Question

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

 

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 

\[a\left( 2 + 1 \right) + b\left( 2 - 2 \right) + c\left( - 1 - 0 \right) = 0\]
\[ \Rightarrow 3a - c = 0 ...............\left( 2 \right)\]

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

\[\frac{x - 1}{1} = \frac{y + \frac{1}{2}}{1} = \frac{z + 1}{- 1}\] 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.

\[\therefore a \times 1 + b \times 1 + c \times \left( - 1 \right) = 0\]
\[ \Rightarrow a + b - c = 0 . . . . . \left( 3 \right)\]

Solving (2) and (3), we get

\[\frac{a}{0 + 1} = \frac{b}{- 1 + 3} = \frac{c}{3 - 0}\]
\[ \Rightarrow \frac{a}{1} = \frac{b}{2} = \frac{c}{3} = \lambda\left( \text{ Say } \right)\]
\[ \Rightarrow a = \lambda, b = 2\lambda, c = 3\lambda\]

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

\[\lambda\left( x + 1 \right) + 2\lambda\left( y - 2 \right) + 3\lambda\left( z - 0 \right) = 0\]

\[ \Rightarrow x + 1 + 2y - 4 + 3z = 0\]

\[ \Rightarrow x + 2y + 3z = 3\]

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