Equation of Plane in Vector form

Here you will learn what is the equation of plane in vector form with examples.

Let begin –

Equation of Plane in Vector form

The vector equation of a plane passing through a point having position vector \(\vec{a}\) and normal to vector \(\vec{n}\) is 

\((\vec{r} – \vec{a}).\vec{n}\) = 0   or,  \(\vec{r}\).\(\vec{n}\) = \(\vec{a}\).\(\vec{n}\).

Note 1 : 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.

Note 2 : 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.

Reduction to Cartesian Form

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 \(\vec{n}\) = \(n_1\hat{i} + n_2\hat{j} + n_3\hat{k}\)

Then, \(\vec{r}\) – \(\vec{a}\) = \((x – a_1)\hat{i} + (y – a_2)\hat{j} + (z – a_3)\hat{k}\)

Substituting the values of (\(\vec{r}\) – \(\vec{a}\)) and \(\vec{n}\) in equation (\(\vec{r}\) – \(\vec{a}\)).\(\vec{n}\) = 0, we get

\(\implies\) \((x – a_1)n_1+ (y – a_2)n_2 + (z – a_3)n_3\) = 0

This is the cartesian equation of a plane passing through \((a_1, a_2, a_3)\).

Example : 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.

Solution : 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, \(\vec{r}\).\(\vec{n}\) = \(\vec{a}\).\(\vec{n}\).

Here, \(\vec{a}\) = \(2\hat{i} + 3\hat{j} – 4\hat{k}\) and \(\vec{n}\) = \(2\hat{i} – \hat{j} + 2\hat{k}\).

So, the equation of the required plane is

\(\vec{r}\).(\(2\hat{i} – \hat{j} + 2\hat{k}\)) = (\(2\hat{i} + 3\hat{j} – 4\hat{k}\)).(\(2\hat{i} – \hat{j} + 2\hat{k}\))

[ By using \(\vec{r}\).\(\vec{n}\) = \(\vec{a}\).\(\vec{n}\) ]

\(\implies\) \(\vec{r}\).(\(2\hat{i} – \hat{j} + 2\hat{k}\)) = 4 – 3 – 8

\(\implies\) \(\vec{r}\).(\(2\hat{i} – \hat{j} + 2\hat{k}\)) = -7                                 ………..(i)

Redution to Cartesian form : 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

(\(x\hat{i} + y\hat{j} + z\hat{k}\)).(\(2\hat{i} – \hat{j} + 2\hat{k}\)) = -7

\(\implies\) 2x – y + 2z = -7

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