First slide

Straight lines in 3D

Question

Equation of the plane through the lines $\frac{x - 1}{2} = \frac{y + 1}{- 2} = \frac{z - 3}{1}$ and $\frac{x - 1}{1} = \frac{y + 1}{2} = \frac{z - 3}{2}$ is

Moderate

A

$2 x + y - 2 z + 5 = 0$
B

$2 x + y - 2 z - 5 = 0$
C

$x + 2 y - 2 z + 7 = 0$
D

$x + 2 y + 2 z - 5 = 0$

Solution

The Equation of the plane through the lines $\frac{x - 1}{2} = \frac{y + 1}{- 2} = \frac{z - 3}{1}$ and $\frac{x - 1}{1} = \frac{y + 1}{2} = \frac{z - 3}{2}$ is $|\begin{matrix} x - x_{1} & y - y_{1} & z - z_{1} \\ l_{1} & m_{1} & n_{1} \\ l_{2} & m_{2} & n_{2} \end{matrix}| = 0$
Hence the equation of the plane required is $|\begin{matrix} x - 1 & y + 1 & z - 3 \\ 2 & - 2 & 1 \\ 1 & 2 & 2 \end{matrix}| = 0$
Simplify
$\begin{matrix} (x - 1) (- 4 - 2) - (y + 1) (3) + (z - 3) (6) = 0 \\ 2 (x - 1) + (y + 1) - 2 (z - 3) = 0 \\ 2 x + y - 2 z + 5 = 0 \end{matrix}$
Therefore, the equation of the required plane is $2 x + y - 2 z + 5 = 0$

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