Here you will learn how to find equation of plane containing two lines with examples.
Letโs begin โ
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
\(\vec{r_1}\).(\(\vec{b_1}\times \vec{b_2}\)) = \(\vec{a_2}\).(\(\vec{b_1}\times \vec{b_2}\))ย
or,ย ย [\(\vec{r}\) \(\vec{b_1}\) \(\vec{b_2}\)] = [\(\vec{a_2}\) \(\vec{b_1}\) \(\vec{b_2}\)]
and the equation of the plane containing them is
\(\vec{r_1}\).(\(\vec{b_1}\times \vec{b_2}\)) = \(\vec{a_1}\).(\(\vec{b_1}\times \vec{b_2}\))ย
or,ย ย \(\vec{r_1}\).(\(\vec{b_1}\times \vec{b_2}\)) = \(\vec{a_2}\).(\(\vec{b_1}\times \vec{b_2}\))ย
(b) Cartesian Form
If the line \(x โ x_1\over l_1\) = \(y โ y_1\over m_1\) = \(z โ z_1\over n_1\) and \(x โ x_2\over l_2\) = \(y โ y_2\over m_2\) = \(z โ z_2\over n_2\) are coplanar then
\(\begin{vmatrix} x_2 โ x_1 & y_2 โ y_1 &ย z_2 โ z_1 \\ย l_1 & m_1 & n_1 \\ย l_2 & m_2 & n_2 \end{vmatrix}\) = 0
and the equation of the plane containing them is
\(\begin{vmatrix} x โ x_1 & y โ y_1 &ย z โ z_1 \\ย l_1 & m_1 & n_1 \\ย l_2 & m_2 & n_2 \end{vmatrix}\) = 0ย
or,ย \(\begin{vmatrix} x โ x_2 & y โ y_2 &ย z โ z_2 \\ย l_1 & m_1 & n_1 \\ย l_2 & m_2 & n_2 \end{vmatrix}\) = 0
Example : Prove that the lines \(x + 1\over 3\) = \(y + 3\over 5\) = \(z + 5\over 7\) and \(x โ 2\over 1\) = \(y โ 4\over 4\) = \(z โ 6\over 7\) are coplanar. Also, find the plane containing these two lines.
Solution : We know that the lines
\(x โ x_1\over l_1\) = \(y โ y_1\over m_1\) = \(z โ z_1\over n_1\) and \(x โ x_2\over l_2\) = \(y โ y_2\over m_2\) = \(z โ z_2\over n_2\) are coplanar if
\(\begin{vmatrix} x_2 โ x_1 & y_2 โ y_1 &ย z_2 โ z_1 \\ย l_1 & m_1 & n_1 \\ย l_2 & m_2 & n_2 \end{vmatrix}\) = 0
and the equation of the plane containing these two lines is
\(\begin{vmatrix} x โ x_1 & y โ y_1 &ย z โ z_1 \\ย l_1 & m_1 & n_1 \\ย l_2 & m_2 & n_2 \end{vmatrix}\) = 0
Here, \(x_1\) = -1, \(y_1\) = -3, \(z_1\) = -5,ย \(x_2\) = 2, \(y_2\) = 4, \(z_2\) = 6,
ย \(l_1\) = 3, \(m_1\) = 5, \(n_1\) = 7,ย \(l_2\) = 1, \(m_1\) = 4, \(n_1\) = 7.
\(\therefore\) \(\begin{vmatrix} x_2 โ x_1 & y_2 โ y_1 &ย z_2 โ z_1 \\ย l_1 & m_1 & n_1 \\ย l_2 & m_2 & n_2 \end{vmatrix}\) = \(\begin{vmatrix} 3 & 7 &ย 11 \\ย 3 & 5 & 7 \\ย 1 & 4 & 7 \end{vmatrix}\) = 21 โ 98 + 77 = 0
So, the given lines are coplanar.
The equation of the plane containing the lines is
\(\begin{vmatrix} x + 1 & y + 3 &ย z + 5 \\ย 3 & 5 & 7 \\ย 1 & 4 & 7 \end{vmatrix}\) = 0
\(\implies\) (x + 1)(35 โ 28) โ (y + 3)(21 โ 7) + (z + 5)(12 โ 5) = 0
\(\implies\) x โ 2y + z = 0
