## Section3.2Unit Vectors

A unit vector is a vector of magnitude $1\text{.}$ Usually we denote a unit vector with a carat symbol $\wedge$ over the letter representing the vector rather than an arrow $\rightarrow$ symbol. For instance, a unit vector $\hat u_{E}$ might denote a unit vector in the direction of East.

Say, you have a position vector $\vec r$ with magnitude, $r=10$ cm, pointed in the direction North. To get unit vector in the direction North, which we can denote by $\hat u_N$ we will just divide $\vec r$ by the magnitude $r=10$ cm.

\begin{equation*} \hat u_N = \dfrac{ (10\text{ cm}, \text{ North}) } { 10 \text{ cm} } = (1, \text{ North}). \end{equation*}

Since multiplying a vector by a number just scales the vector, multiplying a unit vector by a number will give a vector that is as many times bigger or smaller. For example, $2 \hat u_N$ will be a vector that has length $2$ and the direction same as the direction of $\hat u_N\text{.}$ The $\frac{1}{2} \hat u_N$ will be a vector that has length $\frac{1}{2}$ and the direction same as the direction of $\hat u_N\text{.}$ A $-\hat u_N$ will be a vector that has length $1$ but the direction opposite to the direction of $\hat u_N\text{.}$

Thus, you can write any vector along North or South direction as a product of its magnitude and the $\hat u_N$ vector. Say, you want a velocity vector $\vec v$ of magnitude $20$ m/s pointed North.

\begin{equation*} \vec v = (20 \text{ m/s} )\, \hat u_N, \end{equation*}

and if you want a velocity vector of magnitude $20$ m/s pointed South, you would do the following

\begin{equation*} \vec v = -(20 \text{ m/s} )\, \hat u_N, \end{equation*}

where the negative sign will just flip the direction of the unit vector.

There is nothing special about the direction North, and our discussion above using the example of $\hat u_N$ can be generalized to a unit vector in any arbitrary direction, such as East/West vectors, or any other vectors.

Say, you have a unit vector $\hat u$ in some direction, e.g., $40^{\circ}$ North of East. Then, any vector in the $40^{\circ}$ North of East direction or in exactly opposite direction can be written in as a number (with appropriate units and sign) times $\hat u \text{.}$

For instance, if we have a displacement vector $\vec r$ of $20 \text{ m}$ in the direction of the $\hat u$ vector, then $\vec r = 20\text{ m } \hat u\text{.}$ If you have a force vector $\vec F$ of $50 \text{ N}$ in the opposite direction, then $\vec F = -50\text{ N } \hat u\text{.}$

You can see that a unit vector provides a universal way of writing vectors which incorporate the direction information analytically.

### Subsection3.2.1Unit Vectors Along Cartesian Positive Axis Directions

Unit vectors along Cartesian axes play important role in vector analysis. We will often denote these unit vectors by $\hat u_x \text{,}$ $\hat u_y \text{,}$ and $\hat u_z$ respectively. At other times, we will denote them by their traditional symbols $\hat i \text{,}$ $\hat j \text{,}$ and $\hat k$ respectively.

For instance, say, we want a position vector $\vec r = ( x=2 \text{ m}, y=-5 \text{ m})$ in the $xy$-plane. Then, you can see that this vector will be a sum of two vectors: $\vec r_1 = 2 \text{ m} \hat i$ and $\vec r_2 = -5 \text{ m} \hat j \text{.}$ We write this as

\begin{equation*} \vec r = 2 \text{ m} \hat i - 5 \text{ m} \hat j. \end{equation*}

In general, an arbitrary vector $\vec A$ with coordinate components $A_x, A_y, A_z$ is often written as a sum of vectors along the Cartesian axes.

\begin{equation*} \vec A = A_x \hat i + A_y \hat j + A_z \hat k. \end{equation*}

### Subsection3.2.2Direction Cosines

Cosines of the angles a vector makes with unit vectors towards positive Cartesian axes are called its direction cosines of the vector. In Figure 3.2.6, the direction cosines of vector $\vec A$ are $\cos\theta_x\text{,}$ $\cos\theta_y\text{,}$ and $\cos\theta_z\text{.}$ These angles, $\theta_x\text{,}$ $\theta_y\text{,}$ and $\theta_z\text{,}$ are frequently referred to as $\alpha\text{,}$ $\beta\text{,}$ and $\gamma\text{,}$ respectively.

After we discuss dot product of vector I will show that direction cosines are not all independent. Rather, they are related by the following identity.

\begin{equation} \cos^2\theta_x + \cos^2\theta_y + \cos^2\theta_z = 1.\tag{3.2.1} \end{equation}

Suppose you are told that the velocity of a projectile is $25\text{ m/s}$ in the direction of $30^{\circ}$ above the direction of East. Let positive $x$ axis be towards the East and the positive $y$ axis be pointed up.

(a) Find $x$ and $y$ components of a unit vector in the stated direction.

(b) Write the given velocity vector in terms of the unit vector you found.

Hint

Take a vector of magnitude 1 in the direction of the velocity and find its Cartesian components.

(a) $\hat u = \dfrac{\sqrt{3}}{2}\, \hat i + \dfrac{1}{2}\, \hat j\text{,}$ $25\text{ m/s}\, \hat u\text{.}$

Solution 1 (a)

(a) We want a vector $\hat u$ of unit magnitude in the direction described. We need to just find its $x$ and $y$ components.

\begin{align*} \amp u_x = 1\times \cos\, 30^{\circ} = \dfrac{\sqrt{3}}{2}\\ \amp u_x = 1\times \sin\, 30^{\circ} = \dfrac{1}{2}. \end{align*}

Therefore,

\begin{equation*} \hat u = \dfrac{\sqrt{3}}{2}\, \hat i + \dfrac{1}{2}\, \hat j. \end{equation*}
Solution 2 (b)

(b) When we multiple a unit vector by a positive number we get a vector of the new length and the same direction, so, all we have to do to get the velocity in that direcion is just multiply the unit vector by the magnitude of the velocity.

\begin{equation*} \vec v = 25\text{ m/s}\, \hat u. \end{equation*}

Of course, this is same as

\begin{equation*} \vec v = 25\text{ m/s}\, \left( \dfrac{\sqrt{3}}{2}\, \hat i + \dfrac{1}{2}\, \hat j\right) \end{equation*}

Suppose you want a unit vector in the same direction as the velocity vector given to you as $\vec v = (3.0 \text{ m/s})\hat i + (4.0 \text{ m/s})\hat j\text{.}$ Find this unit vector.

Hint

Divide by the magnitude.

$0.6\hat i + 0.8\hat j$

Solution

To get a unit vector, we just divide out the magnitude.

\begin{equation*} \hat u = \dfrac{ 3\hat i + 4 \hat j}{\sqrt{3^2 + 4^2}} = 0.6\hat i + 0.8\hat j. \end{equation*}

Suppose you want a unit vector in the direction of a star that is $30^\cric$ above horizon above $40^\circ$ North of East direction. Find a unit vector in the direction in terms of unit vectors along East, North, and Up.

Hint

Project unit magnitude on the East-North Plane and Up directions first.

$\hat u = 0.66 \hat i + 0.56 \hat j + 0.5 \hat k.$

Solution

Let us use $\hat i$ towards East, $\hat j$ towards North, and $\hat k$ towards Up. First we project a unit vector in the plane containing Up and “$40^\circ$ N of E” directions. We will call them as $u_z$ and $u_r\text{,}$ respectively. We get

\begin{align*} \amp u_z = 1\, \sin(30^\circ) = \frac{1}{2}.\\ \amp u_r = 1\, \cos(30^\circ) = \frac{\sqrt{3}}{2}. \end{align*}

Now, we project $u_r$ on $x$ and $y$ directions to get

\begin{align*} \amp u_x = u_r\, \cos(40^\circ) = 0.66.\\ \amp u_y = u_r\, \sin(40^\circ) = 0.56. \end{align*}

Therefore, our unit vector has the following representation.

\begin{equation*} \hat u = 0.66 \hat i + 0.56 \hat j + 0.5 \hat k. \end{equation*}

We should verify that it has magnitude of 1 by

\begin{equation*} \sqrt{0.66^2 + 0.56^2 + 0.5^2}, \end{equation*}

which is indeed 1 within the tolerance of the rounding. My answer is a little different than exact 1 since I had rounded off earlier.