# 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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