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Question: If the straight lines \(x = 1 + s , y = 3 - \lambda s , z = 1 + \lambda s\) and \(x = \frac { t } {...

If the straight lines x=1+s,y=3−λs,z=1+λsx = 1 + s , y = 3 - \lambda s , z = 1 + \lambda s and

x=t2,y=1+t,z=2−tx = \frac { t } { 2 } , y = 1 + t , z = 2 - t, with parameters s and t respectively, are co-planar, then λ equals

A

0

B

–1

C

−12- \frac { 1 } { 2 }

D

–2

Answer

–2

Explanation

Solution

We have

x−11=y+3−λ=z−1λ=s\frac { x - 1 } { 1 } = \frac { y + 3 } { - \lambda } = \frac { z - 1 } { \lambda } = sand 2x1=y−11=z−2−1=t\frac { 2 x } { 1 } = \frac { y - 1 } { 1 } = \frac { z - 2 } { - 1 } = t

i.e. x−01=y−12=z−2−2=t2\frac { x - 0 } { 1 } = \frac { y - 1 } { 2 } = \frac { z - 2 } { - 2 } = \frac { t } { 2 }

Since, lines are co-planar,

Then, ∣x2−x1y2−y1z2−z1l1m1n1l2m2n2∣=0\left| \begin{array} { c c c } x _ { 2 } - x _ { 1 } & y _ { 2 } - y _ { 1 } & z _ { 2 } - z _ { 1 } \\ l _ { 1 } & m _ { 1 } & n _ { 1 } \\ l _ { 2 } & m _ { 2 } & n _ { 2 } \end{array} \right| = 0

⇒ ∣−1411−λλ12−2∣=0\left| \begin{array} { c c c } - 1 & 4 & 1 \\ 1 & - \lambda & \lambda \\ 1 & 2 & - 2 \end{array} \right| = 0

On solving, λ=−2\lambda = - 2.