Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Let the shortest distance between the lines  <img alt="L: \frac{x-5}{-2}=\frac{y-\lambda}{0}=\frac{z+\lambda}{1}, \lambda \geq 0" src="https://entrancecorner.oncodecogs.com/gif.latex?L%3A

Let the shortest distance between the lines  L: \frac{x-5}{-2}=\frac{y-\lambda}{0}=\frac{z+\lambda}{1}, \lambda \geq 0    and L_1: x+1=y-1=4-z \text { be } 2 \sqrt{6}  . If  (\alpha ,\beta ,\gamma )  lies on L, then which of the following is
NOT possible ?

Option: 1

α − 2γ = 19


Option: 2

2α + γ = 7


Option: 3

2α − γ = 9


Option: 4

α + 2γ = 24


Answers (1)

best_answer

Let    \begin{aligned} & \overrightarrow{\mathrm{b}}_1=\langle-2,0,1\rangle \overrightarrow{\mathrm{a}}_1=(5, \lambda,-\lambda) \\ & \end{aligned}

         \begin{aligned} & \overrightarrow{\mathrm{b}}_2=\left\langle 1,1,-1>\vec{a}_2=(-1,1,4)\right. \end{aligned}

Normal vector of both line is    \overrightarrow{\mathrm{b}}_1 \times \overrightarrow{\mathrm{b}}_2=\left|\begin{array}{ccc} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ -2 & 0 & 1 \\ 1 & 1 & -1 \end{array}\right|  

\begin{aligned} & \hat{\mathrm{i}}(-1)-\hat{\mathrm{j}}(1)+\hat{\mathrm{k}}(-2) \\ & \overrightarrow{\mathrm{b}}_1 \times \overrightarrow{\mathrm{b}}_2=<-1,-1,-2> \\ & \overrightarrow{\mathrm{a}}_1-\overrightarrow{\mathrm{a}}_2=\langle 6, \lambda-1,-\lambda-4> \end{aligned}

Shortest distance d =\left|\frac{\left(\vec{a}_2-\overrightarrow{\mathrm{a}}_1\right) \times\left(\overrightarrow{\mathrm{b}}_1 \times \overrightarrow{\mathrm{b}}_2\right)}{\left|\overrightarrow{\mathrm{b}}_1 \times \overrightarrow{\mathrm{b}}_2\right|}\right| 

\begin{aligned} & 2 \sqrt{6}=\left|\frac{\langle 6, \lambda-1,-\lambda-4>\times<-1,-1,-2\rangle}{\sqrt{(1)^2+(1)^2+(2)^2}}\right| \\ & 12=|-6-\lambda+1+2 \lambda+8| \\ & |\lambda+3|=12 \end{aligned}

 $$ \begin{aligned} & \lambda=9\, ,-15 \end{aligned}

$$ \begin{aligned} & \lambda=9\, (\because \lambda\geq 0) \end{aligned}

\because (\alpha ,\beta ,\gamma ) lies on line L then 

\frac{\alpha -5}{-2}=\frac{\beta -9}{0}=\frac{\gamma +9}{1}=K

$$ \begin{aligned} & \alpha=5-2 \mathrm{~K}, \beta=9 \mathrm{~K}, \gamma=-9+\mathrm{K} \\ & \alpha+2 \gamma,=5-2 \mathrm{~K}-18+2 \mathrm{~K}=-13 \neq 24 \end{aligned}

Therefore \alpha + 2\gamma  =24 is not possible.

Posted by

Gautam harsolia

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2025: 11 from Kota coaching centres among 24 students with 100 percentile
anam-khan

Low JEE Main rank? Explore alternative paths for engineering admission
anam-khan

Gender-gap in JEE Main 2025: Female participation remains around 31%, overall pool drops by 7.09%

Latest Question