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

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