Learn to derive the equation of a plane in normal form through this lesson. Both, Vector and Cartesian equations of a plane in normal form are covered and explained in simple terms for your understanding. Solved examples at the end of the lesson help you quickly glance through how to tackle exam questions on this topic.

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**The Equation of a Plane in Normal Form**

The concept of planes is integral to three-dimensional geometry. One of the important aspects of learning about planes is to understand what it means to write or express the equation of a plane in **normal form**.

You must note that to be able to write the equation of a plane in normal form, two things are required – you must know the normal to the plane as well as the distance of the plane from the origin. So we can proceed to look at how the equation of a plane is expressed in normal form. As usual, we will look at the Vector form as well as Cartesian form in the following sections.

**Browse more Topics under Three Dimensional Geometry**

- Angle Between a Line and a Plane
- Angle Between Two Lines
- Coplanarity of Two Lines
- Angle Between Two Planes
- Direction Cosines and Direction Ratios of a Line
- Distance Between Parallel Lines
- The Distance Between Two Skew Lines
- Distance of a Point from a Plane
- Equation of a Plane Perpendicular to a Given Vector and Passing Through a Given Point
- The Equation of Line for Space
- Equation of Plane Passing Through Three Non Collinear Points
- Intercept Form of the Equation of a Plane
- Plane Passing Through the Intersection of Two Given Planes

**Vector form**

The vector equation of a plane in normal form is

$$ \vec{r} . \hat{n} = d $$

Here \( \vec{r} \) is the position vector of a point lying on the said plane; \( \hat{n} \) is a unit normal vector parallel to the normal that joins the origin to the plane *(a unit vector is a vector whose magnitude is unity)* and, ‘d’ is the perpendicular distance of the plane of the plane from the origin.

So, if you are given a point A (x, y, z) that lies on a plane, then the vector \( \vec{OA} \) where O is the origin, can be written as –

$$ \vec{OA} = x \hat{i} + y \hat{j} + z \hat{k} $$

We can write \( \hat{n} \) as –

$$ \hat{n} = l \hat{i} + m \hat{j} + n \hat{k} $$ where l, m, n are the direction cosines of the unit vector. Combining the above two equations, we can write the equation of a plane in the normal form as under –

$$ (x \hat{i} + y \hat{j} + z \hat{k} ) . ( l \hat{i} + m \hat{j} + n \hat{k} ) = d $$

This is the equation in vector form.

**Cartesian Form**

The Cartesian equation of a plane in normal form is

lx + my + nz = d

where l, m, n are the direction cosines of the unit vector parallel to the normal to the plane; (x,y,z) are the coordinates of the point on a plane and, ‘d’ is the distance of the plane from the origin. You can now follow a worked out problem as shown below to understand how to solve questions on the topic.

**Solved Example for You**

**Question 1: Find the vector equation of a plane whose normal is by \( 5 \hat{i} + 3 \hat{j} – 2 hat{k} \) and is at a distance of 9 / \( \sqrt{38} \) from the origin.**

**Answer:** We have the normal vector as

$$ \vec{n} = 5 \hat{i} + 3 \hat{j} – 2 \hat{k} $$

To find the unit vector –

$$ \hat{n} = \vec{n} / | vec{n} | = 5 \hat{i} + 3 \hat{j} – 2 \hat{k} / \sqrt{ (25 + 9 + 4) } = 5 \hat{i} + 3 \hat{j} – 2 \hat{k} / \sqrt{38} $$

Thus, we shall now use the formula for the vector equation of a plane in normal form and derive at the required equation –

$$ \vec{n} = \vec{r} . ( \frac{5}{ \sqrt{38} } \hat{i} + \frac{3}{ \sqrt{38} } \hat{j} + \frac{-2}{ \sqrt{38} } \hat{k} ) = \frac{9}{ \sqrt{38} } $$

**Question 2: What is the equation of a plane in 3D?**

**Answer:** We can define it as the space that has three points (that don’t lie on the same line) or by a point and a normal vector to the plane. We also know it as the vector equation of a plane. In addition, the general equation of a plane in 3D space is A ∙ 0 + B ∙ 0 + C ∙ 0 + D = 0 => D = 0.

**Question 3: What is the equation of the XY plane?**

**Answer:** In a z-coordinate of any point on the x-y plane is always 0. Therefore, the equation z = 0 can denote every point that has its z-coordinate equal to 0. Hence the equation z = 0 represents the entire x-y plane.

**Question 4: What is the general equation of a plane?**

**Answer:** When you know the normal vector of a plane and a point passing through the plane, the equation of the plane is established as a (x – x1) + b (y– y1) + c (z –z1) = 0.

**Question 5: What does the XY plane mean?**

**Answer:**The XY plane refers to a plane that contains the x- and y-axis. Moreover, the yz-plane contains the y- and z-axis, and xz-plane contains the x- and z-axis.

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