{"id":6388,"date":"2021-10-13T18:21:46","date_gmt":"2021-10-13T12:51:46","guid":{"rendered":"https:\/\/mathemerize.com\/?p=6388"},"modified":"2021-10-14T22:49:59","modified_gmt":"2021-10-14T17:19:59","slug":"equation-of-plane-containing-two-lines","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/equation-of-plane-containing-two-lines\/","title":{"rendered":"Equation of Plane Containing Two Lines"},"content":{"rendered":"
Here you will learn how to find equation of plane containing two lines with examples.<\/p>\n
Let’s begin –<\/p>\n
If the lines \\(\\vec{r}\\) = \\(a_1 + \\lambda\\vec{b_1}\\) and \\(\\vec{r}\\) = \\(a_2 + \\mu\\vec{b_2}\\) are coplanar, then<\/p>\n
\n\\(\\vec{r_1}\\).(\\(\\vec{b_1}\\times \\vec{b_2}\\)) = \\(\\vec{a_2}\\).(\\(\\vec{b_1}\\times \\vec{b_2}\\)) <\/p>\n
or, [\\(\\vec{r}\\) \\(\\vec{b_1}\\) \\(\\vec{b_2}\\)] = [\\(\\vec{a_2}\\) \\(\\vec{b_1}\\) \\(\\vec{b_2}\\)]<\/p>\n<\/blockquote>\n
and the equation of the plane containing them is<\/p>\n
\n\\(\\vec{r_1}\\).(\\(\\vec{b_1}\\times \\vec{b_2}\\)) = \\(\\vec{a_1}\\).(\\(\\vec{b_1}\\times \\vec{b_2}\\)) <\/p>\n
or, \\(\\vec{r_1}\\).(\\(\\vec{b_1}\\times \\vec{b_2}\\)) = \\(\\vec{a_2}\\).(\\(\\vec{b_1}\\times \\vec{b_2}\\)) <\/p>\n<\/blockquote>\n
(b) Cartesian Form<\/h3>\n
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<\/p>\n
\n\\(\\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<\/p>\n<\/blockquote>\n
and the equation of the plane containing them is<\/p>\n
\n\\(\\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 <\/p>\n
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<\/p>\n<\/blockquote>\n
Example<\/strong><\/span> : 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.<\/p>\n
Solution<\/span><\/strong> : We know that the lines<\/p>\n
\\(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<\/p>\n
\\(\\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<\/p>\n
and the equation of the plane containing these two lines is<\/p>\n
\\(\\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<\/p>\n
Here, \\(x_1\\) = -1, \\(y_1\\) = -3, \\(z_1\\) = -5, \\(x_2\\) = 2, \\(y_2\\) = 4, \\(z_2\\) = 6,<\/p>\n
\\(l_1\\) = 3, \\(m_1\\) = 5, \\(n_1\\) = 7, \\(l_2\\) = 1, \\(m_1\\) = 4, \\(n_1\\) = 7.<\/p>\n
\\(\\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<\/p>\n
So, the given lines are coplanar.<\/p>\n
The equation of the plane containing the lines is<\/p>\n
\\(\\begin{vmatrix} x + 1 & y + 3 & z + 5 \\\\ 3 & 5 & 7 \\\\ 1 & 4 & 7 \\end{vmatrix}\\) = 0<\/p>\n
\\(\\implies\\) (x + 1)(35 – 28) – (y + 3)(21 – 7) + (z + 5)(12 – 5) = 0<\/p>\n
\\(\\implies\\) x – 2y + z = 0<\/p>\n\n\n