First slide

Straight lines 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)

Moderate

A

$y + 2 z = - 1$
B

$y + z = - 1$
C

$y - z = - 1$
D

$y - 2 z = - 1$

Solution

For given lines to be coplanar, we should have $|\begin{matrix} 2 & 0 & 0 \\ 2 & k & 2 \\ 5 & 2 & k \end{matrix}| = 0 \Rightarrow k = \pm 2$
For k = 2, obviously the plane $y + 1 = z$ is common in both lines.
For k = -2, the plane is given by $|\begin{matrix} x - 1 & y + 1 & z \\ 2 & - 2 & 2 \\ 5 & 2 & - 2 \end{matrix}| = 0 \Rightarrow y + z + 1 = 0$

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