Physics Formulas

Unit Vector Formula

A unit vector is something that we use to have both direction and magnitude. Moreover, it denotes direction and uses a 2-D (2 dimensional) vector because it is easier to understand. In addition, we can plot it on a graph. Besides, in this topic, we will discuss unit vector and unit vector formula, its derivation and solved examples.

Unit Vector

In mathematics, unit vector refers to the normal vector space (often a spatial vector) of length 1. Moreover, we use a lowercase letter with a circumflex, or ‘hat’ (Pronunciation “i-hat”). Usually, we use the term direction vector to describe a unit vector to represent the spatial vector. Besides, we denote them as d. Moreover, 2D spatial directions represent this way are numerically equivalent to the points on the unit circle.

Unit vector Formula

Furthermore, unit vector often forms the basis of a vector space and every vector in the space can be written as a linear combination of unit vectors.

Most noteworthy, in Euclidean space the dot product of two unit vector is a scalar value amounting to the cosine of the smaller subtended angle. Also, in a 3 dimensional (3D) Euclidean space, the cross product of two random unit vectors is a third vector orthogonal to both of them having a length equal to the sine of the smaller subtended angle.

Also, regularized cross-product exacts for this varying span, and gives way to the jointly orthogonal unit vector to the two inputs, concerning the right-hand rule to determine one of two possible directions.

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Unit Vector Formula

Vectors are labeled with arrows like this \(\vec{a}\). Also, a unit vector has a magnitude of 1 and they are labeled with a “^” such as \(\hat{b}\). Furthermore, any vector can become a unit vector by dividing it by the vector’s magnitude. Besides, they are often written in XYZ coordinates. Moreover, we can do it in two ways:

The first one is to write the coordinates in brackets: \(\vec{v}\) = (x, y, z).
Furthermore the other way is to use three unit vectors, in which each point along one of the axes: \(\vec{v}\) = \(x\hat{i}\) + \(y\hat{j}\) + \(z\hat{k}\). Moreover, the magnitude of vector is:

\(\mid \vec{v} \mid\) = \(\sqrt{x^{2} + y^{2} + z^{2}}\)

Unit vector = \(\frac{vector}{magnitude of the vector}\)

If we write it in bracket format then:

\(\hat{v}\) = \(\frac{\vec{v}}{\mid\vec{v} \mid}\) = \(\frac{(x, y, z)}{\sqrt{x^{2}+y^{2}+z^{2}}}\) = \((\frac{x}{\sqrt{x^{^{2}+ y^{2}+ z^{2}}}}, \frac{y}{\sqrt{x^{2}+ y^{2}+ z^{2}}}, \frac{z}{\sqrt{x^{2}+ y^{2}+z^{2}}})\)

If we write it in unit vector component format:

\(\hat{v}\) = \(\frac{\vec{v}}{\mid\vec{v} \mid}\) = \(\frac{(x\hat{i}, y\hat{j}, z\hat{k})}{\sqrt{x^{2}+y^{2}+z^{2}}} = \frac{x}{\sqrt{x^{^{2}+ y^{2}+ z^{2}}}}\hat{i}+ \frac{y}{\sqrt{x^{2}+ y^{2}+ z^{2}}}\hat{j}+ \frac{z}{\sqrt{x^{2}+ y^{2}+z^{2}}}\hat{k}\)

Derivation of the unit vector formula

\(\hat{v}\) = Refers to a unit vector with magnitude of 1 and direction
\(\vec{v}\) = refers to a vector with any direction and magnitude
\(\mid\vec{v} \mid\) = refers to the magnitude of the vector \(\vec{v}\)
x = refers to the value of the vector in x axis
y = refers to the value of the vector in y axis
z = refers to the value of the vector in z axis
\(\hat{i}\) = refers to the unit vector bound for the positive axis x
\(\hat{j}\) = refers to the unit vector bound for the positive axis y
\(\hat{k}\) = refers to the unit vector bound for the positive axis z

Solved Example on Unit Vector Formula 

Example 1

There is a vector \(\vec{r}\) = \(12\hat{i}-3\hat{j}-4\hat{k}\). Calculate the unit vector \(\vec{r}\). Also, express it in both unit vector component format and bracket format.


\(\mid \vec{r}\mid\) = \(\sqrt{x^{2}+ y^{2}+ z^{2}}\)
\(\mid \vec{r}\mid\) = \(\sqrt{12^{2}+ (-3)^{2}+ (-4)^{2}}\)
\(\mid \vec{r}\mid\) = \( \sqrt{144 + 9 + 16}\)
\(\mid \vec{r}\mid\) = \(\sqrt{169}\)
\(\mid \vec{r}\mid\) = 13

So, now we can use the magnitude to find the unit vector \(\hat{b}\):

\(\hat{r}\) = \(\frac{\vec{r}}{\mid \vec{r}\mid}\) = \(\frac{(x\hat{i}, y\hat{j}, z\hat{k})}{\sqrt{x^{2}+y^{2}+z^{2}}}\)
\(\hat{r}\) = \(\frac{12\hat{i}-3 \hat{j}- 4\hat{k}}{13}\)
\(\hat{r}\) = \(\frac{12}{13}\hat{i}-\frac{3}{13}\hat{j}-\frac{4}{13}\hat{k}\)

Besides, in bracket format the unit vector is:

\(\hat{r}\) = \(\frac{(12,-3,-4)}{13}\)
\(\hat{r}\) = \((\frac{12}{13},-\frac{3}{13},-\frac{4}{13})\)

Moreover, the answer does not ask for decimal numbers so we leave the numbers in fractions.

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5 responses to “Spring Potential Energy Formula”

  1. Typo Error>
    Speed of Light, C = 299,792,458 m/s in vacuum
    So U s/b C = 3 x 10^8 m/s
    Not that C = 3 x 108 m/s
    to imply C = 324 m/s
    A bullet is faster than 324m/s

  2. Malek safrin says:

    I have realy intrested to to this topic

  3. umer says:

    m=f/a correct this

  4. Kwame David says:

    Interesting studies

  5. Yashdeep tiwari says:

    It is already correct f= ma by second newton formula…

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