{"id":6378,"date":"2021-10-13T18:13:41","date_gmt":"2021-10-13T12:43:41","guid":{"rendered":"https:\/\/mathemerize.com\/?p=6378"},"modified":"2021-10-14T22:47:20","modified_gmt":"2021-10-14T17:17:20","slug":"equation-of-plane-passing-through-intersection-of-two-planes","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/equation-of-plane-passing-through-intersection-of-two-planes\/","title":{"rendered":"Equation of Plane Passing Through Intersection of Two Planes"},"content":{"rendered":"
Here you will learn what is the equation of plane passing through intersection of two planes with examples.<\/p>\n
Let’s begin –<\/p>\n
The equation of a plane passing through the intersection of the planes \\(\\vec{r}.\\vec{n_1}\\) = \\(d_1\\) and \\(\\vec{r}.\\vec{n_2}\\) = \\(d_2\\) is given by<\/p>\n
\n(\\(\\vec{r}.\\vec{n_1}\\) – \\(d_1\\)) + \\(\\lambda\\)(\\(\\vec{r}.\\vec{n_2}\\) – \\(d_2\\)) = 0<\/p>\n
or, \\(\\vec{r}\\).(\\(\\vec{n_1} + \\lambda\\vec{n_2}\\)) = \\(d_1 + \\lambda d_2\\),<\/p>\n
where \\(\\lambda\\) is an arbitrary constant.<\/p>\n<\/blockquote>\n
Example<\/span><\/strong> : Find the equation of the plane containing the line of intersection of the planes \\(\\vec{r}\\).(\\(\\hat{i} + 3\\hat{j} – \\hat{k}\\)) = 5 and \\(\\vec{r}\\).(\\(2\\hat{i} – \\hat{j} + \\hat{k}\\)) = 3 and passing through the point (2, 1, -2),<\/p>\n
Solution<\/span><\/strong> : The equation of the plane through the line of intersection of the given planes is<\/p>\n
[\\(\\vec{r}\\).(\\(\\hat{i} + 3\\hat{j} – \\hat{k}\\)) – 5] + \\(\\lambda\\)[\\(\\vec{r}\\).(\\(2\\hat{i} – \\hat{j} + \\hat{k}\\)) – 3] = 0 <\/p>\n
\\(\\implies\\) \\(\\vec{r}\\).[\\(1 + 2\\lambda)\\hat{i} + (3 – \\lambda)\\hat{j} + (-1 + \\lambda)\\hat{k}\\)] – 5 – 3\\(\\lambda\\) = 0 ………..(i)<\/p>\n
If plane in (i) passes through (2, 1, -2), then the vector \\(2\\hat{i} + \\hat{j} – 2\\hat{k}\\) should satisfy it<\/p>\n
(\\(2\\hat{i} + \\hat{j} – 2\\hat{k}\\)).[\\(1 + 2\\lambda)\\hat{i} + (3 – \\lambda)\\hat{j} + (-1 + \\lambda)\\hat{k}\\)] – (5 + 3\\(\\lambda\\)) = 0<\/p>\n
\\(\\implies\\) \\(-2\\lambda + 2\\) = 0 \\(\\implies\\) \\(\\lambda\\) = 1 <\/p>\n
Putting \\(\\lambda\\) = 1 in equation (i), we obtain the equation of the required plane as<\/p>\n
\\(\\vec{r}\\).(\\(3\\hat{i} + 2\\hat{j} + 0\\hat{k}\\)) = 8.<\/p>\n
(b) Cartesian Form<\/h3>\n
The equation of a plane passing through the intersection of planes \\(a_1x + b_1y + c_1z + d_1\\) = 0 and \\(a_2x + b_2y + c_2z + d_2\\) = 0 is<\/p>\n
\n(\\(a_1x + b_1y + c_1z + d_1\\)) + \\(\\lambda\\)(\\(a_2x + b_2y + c_2z + d_2\\)) = 0,<\/p>\n
where \\(\\lambda\\) is a constant.<\/p>\n<\/blockquote>\n
Example<\/span><\/strong> : Find the equation of the plane containing the line of intersection of the planes x + y + z – 6 = 0 and 2x + 3y + 4z + 5 = 0 and passing through the point (1, 1, 1),<\/p>\n
Solution<\/span><\/strong> : The equation of the plane through the line of intersection of the given planes is<\/p>\n
(x + y + z – 6) + \\(\\lambda\\)(2x + 3y + 4z + 5) = 0 ………..(i)<\/p>\n
If (i) passes through (1, 1, 1), then<\/p>\n
-3 + 14 \\(\\lambda\\) = 0 \\(\\implies\\) \\(\\lambda\\) = 3\/14<\/p>\n
Putting \\(\\lambda\\) = 3\/14 in equation (i), we obtain the equation of the required plane as<\/p>\n
(x + y + z – 6) + \\(3\\over 14\\) (2x + 3y + 4z + 5) = 0<\/p>\n
or, 20x + 23y + 26z – 69 = 0.<\/p>\n\n\n