# Unit Vector

A vector is a quantity which has both magnitudes, as well as direction, is a unit vector. As a unit means 1, This vector has a magnitude of 1. It is also known as Direction Vector.

Unit Vector Symbol – $\hat{a}$ , (called as a hat). It is given by $\hat{a}\hat{\equiv }\frac{a}{[\left |a \right |}$

Where $\left |a \right |$ is for norm of a.

It can be calculated using a Unit vector calculator or using its Formula.

## Unit Vector Formula

As explained above vectors have both magnitude (Value) and a direction. They are shown with an arrow $\vec{a}$. $\hat{a}$ denotes a unit vector. If we want to change any vector in unit vector, divide it by the vector’s magnitude. Usually xyz coordinates are used to write any vector.

It can be done in two ways:

1. $\vec{a}$ = (x, y, z) using the brackets.
2. $\vec{a}$ = x$\hat{i}$ + y $\hat{j}$ +z $\hat{k}$

Formula the magnitude of a vector is:

 $\left | \vec{a} \right |=\sqrt{x^{2}+y^{2}+z^{2}}$ Unit Vector = $\frac{Vector}{Vector’s magnitude}$

The above is a unit vector formula.

### How to represent Vector in a bracket format:

$\hat{a} \hat{\equiv }\frac{a}{\left|a \right |} =\frac{(x,y,z)}{\sqrt{x^{2}+y^{2}+z^{2}}} =\frac{x}{\sqrt{x^{2}+y^{2}+z^{2}}},\frac{y}{\sqrt{x^{2}+y^{2}+z^{2}}},\frac{z}{\sqrt{x^{2}+y^{2}+z^{2}}}$

### How to represent Vector in a unit vector component format:

$\hat{a}\hat{\equiv }\frac{a}{\left |a \right | } =\frac{x\hat{i}+ y\hat{j} +z \hat{k}}{\sqrt{x^{2}+y^{2}+z^{2}}} =(\frac{x}{\sqrt{x^{2}+y^{2}+z^{2}}}\hat{i},\frac{y}{\sqrt{x^{2}+y^{2}+z^{2}}}\hat{j},\frac{z}{\sqrt{x^{2}+y^{2}+z^{2}}}\hat{k})$

Where x, y, z are the value of the vector in the x, y, z axis respectively and

$\hat{a}$ is a unit vector, $\vec{a}$ is a vector, $\left | \vec{a} \right |$ is the magnitude of the vector $\vec{a}, \hat{i}, \hat{j}, \hat{k}$ are the directed unit vectors along the x , y , z axis.

## Unit Vector Examples

Question 1:

Find the unit vector $\vec{p}$ for the given vector, 12$\hat{i}$ – 3$\hat{j}$ – 4 $\hat{k}$. Show it in both the formats – Bracket and Unit vector component.

Solution: Let’s find the magnitude of the given vector first, $\vec{p}$ is :

$\left |p \right | = \sqrt{x^{2}+y^{2}+z^{2}} \left |p \right | = \sqrt{x^{2}+y^{2}+z^{2}} \left |p \right | = \sqrt{x^{2}+y^{2}+z^{2}} \left |p \right | = \sqrt{144 + 9 + 16} \left |p \right | = \sqrt{169} \left |p \right | = 13$

Let’s use this magnitude to find the unit vector now:

$\hat{p}\hat{\equiv }\frac{p}{\left |p \right |} = \frac{x\hat{i}+y \hat{j} +z\hat{k}}{\sqrt{x^{2}+y^{2}+z^{2}}}$=

=$\hat{p}=\frac{12\hat{i}-3 \hat{j} – 4\hat{j}}{13}$ $\hat{p} = \frac{12}{13}\hat{i} -\frac{3}{13}\hat{j}-\frac{4}{13}\hat{k}$

The unit vector in Bracket form is:

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

## Unit Vector Problem:

Question 2:

Find the unit vector $\vec{q}$ for the given vector, $-2\hat{i} + 4\hat{j} – 4 \hat{k}.$. Show it in both the formats – Bracket and Unit vector component.

Solution: Let’s find the magnitude of the given vector first, $\vec{q}$ is :

$\left |q \right |$ = $\sqrt{x^{2}+y^{2}+z^{2}}$ $\left |q \right | = \sqrt{-2^{2}+(4)^{2}+(-4)^{2}}$ $\left |q \right |$ = $\sqrt{4 + 16 + 16}$ $\left |q \right |$ = $\sqrt{36}$ $\left |q \right |$ = 6

Let’s use this magnitude to find the unit vector now:

$\hat{q}\hat{\equiv }\frac{q}{\left |q \right |} ={\frac{x\hat{i} + y\hat{j} +z \hat{k}}{\sqrt{x^{2}+y^{2}+z^{2}}} \hat{q} = \frac{-2\hat{i} +4 \hat{j} – 4\hat{q}}{6}}= \frac{-2}{6}\hat{i} + \frac{4}{6} \hat{j} -\frac{4}{6}\hat{k}$ $\hat{q}= \frac{-2\hat{i} +4 \hat{j}- 4 \hat{k}} 6$ $\hat{p} = \frac{-2}{6}\hat{i} + \frac{4}{6} \hat{j} -\frac{4}{6}\hat{k}$

The unit vector in Bracket form is:

$\hat{p} = \frac{(-2, 4, -4)}{6} \hat{p} = \frac{(-2)}{6}, \frac{(4)}{6}, \frac{(4)}{6}= \frac{(-1)}{3}, \frac{(2)}{4}, \frac{(2)}{3}$

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