Answer to Question #200875 in Analytic Geometry for Mohamed

Question

1. X-2/2=y-3/-1=z+4/3
2. X-3=y+1/3=z-1/-2

Find the equation of plane which containing the two lines

Answer 1

Equation of any plane through the first line is 

$$a(x-2)+b(y-3)+c(z+4)=0,$$

where

$$2a-b+3c=0$$

This plane will pass through the second line if a point on the second line $(3, -1, 1)$ lies on it i.e., if 

$$a(3-2)+b(-1-3)+c(1+4)=0.$$

  Then

$$\begin{matrix} 2a-b+3c=0 \\ a-4b+5c=0 \end{matrix}$$

$$\begin{matrix} -7b+7c=0 \\ a=4b-5c \end{matrix}$$

$$\begin{matrix} a=-c \\ b=c \end{matrix}$$

Hence, the euation of required plane is

$$-(x-2)+(y-3)+(z+4)=0$$

$$(x-2)-(y-3)-(z+4)=0$$

Or

$$x-y-z-3=0$$