{"id":6370,"date":"2021-10-13T18:00:41","date_gmt":"2021-10-13T12:30:41","guid":{"rendered":"https:\/\/mathemerize.com\/?p=6370"},"modified":"2021-10-14T22:45:10","modified_gmt":"2021-10-14T17:15:10","slug":"equation-of-plane-in-vector-form","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/equation-of-plane-in-vector-form\/","title":{"rendered":"Equation of Plane in Vector form"},"content":{"rendered":"
Here you will learn what is the equation of plane in vector form with examples.<\/p>\n
Let begin –<\/p>\n
\nThe vector equation of a plane passing through a point having position vector \\(\\vec{a}\\) and normal to vector \\(\\vec{n}\\) is <\/p>\n
\\((\\vec{r} – \\vec{a}).\\vec{n}\\) = 0 or, \\(\\vec{r}\\).\\(\\vec{n}\\) = \\(\\vec{a}\\).\\(\\vec{n}\\).<\/p>\n<\/blockquote>\n
Note 1<\/strong> : It is to note here that vector equation of a plane means a relation involving the position vector \\(\\vec{r}\\) of an arbitrary point on the plane.<\/p>\n
Note 2<\/strong> : The above equation can also be written as \\(\\vec{r}\\).\\(\\vec{n}\\) = \\(\\vec{d}\\), where \\(\\vec{d}\\) = \\(\\vec{a}\\).\\(\\vec{n}\\). This is also known as scalar product form of a plane<\/strong>.<\/p>\n
Reduction to Cartesian Form<\/h3>\n
If \\(\\vec{r}\\) = \\(x\\hat{i} + y\\hat{j} + z\\hat{k}\\), \\(\\vec{a}\\) = \\(a_1\\hat{i} + a_2\\hat{j} + a_3\\hat{k}\\) and \\(\\vec{n}\\) = \\(n_1\\hat{i} + n_2\\hat{j} + n_3\\hat{k}\\)<\/p>\n
Then, \\(\\vec{r}\\) – \\(\\vec{a}\\) = \\((x – a_1)\\hat{i} + (y – a_2)\\hat{j} + (z – a_3)\\hat{k}\\)<\/p>\n
Substituting the values of (\\(\\vec{r}\\) – \\(\\vec{a}\\)) and \\(\\vec{n}\\) in equation (\\(\\vec{r}\\) – \\(\\vec{a}\\)).\\(\\vec{n}\\) = 0, we get<\/p>\n
\\(\\implies\\) \\((x – a_1)n_1+ (y – a_2)n_2 + (z – a_3)n_3\\) = 0<\/p>\n
This is the cartesian equation of a plane passing through \\((a_1, a_2, a_3)\\).<\/p>\n
Example<\/strong><\/span> : Find the vector equation of a plane passing through a point having position vector \\(2\\hat{i} + 3\\hat{j} – 4\\hat{k}\\) and perpendicular to the vector \\(2\\hat{i} – \\hat{j} + 2\\hat{k}\\). Also reduce it to cartesian form.<\/p>\n
Solution<\/strong><\/span> : We know that the cartesian equation of a plane passing through a point \\(\\vec{a}\\) and normal to \\(\\vec{n}\\) is \\((\\vec{r} – \\vec{a}).\\vec{n}\\) = 0 or, \\(\\vec{r}\\).\\(\\vec{n}\\) = \\(\\vec{a}\\).\\(\\vec{n}\\).<\/p>\n
Here, \\(\\vec{a}\\) = \\(2\\hat{i} + 3\\hat{j} – 4\\hat{k}\\) and \\(\\vec{n}\\) = \\(2\\hat{i} – \\hat{j} + 2\\hat{k}\\).<\/p>\n
So, the equation of the required plane is<\/p>\n
\\(\\vec{r}\\).(\\(2\\hat{i} – \\hat{j} + 2\\hat{k}\\)) = (\\(2\\hat{i} + 3\\hat{j} – 4\\hat{k}\\)).(\\(2\\hat{i} – \\hat{j} + 2\\hat{k}\\))<\/p>\n
[ By using \\(\\vec{r}\\).\\(\\vec{n}\\) = \\(\\vec{a}\\).\\(\\vec{n}\\) ]<\/p>\n
\\(\\implies\\) \\(\\vec{r}\\).(\\(2\\hat{i} – \\hat{j} + 2\\hat{k}\\)) = 4 – 3 – 8<\/p>\n
\\(\\implies\\) \\(\\vec{r}\\).(\\(2\\hat{i} – \\hat{j} + 2\\hat{k}\\)) = -7 ………..(i)<\/p>\n
Redution to Cartesian form<\/strong> : Since \\(\\vec{r}\\) denotes the position vector of an arbitrary point (x, y, z) on the plane. Therefore, putting \\(\\vec{r}\\) = \\(x\\hat{i} + y\\hat{j} + z\\hat{k}\\) in equation (i), we obtain<\/p>\n
(\\(x\\hat{i} + y\\hat{j} + z\\hat{k}\\)).(\\(2\\hat{i} – \\hat{j} + 2\\hat{k}\\)) = -7<\/p>\n
\\(\\implies\\) 2x – y + 2z = -7<\/p>\n\n\n