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