+\left(a_{1} b_{2}-a_{2} b_{1}\right)^{2}}}\right|\neq 0

So lines are not intersecting. 

Posted by

Ravindra Pindel

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