# Equation of a Plane Passing through Two Parallel Lines

## Equation of a Plane Passing through Two Parallel Lines

Let the plane pass through parallel lines $$\vec{r}=\vec{a}+\lambda \vec{b}$$ and $$\vec{r}=\vec{c}+\mu \vec{b}$$.

As shown in the diagram, for any position of R in the plane, vectors $$\overrightarrow{RA}$$, $$\overrightarrow{AC}$$ and $$\vec{b}$$ are coplanar. Then $$\left[ \vec{r}-\vec{a}\vec{c}-\vec{a}\vec{b} \right]=0$$, which is the required equation of the plane.

Equation of Plane Parallel to a Given Plane:

The general equation of the plane parallel to the plane ax + by + cz + d = 0 is ax + by + cz + k = 0, where k is any scalar, as normal to the both planes is $$a\hat{i}+b\hat{j}+c\hat{k}$$.

Example: Find the equation of the plane passing through (3, 4, -1), which is parallel to the plane $$\vec{r}.\left( 2\hat{i}-3\hat{j}+5\hat{k} \right)+7=0$$.

Solution: The equation of any plane which is parallel to $$\vec{r}.\left( 2\hat{i}-3\hat{j}+5\hat{k} \right)+7=0$$ is

$$\vec{r}.\left( 2\hat{i}-3\hat{j}+5\hat{k} \right)+\lambda =0$$ … (i)

2x – 3y + 5z + λ = 0

(3, 4, -1) is passing through equation (i)

2 (3) – 3 (4) + 5 (-1) + λ = 0

6 – 12 – 5 + λ = 0

λ = 11

Required equation is  $$\vec{r}.\left( 2\hat{i}-3\hat{j}+5\hat{k} \right)+11=0$$.