# Equation of Plane Containing Two Lines

Here you will learn how to find equation of plane containing two lines with examples.

Let’s begin –

## Equation of Plane Containing Two Lines

### (a) Vector Form

If the lines $$\vec{r}$$ = $$a_1 + \lambda\vec{b_1}$$ and $$\vec{r}$$ = $$a_2 + \mu\vec{b_2}$$ are coplanar, then

$$\vec{r_1}$$.($$\vec{b_1}\times \vec{b_2}$$) = $$\vec{a_2}$$.($$\vec{b_1}\times \vec{b_2}$$)

or,   [$$\vec{r}$$ $$\vec{b_1}$$ $$\vec{b_2}$$] = [$$\vec{a_2}$$ $$\vec{b_1}$$ $$\vec{b_2}$$]

and the equation of the plane containing them is

$$\vec{r_1}$$.($$\vec{b_1}\times \vec{b_2}$$) = $$\vec{a_1}$$.($$\vec{b_1}\times \vec{b_2}$$)

or,   $$\vec{r_1}$$.($$\vec{b_1}\times \vec{b_2}$$) = $$\vec{a_2}$$.($$\vec{b_1}\times \vec{b_2}$$)

### (b) Cartesian Form

If the line $$x – x_1\over l_1$$ = $$y – y_1\over m_1$$ = $$z – z_1\over n_1$$ and $$x – x_2\over l_2$$ = $$y – y_2\over m_2$$ = $$z – z_2\over n_2$$ are coplanar then

$$\begin{vmatrix} x_2 – x_1 & y_2 – y_1 & z_2 – z_1 \\ l_1 & m_1 & n_1 \\ l_2 & m_2 & n_2 \end{vmatrix}$$ = 0

and the equation of the plane containing them is

$$\begin{vmatrix} x – x_1 & y – y_1 & z – z_1 \\ l_1 & m_1 & n_1 \\ l_2 & m_2 & n_2 \end{vmatrix}$$ = 0

or,  $$\begin{vmatrix} x – x_2 & y – y_2 & z – z_2 \\ l_1 & m_1 & n_1 \\ l_2 & m_2 & n_2 \end{vmatrix}$$ = 0

Example : Prove that the lines $$x + 1\over 3$$ = $$y + 3\over 5$$ = $$z + 5\over 7$$ and $$x – 2\over 1$$ = $$y – 4\over 4$$ = $$z – 6\over 7$$ are coplanar. Also, find the plane containing these two lines.

Solution : We know that the lines

$$x – x_1\over l_1$$ = $$y – y_1\over m_1$$ = $$z – z_1\over n_1$$ and $$x – x_2\over l_2$$ = $$y – y_2\over m_2$$ = $$z – z_2\over n_2$$ are coplanar if

$$\begin{vmatrix} x_2 – x_1 & y_2 – y_1 & z_2 – z_1 \\ l_1 & m_1 & n_1 \\ l_2 & m_2 & n_2 \end{vmatrix}$$ = 0

and the equation of the plane containing these two lines is

$$\begin{vmatrix} x – x_1 & y – y_1 & z – z_1 \\ l_1 & m_1 & n_1 \\ l_2 & m_2 & n_2 \end{vmatrix}$$ = 0

Here, $$x_1$$ = -1, $$y_1$$ = -3, $$z_1$$ = -5,  $$x_2$$ = 2, $$y_2$$ = 4, $$z_2$$ = 6,

$$l_1$$ = 3, $$m_1$$ = 5, $$n_1$$ = 7,  $$l_2$$ = 1, $$m_1$$ = 4, $$n_1$$ = 7.

$$\therefore$$ $$\begin{vmatrix} x_2 – x_1 & y_2 – y_1 & z_2 – z_1 \\ l_1 & m_1 & n_1 \\ l_2 & m_2 & n_2 \end{vmatrix}$$ = $$\begin{vmatrix} 3 & 7 & 11 \\ 3 & 5 & 7 \\ 1 & 4 & 7 \end{vmatrix}$$ = 21 – 98 + 77 = 0

So, the given lines are coplanar.

The equation of the plane containing the lines is

$$\begin{vmatrix} x + 1 & y + 3 & z + 5 \\ 3 & 5 & 7 \\ 1 & 4 & 7 \end{vmatrix}$$ = 0

$$\implies$$ (x + 1)(35 – 28) – (y + 3)(21 – 7) + (z + 5)(12 – 5) = 0

$$\implies$$ x – 2y + z = 0