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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)

R Ravindra Pindel

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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