Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • If the straight lines   <img alt="x=1+s,y=-3-\lambda s,z=1+\lambda s\; and\; x=\frac{t}{2},y=1+t,z=2-t," src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cdpi%7B100%7D%20x%3D1&plus;s

If the straight lines   x=1+s,y=-3-\lambda s,z=1+\lambda s\; and\; x=\frac{t}{2},y=1+t,z=2-t,   with parameters s and t respectively, are co­planar, then \lambda equals

Option: 1

–1/2


Option: 2

–1


Option: 3

–2


Option: 4

0


Answers (1)

best_answer

 

Cartesian eqution of a line -

The equation of a line passing through two points A\left ( x_{0},y_{0},z_{0} \right )and parallel to vector having direction ratios as \left ( a,b,c \right )is given by

\frac{x-x_{0}}{a}= \frac{y-y_{0}}{b}= \frac{z-z_{0}}{c}

The equation of a line passing through two points A\left ( x_{1},y_{1},z_{1} \right )\, and \, B\left ( x_{2},y_{2},z_{2} \right ) is given by

\frac{x-x_{1}}{x_{2}-x_{1}}= \frac{y-y}{y_{2}-y_{1}}=\frac{z-z_{1}}{z_{2}-z_{1}}

- wherein

 

 

x-1= \frac{y+3}{-\lambda }= \frac{z-1}{\lambda }..........(i)

& \frac{x}{\left ( 1/2 \right )}=y-1= \frac{z-2}{-1}...........(ii)

For Coplanarity

 

\left [ \left ( \overrightarrow{a_2}-\overrightarrow{a_1}\right )\overrightarrow{b} \overrightarrow{c} \right ]= 0

\Rightarrow \begin{vmatrix} 1 &-4 & -1\\1 &-\lambda & \lambda \\ 1/2 & 1 & -1 \end{vmatrix}= 0

\Rightarrow 4\left ( -1-\frac{\lambda }{2} \right )-\left ( 1+\lambda /2 \right )= 0

\Rightarrow 5= \frac{-5\lambda }{2}\Rightarrow \lambda = -2

Posted by

Ritika Kankaria

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

BTech Admission 2025 Live: JEE Main correction window closes at 11:50 pm; VITEEE, SRMJEE, AEEE dates out
anam-khan

JEE Main 2025 application form correction begins at jeemain.nta.nic.in; editable fields
anam-khan

JEE Main 2025 application form correction window opens today; apply at jeemain.nta.nic.in

Latest Question