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# Normal Vector

In two and three-dimensional geometry, we commonly come across with the term "normal vector" or simply "normal" to a line or a plane. The normal vectors are also being utilized in order to find light orientation towards a surface for shading purpose in computer graphics. In this article, we are going to learn about a normal vector in detail. We shall understand how to find the normal vector to a line and a plane.

 Related Calculators Normality Calculations Calculate Vector Area under Normal Curve Calculator Cumulative Normal Distribution Calculator

## Definition

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The normal vector is a vector that is perpendicular at a given point on something which can be a line, a plane or surface of any object. A vector that is inclined at right angles to a point on a surface is said to be a normal vector. In the image below, two normal vectors on a flat surface are denoted by $n_{1}$ and $n_{2}$. In the case of closed surfaces, a normal vector pointing inwards, and a normal vector pointing outwards are generally two different vectors.

The "unit normal" is a vector that is determined by dividing the normal vector by the value of its norm which is called normalizing a normal vector.

## Normal Vector to a Line

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A normal vector $\vec{n}$ to a line ax + by = c at a given point is shown in the following diagram. The normal vector to the given line would be represented by $\vec{n}$ = (a, b) and $\vec{n} = (-a, -b). The corresponding unit normal vector is then denoted by the following :$(\frac{a}{\sqrt{a^{2}+b^{2}}}, \frac{b}{\sqrt{a^{2}+b^{2}}})$and$(-\frac{a}{\sqrt{a^{2}+b^{2}}}, -\frac{b}{\sqrt{a^{2}+b^{2}}})$## Normal Vector to a Plane Equation Back to Top The normal vector$\vec{n}$to a plane given by ax + by + cz = d is always represented by$\vec{n}$= (a, b, c). The unit normal vector would eventually be denoted by the following :$(\frac{a}{\sqrt{a^{2}+b^{2} + c^{2}}}, \frac{b}{\sqrt{a^{2}+b^{2} + c^{2}}}, \frac{c}{\sqrt{a^{2}+b^{2} + c^{2}}})$## How to Find Normal Vector of a Plane With 3 Point Back to Top In order to find the normal vector of a plane using three points say P =$(x_{1}, y_{1}, z_{1})$, Q =$(x_{2}, y_{2}, z_{2})$and R =$(x_{3}, y_{3}, z_{3})$lying on the plane, one should follow the steps mentioned below. Step 1 : Create vector PQ and find its value$\vec{PQ} = (x_{2} - x_{1}, y_{2} - y_{1}, z_{2} - z_{1})$. Step 2 : Create vector PR and determine its value$\vec{PQ} = (x_{3} - x_{1}, y_{3} - y_{1}, z_{3} - z_{1})$. Step 3 : Calculate the cross product$\vec{PQ} \times \vec{PR}$which is a vector orthogonal to both$\vec{PQ}$and$\vec{PR}$. This would attain the form a i + b j + c k. Step 4 : Write the vector (a, b, c) which is the required normal vector to the plane passing through given 3 points. ## Normal Vectors in the Higher Dimensional Spaces Back to Top A normal vector in higher dimensional space can be referred to a vector that is perpendicular to a shape that is an analogue to a plane through the origin in that number of dimensions. Such as, in three-dimensional space, a normal vector to two vectors is constructed by drawing a plane through the origin and given two vectors. Then, the normal vector is a vector perpendicular to this plane. Similarly, the normal vectors in the higher dimensions are the vectors which are perpendicular to a set of other vectors. ## Examples Back to Top Take a look at the examples below. Example 1 : Find the normal vector to the line 3x + 4 y = 13. Solution : Here, a = 3 and b = 4. Thus, the normal vector to the given line is (3, 4) and (-3, -4). Example 2 : Determine the unit normal to the plane x + 2y - 2z = 4. Solution : a = 1, b = 2, c = -2 The normal vector to the given plane is (1, 2, -2). Unit normal vector is given as under :$(\frac{a}{\sqrt{a^{2}+b^{2} + c^{2}}}, \frac{b}{\sqrt{a^{2}+b^{2} + c^{2}}}, \frac{b}
{\sqrt{a^{2}+b^{2} + c^{2}}})$=$(\frac{1}{\sqrt{1^{2}+2^{2} + (-2)^{2}}}, \frac{2}{\sqrt{1^{2}+2^{2} + (-2)^{2}}}, \frac{-2}{\sqrt{1^{2}+2^{2} + (-2)^{2}}})$=$(\frac{1}{3}, \frac{2}{3}, \frac{-2}{3})$Example 3 : Find the normal vector to a plane passing through three points (4, 4, 1), (3, 1, -8) and (-5, 7, 2). Solution : Let P = (4, 4, 1), Q = (3, 1, -8) and R = (-5, 7, 2)$\vec{PQ}$= (3 - 4, 1 - 4, -8 - 1) = (-1, -3, -9)$\vec{PR}$= (-5 - 4, 7 - 4, 2 - 1) = (-9, 3, 1)$\vec{PQ} \times \vec{PR} = \begin{vmatrix}i & j & k\\ -1 & -3 & -9\\ -9 & 3 & 1\end{vmatrix}\$

= i (-3 + 27) - j (-1 - 81) + k (-3 - 27)

= 24 i + 82 j - 30 k

Normal vector = (24, 82, -30)
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