If the straight lines \frac{x - 1}{2} = \frac{y + 1}{k} = \frac{z}{2} and \frac{x + 1}{5} = \frac{y + 1}{2} = \frac{z}{k} are coplanar, then the plane(s) containing these two lines is(are)

Engineering

Mathematics

Equation of Straight Line in 3D

Question

If the straight lines $\frac{x - 1}{2} = \frac{y + 1}{k} = \frac{z}{2}$ and $\frac{x + 1}{5} = \frac{y + 1}{2} = \frac{z}{k}$ are coplanar, then the plane(s) containing these two lines is(are)

y + 2z = – 1

y + z = – 1

y – z = – 1

y – 2z = – 1

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Plane containing the given lines is $| \begin{matrix} x - 1 & y + 1 & z \\ 2 & k & 2 \\ 5 & 2 & k \end{matrix} |$ = 0

$\Rightarrow$ (x – 1) (k² – 4) – (y + 1) (2k – 10) + z (4 – 5k) = 0 .......(1)

which also contains the point (–1, – 1, 0)

– 2 (k² – 4) – 0 + 0 = 0 $\Rightarrow$ k = ± 2

Putting k = 2 in (1)

0 – (y + 1) (– 6) + z (–6) = 0 $\Rightarrow$ y + 1 – z = 0

Again putting k = – 2 in (1)

0 – (y + 1) (–14) + z (14) = 0 $\Rightarrow$ y + 1 + z = 0.