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Normal Vector – Explanation and Examples
The world of vector geometry does not end at directed vectors emerging out or into two-dimensional or three-dimensional planes. The most important type of vectors that make up for most of the vector geometry concepts is a normal vector.
Normal vector can be defined as:
“A normal vector is a vector that is perpendicular to another surface, vector, or axis, in short, making an angle of 90° with the surface, vector, or axis.”
In this section of normal vectors, we will be covering the following topics:
- What is a normal vector?
- How to find a normal vector?
- What is the formula of normal vectors?
- Examples
- Practice problems
What is a Normal Vector?
A normal vector is a vector inclined at 90° in a plane or is orthogonal to all the vectors.
Before we indulge in the concept of normal vectors, let’s first get an overview of the term ‘normal.’
In mathematical terms, or more specifically in geometrical terms, the term ‘normal’ is defined as being perpendicular to any stated surface, plane, or vector. We can also state that being normal means that the vector or any other mathematical object is directed at 90° to another plane, surface, or axis.
Now that we know what the term ‘normal’ refers to in the mathematical domain let’s analyze normal vectors.
Normal vectors are inclined at an angle of 90° from a surface, plane, another vector, or even an axis. Its representation is as shown in the following figure:
The concept of normal vectors is usually applied to unit vectors.
Normal vectors are the vectors that are perpendicular or orthogonal to the other vectors. If we talk about the technical aspect of the matter, there are an infinite number of normal vectors to any given vector as the only standard for any vector to be regarded as a normal vector is that they are inclined at an angle of 900 to the vector. If we consider the dot product of a normal vector and any given vector, then the dot product is zero.
a . n = |a| |n| cos (90)
a . n = 0
Similarly, if we consider the cross product of the normal vector and the given vector, then that is equivalent to the product of magnitudes of both the vectors as sin (90) = 1.
a x n = |a| |n| sin (90)
a x n = |a| |n|
The realm of vector geometry is all about different vectors and how we can practically incorporate these directional mathematical objects in our daily lives. Whether it be from the engineering, architectural, aeronautical, or even medical sector, every real-life problem cannot be solved without implementing vectors’ concepts. In short, we can conclude that every practical problem requires a vector solution.
Due to such significance of vectors in our everyday lives, understanding every vector’s role and concept becomes a top priority for mathematicians and students. Amongst these vectors, the normal vector is of prime importance.
Every vector has some magnitude and direction. In mathematics, the vector’s magnitude is the most important factor, but in some cases, magnitude is not that significant. It completely depends upon the requirement. In some cases, we only require direction. That is why magnitude is not necessary in such cases. Hence, we can say the direction of a vector is unique. We can view this concept geometrically as well; the normal vector to the plane resides on the line, and there exist several vectors on that line that are perpendicular to the plane. So, direction introduces uniqueness in the system.
Now, let us solve an example to have a better concept of normal vectors.
Example 1
Find out the normal vectors to the given plane 3x + 5y + 2z.
Solution
For the given equation, the normal vector is,
N = <3, 5, 2>
So, the n vector is the normal vector to the given plane.
We stated earlier in our previous topic of ‘Unit Vectors’ that these vectors have the magnitude 1 and are perpendicular to the plane’s remaining axes. Since the unit vector along an axis is perpendicular to the remaining axes, the unit vector can also fall into the domain of normal vectors. This concept is elaborated below:
Unit Normal Vector
A unit normal vector is defined as:
“A vector that is perpendicular to the plane or a vector and has a magnitude 1 is called a unit normal vector.”
As we stated above, normal vectors are directed at 90° angles. We have already discussed that unit vectors are also perpendicular or directed at 90° to the remaining axes; hence, we can mix these two terms up. The joint concept is termed as Unit Normal Vector, and it is actually a subcategory of Normal Vectors.
We can distinguish unit normal vectors from any other normal vector by stating any normal vector with a magnitude of 1 can be declared a unit normal vector. Such vectors would have magnitude 1 and would also be directed at exactly an angle of 90° from any specific surface, plane, vector, or corresponding axis. The representation of such a vector can be depicted by placing a hat (^) on top of the vector n, n(^).
Another thing to note here is the common misconception and confusion some mathematicians and students encounter while validating this concept. If we have a vector v, then one thing to note is not to intermix the concept of a unit vector and a normal vector. The unit vectors of vector v will be directed along the axes of the plane in which the vector v exists. In contrast, the normal vector would be a vector that would be particular to the vector v. The unit normal vector, in this case, are the unit vectors of the vector v, not the normal vector, which is at 90° from the vector v.
For example, let’s consider a vector r <a, b, c> which indicates an x-coordinate, b as y-coordinate, and c as the vector’s z-coordinate. The unit vector is a vector whose direction is the same as the vector a, and its magnitude is 1.
The unit vector is given as,
u = a / |a|
u = .
Where |r| is the magnitude of the vector and u is the unit vector.
Let’s discuss the concept of unit normal vectors with the help of an example.
Example 2
Find the normal unit vector when the vector is given as v = <2, 3, 5>
Solution
As we know, the unit vector is a vector with a magnitude equals to 1 and direction along the given vector’s direction.
So, unit vector is given as,
u = 1 . ( v / |v| )
Hence, magnitude of the vector is given as
|v| = √ ( (2)^2 + (3)^2 + (5)^2 )
|v| = √ ( 4 + 9 + 25 )
|v| = √ ( 38 )
Now, putting the values in the above-mentioned formula gives,
u = 1 . ( < (2 / √ (38) ) + (3 / √ (38) ) + (5 / √ (38) ) >)
u = < (2 / √ (38) ) + (3 / √ (38) ) + (5 / √ (38) ) >
Normal Vector And Cross Product
As we know that cross product gives a vector that is perpendicular to both the vectors A and B. Its direction is specified by the right-hand rule. Hence, this concept is very useful for generating the normal vector. So, it can be stated that a normal vector is the cross product of two given vectors A and B.
Let’s understand this concept with the help of an example.
Example 3
Let’s consider two vectors PQ = <0, 1, -1> and RS = <-2, 1, 0> . Calculate the normal vector to the plane containing these two vectors.
Solution:
Since we know that the cross product of two vectors gives the normal vector so,
| PQ x RS | = i j k
1 1 -1
-2 1 0
| PQ x RS | = i ( 0 + 1 ) – j ( 0 – 2 ) + k ( 0 + 2 )
| PQ x RS | = 1i + 2j + 2k
Hence, this is the normal vector.
Conditions For A Normal Vector
As we know that we can find out the normal vector using the cross product. Similarly, there exist two conditions for vectors to be orthogonal or perpendicular.
- Two vectors are said to be perpendicular if their dot product is equal to zero.
- Two vectors are said to be perpendicular if their cross product is equal to 1.
To verify our result, we can use the above-mentioned two conditions.
Let us verify this with the help of examples.
Example 4
Show that the two vectors v = <1, 0, 0> and u = <0, -2, -3> are perpendicular to each other.
Solution
If the dot product of two vectors is equal to zero, then the two vectors are perpendicular to each other.
So, the dot product of the vectors u and v is given as,
u . v = <1, 0, 0> . <0, -2, -3> = 0
u . v = 1 – 0 – 0
u . v = 0
Hence, proved that two vectors are perpendicular to each other.
Unit Tangent Vectors
When we discuss the unit normal vectors, there comes another type called unit tangent vectors. To understand the concept, let’s consider a vector r(t) to be a differentiable vector-valued function and v(t) = r’(t) then the unit tangent vector with the direction in the direction of the velocity vector is given as,
t (t) = v (t) / |v (t)|
where |v(t)| is the magnitude of the velocity vector.
Let us have a better understanding of this concept with the help of an example.
Example 5
Consider r (t) = t2i + 2tj + 5k, find out the unit tangent vector. Also calculate the value of the tangent vector at t = 0.
Solution
According to the formula, unit tangent vector is given as,
t (t) = v (t) / |v (t) |
where v (t) = r’ (t)
Let’s calculate the value of v (t)
v (t) = 2ti + 2j
now, calculating the value of the magnitude of the vector v (t) that is given as,
|v| = √ ( 4t^2 + 4 )
Putting the values in the formula of unit tangent vector gives,
t (t) = ( 2ti + 2j ) / ( √ ( 4t^2 + 4 ) )
Now, finding the value of t (0),
t (0) = 2j / ( √(4) )
t (0) = 2j / ( 2)
t (0) = 1j
Example 6
Consider r (t) = e t i + 2t 2 j + 2t k, find out the unit tangent vector. Also calculate the value of the tangent vector at t = 1.
Solution
According to the formula, unit tangent vector is given as,
t (t) = v (t) / |v (t)|
where v (t) = r’ (t)
Let’s calculate the value of v (t)
v (t) = e ^t i + 4t j + 2 k
now, calculating the value of the magnitude of the vector v (t) that is given as,
|v| = √ ( e ^2t + 16t^2 + 4 )
Putting the values in the formula of unit tangent vector gives,
t (t) = ( e ^t i + 4t j + 2 k ) / ( √ ( e ^2t + 16t^2 + 4 ) )
Now, finding the value of t (1),
t (1) = ( e ^1 i + 4 (1) j + 2 k ) / ( √ ( e ^2(1) + 16 (1)^2+ 4 ) )
t (1) = ( e^ 1 i + 4 j + 2 k ) / ( √ ( e ^2 + 16 + 4 ) )
t (1) = ( e i + 4 j + 2 k ) / ( √ ( e^ 2 + 20 ) )
Practice Problems
- Find the normal unit vector when the vector is given as v = <1, 0, 5>
- Consider r (t) = 2x2 i + 2x j + 5 k, find out the unit tangent vector. Also calculate the value of the tangent vector at t = 0.
- Let r(t) = t i + et j – 3t2 k. Find the T(1) and T(0).
- Find out the normal vectors to the given plane 7x + 2y + 2z = 9.
Answers
- <1, 0, 5>/ ( √(26)
- (4x + 2)/( √(16x2 + 2)
- (1 + et – 6t) / √(1 + e2t + 36t2)
- <7, 2, 2>
All the images are constructed using GeoGebra.