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  • If the shortest distance between the lines <img alt="\mathrm{\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{\lambda}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Cfrac%7Bx-1%7D

If the shortest distance between the lines \mathrm{\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{\lambda}} and \mathrm{\frac{x-2}{1}=\frac{y-4}{4}=\frac{z-5}{5}} is \mathrm{\frac{1}{\sqrt{3}}} then the sum of all possible values of  \mathrm{\lambda } is

Option: 1

16


Option: 2

6


Option: 3

12


Option: 4

15


Answers (1)

best_answer

\begin{matrix} \mathrm{\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{\lambda}} & \begin{matrix} \mathrm {\vec{a}=(1,2,3)}\\ \mathrm{\vec{c}=(2,3,\lambda)} \end{matrix} & \\ \mathrm{\frac{x-2}{1}=\frac{y-4}{4}=\frac{z-5}{5}} & \begin{matrix} \mathrm {\vec{b}=(2,4,5)}\\ \mathrm{\vec{d}=(1,4,5)} \end{matrix} \end{matrix}

Shortest distancee=\mathrm{\frac{[(\bar{a}-\bar{b}) \quad \bar{c} \bar{d}]}{|\bar{c} \times \bar{d}|}}

\mathrm{\left [ (\vec{a}-\vec{b}) \;\;\; \vec{c}\;\;\; \vec{d} \right ]}= \mathrm{\begin{vmatrix} -1 & -2 & -2 \\ 2& 3 & \lambda \\ 1 & 2 & 5 \end{vmatrix}}

\mathrm {=4 \lambda-15+20-2 \lambda-10 }\\

\mathrm{=2 \lambda-5}

\mathrm {\bar{c} \times \bar{d}=(15-4 \lambda) \hat{u}+j(\lambda-10)+\hat{x}(5)}

\begin{aligned} \therefore \mathrm{|\vec{c}\times \vec{d}|}=& \mathrm{\sqrt{225+16 x^{2}-120 \lambda+\lambda^{2}+100-20 \lambda+25}} \\ &=\mathrm{\sqrt{17 \lambda^{2}-140 \lambda+350} }\end{aligned}

\therefore \mathrm{\frac{1}{\sqrt{3}}=\pm \frac{2 \lambda-5}{\sqrt{17 \lambda^{2}-14 0 \lambda+350}} }

\mathrm{17 \lambda ^{2}-140 \lambda +350=3\left(4 \lambda ^{2}+25-20 \lambda \right) }

\begin{gathered} \mathrm {5 \lambda^{2}-80 \lambda+275=0 }\\ \mathrm {\lambda=5,11 }\end{gathered}

\therefore sum of all values of \lambda =16

 

Posted by

seema garhwal

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