Question

Find the equation of the plane passing through the points (−1, 2, 0), (2, 2, −1) and parallel to the line $\frac{x-1}{1}=\frac{2y+1}{2}=\frac{z+1}{-1}$. [CBSE 2015]

Answer 1

The general equation of the plane passing through the point (−1, 2, 0) is given by

$a\left(x+1\right)+b\left(y-2\right)+c\left(z-0\right)=0$ .....(1)

If this plane passes through the point (2, 2, −1), we have

$$\begin{gathered} a\left(2+1\right)+b\left(2-2\right)+c\left(-1-0\right)=0 \\ \Rightarrow 3a-c=0.....\left(2\right) \end{gathered}$$

Direction ratio's of the normal to the plane (1) are a, b, c.

The equation of the given line is $\frac{x-1}{1}=\frac{2y+1}{2}=\frac{z+1}{-1}$. This can be re-written as
$\frac{x-1}{1}=\frac{y+{\displaystyle \frac{1}{2}}}{1}=\frac{z+1}{-1}$

Direction ratio's of the line are 1, 1, −1.

The required plane is parallel to the given line when the normal to this plane is perpendicular to this line.

$$\begin{gathered} \therefore a\times 1+b\times 1+c\times \left(-1\right)=0 \\ \Rightarrow a+b-c=0.....\left(3\right) \end{gathered}$$

Solving (2) and (3), we get

$$\begin{gathered} \frac{a}{0+1}=\frac{b}{-1+3}=\frac{c}{3-0} \\ \Rightarrow \frac{a}{1}=\frac{b}{2}=\frac{c}{3}=\lambda \left(\mathrm{Say}\right) \\ \Rightarrow a=\lambda ,b=2\lambda ,c=3\lambda \end{gathered}$$

Putting these values of a, b, c in (1), we have

$$\begin{gathered} \lambda \left(x+1\right)+2\lambda \left(y-2\right)+3\lambda \left(z-0\right)=0 \\ \Rightarrow x+1+2y-4+3z=0 \\ \Rightarrow x+2y+3z=3 \end{gathered}$$

Thus, the equation of the required plane is x + 2y + 3z = 3.