If the straight lines x $$=$$ $$1$$ $$+$$ s, y $$=$$ –$$3$$ –λ s, z $$=$$ $$1$$ $$+$$λ s and $$x=t2$$,$$y=1+t$$,$$z=2$$−t with parameters ‘s’ and ‘t’ respectively are coplanar then λ $$=$$

Given lines are x−$$11=y+3$$−λ$$=z$$−$$1$$λ,$$x12=y$$−$$11=z$$−$$2$$−$$1Given$$ lines are coplanar⇒$$1$$−$$4$$−$$11$$−λλ$$121$$−$$1=0$$, ⇒λ$$=$$−$$2$$

Straight lines in 3D

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Question

If the straight lines x = 1 + s, y = –3 – $λ$ s, z = 1 + $λ$ s and $x = \frac{t}{2}, y = 1 + t, z = 2 - t$ with parameters ‘s’ and ‘t’ respectively are coplanar then
$λ$ =

Moderate

Solution

Given lines are $\frac{x - 1}{1} = \frac{y + 3}{- λ} = \frac{z - 1}{λ},$
$\frac{x}{\frac{1}{2}} = \frac{y - 1}{1} = \frac{z - 2}{- 1}$
Given lines are coplanar
$\Rightarrow |\begin{matrix} 1 & - 4 & - 1 \\ 1 & - λ & λ \\ \frac{1}{2} & 1 & - 1 \end{matrix}| = 0$ , $\Rightarrow λ = - 2$

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