First slide

Straight lines in 3D

Question

Which of the following lines lies on the plane $x + 2 y - z + 4 = 0$

Moderate

A

$\frac{x - 1}{1} = \frac{y}{- 1} = \frac{z - 5}{- 1}$
B

$x - y + z = 0 = 2 x + y - z$
C

$\frac{x - 2}{1} = \frac{y}{2} = \frac{z - 5}{3}$
D

$\frac{x - 2}{1} = \frac{y}{- 2} = \frac{z - 5}{- 3}$

Solution

The condition that the line $\frac{x - x_{1}}{l} = \frac{y - y_{1}}{l} = \frac{z - z_{1}}{n}$ lies on the plane $a x + b y + c z + d = 0$ is $a l + b m + c n = 0$ and $a x_{1} + b y_{1} + c z_{1} + d = 0$
Verify each option and then finalize the answer.
The given plane is $x + 2 y - z + 4 = 0$
For the option(1), $1 (1) + 2 (- 1) - 1 (- 1) = 1 - 2 + 1 = 0$
and $1 (1) + 2 (0) - 1 (5) + 4 = 1 - 5 + 4 = 0$
Therefore, the line $\frac{x - 1}{1} = \frac{y}{- 1} = \frac{z - 5}{- 1}$ lies on the plane $x + 2 y - z + 4 = 0$ , , the option (1) is the answer.

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