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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

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

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

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

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

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Ritika Kankaria

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