Find the value of \lambda, so that the lines \frac{1-x}{3}=\frac{7y-14}{\lambda }=\frac{z-3}{2}  and \frac{7-7x}{3\lambda }=\frac{y-5}{1}=\frac{6-z}{5} are at right angles. Also, find whether the lines are intersecting or not. 

 

 

 

 
 
 
 
 

Answers (1)

Equation of 1st line 

\frac{1-x}{3}=\frac{7y-14}{\lambda }=\frac{z-3}{2}

\frac{x-1}{-3}=\frac{y-2}{\frac{\lambda}{7} }=\frac{z-3}{2}

D.R  of 1st line     \rightarrow -3,\frac{\lambda }{7},2

Equation of 2nd line : 

\frac{7-7x}{3\lambda }=\frac{y-5}{1}=\frac{6-z}{5}

\frac{x-1}{\frac{-3\lambda }{7} }=\frac{y-5}{1}=\frac{z-6}{-5}

D.R of 2nd line= \left ( \frac{-3\lambda }{7},1,-5 \right )

Since given lines are at right angles,

-3\left ( \frac{-3\lambda }{7} \right )+\frac{\lambda }{7}(1)+2(-5)=0

\frac{9\lambda }{7}+\frac{\lambda }{7}-10=0

10\lambda -70=0

10\lambda=70

\lambda=7

Now, equation of 1st line \frac{x-1}{-3}=\frac{y-2}{1}=\frac{z-3}{2}

 

and equation of 2nd line \frac{x-1}{-3}=\frac{y-5}{1}=\frac{z-6}{-5}

Here, \overrightarrow{a_1}=\hat{i}+2\hat{j}+3\hat{k},\: \overrightarrow{a_2}=\hat{i}+5\hat{j}+6\hat{k}

          \overrightarrow{b_1}=-3\hat{i}+\hat{j}+2\hat{k},\: \overrightarrow{b_2}=\hat{i}+\hat{j}-5\hat{k}

SD=\left | \frac{(\vec{a_2}-\vec{a_1})\cdot [\vec{b_1}\times \overrightarrow{b_2}]}{|\overrightarrow{b_1}\times \overrightarrow{b_2}|} \right |

(\overrightarrow{a_2}-\overrightarrow{a_1})=0\hat{i}+3\hat{j}+3\hat{k}

and  (\overrightarrow{b_1}\times\overrightarrow{b_2})=\begin{vmatrix} \hat{i} &\hat{j} &\hat{k} \\ -3 &1 & 2\\ -3 &1 &-5 \end{vmatrix}

                                                                                                       =\hat{i}(-5-2)-\hat{j}(15+6)+\hat{k}(-3+3)

           =-7\hat{i}-11\hat{j}

\therefore SD=\left | \frac{3\hat{i}+3\hat{k}(-7\hat{k}-11\hat{j})}{\sqrt{49+121}} \right |

\therefore SD=\left | \frac{-21-33}{\sqrt{170}} \right |=\frac{54}{\sqrt{170}}\neq 0

Hence, the given two lines do not intersect with each other.

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