## Section3.8Working with Cartesian Unit Vectors

###### Unit 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*}

### Subsection3.8.1Dot and Cross Products of Cartesian Unit Vectors

Since $\hat i\text{,}$ $\hat j\text{,}$ and $\hat k\text{,}$ has unit magnitude, and are perpendicular to each other, we get the following dot and cross products.

\begin{align} \text{Dot products: }\amp\hat i \cdot \hat i = 1,\ \ \hat j \cdot \hat j = 1,\ \ \hat k \cdot \hat k = 1,\tag{3.8.1}\\ \amp \hat i \cdot \hat j = 0,\ \ \hat j \cdot \hat k = 0,\ \ \hat k \cdot \hat i = 0,\tag{3.8.2}\\ \text{Cross products: }\amp\hat i \times \hat i = 0,\ \ \hat j \times \hat j = 0,\ \ \hat k \times \hat k = 0,\tag{3.8.3}\\ \amp \hat i \times \hat j = \hat k,\ \ \hat j \times \hat k = \hat i,\ \ \hat k \times \hat k = \hat j,\tag{3.8.4} \end{align}

With these, you can easily see how the components of an arbitrary vector $\vec A = \left( A_x, A_y, A_z\right)$ are just the dot product of $\vec A$ with the unit vectors along the axex.

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

The magnitude of $\vec A$ is same as before

\begin{equation*} A = \sqrt{\vec A \cdot \vec A} = \sqrt{A_x^2 + A_y^2 + A_z^2}. \end{equation*}

Cosines of the angles $\vec A$ makes with positive axes, called direction cosines of the vector, are

\begin{equation*} \cos\,\theta_x = \dfrac{A_x}{A},\ \ \cos\,\theta_y = \dfrac{A_y}{A},\ \ \cos\,\theta_z = \dfrac{A_z}{A}, \end{equation*}

of which you only need two to denote direction of $\vec A$ since

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

Proof: It is easy to prove relation between direction cosines.

\begin{align*} \cos^2\theta_x \amp + \cos^2\theta_y + \cos^2\theta_z \\ \amp = \left( \frac{A_x}{A} \right)^2 + \left( \frac{A_y}{A} \right)^2 + \left( \frac{A_z}{A} \right)^2 \\ \amp = \frac{A_x^2 + A_y^2 + A_z^2}{A^2} = \frac{A^2}{A^2} = 1. \end{align*}
Dot Product of Two Vectors Using Cartesian Unit Vectors: A dot product of two vectors is rather simple to obtain from their Cartesian components. Let
\begin{equation*} \vec A = A_x \hat i + A_y \hat j + A_z \hat k,\ \ \vec B = B_x \hat i + B_y \hat j + B_z \hat k. \end{equation*}
Then, we will havbe
\begin{align*} \vec A \cdot \vec B \amp = (A_x \hat i + A_y \hat j + A_z \hat k) \cdot (B_x \hat i + B_y \hat j + B_z \hat k)\\ \amp = A_x B_x (\hat i \cdot \hat i) + A_x B_y (\hat i \cdot \hat j) + A_x B_z (\hat i \cdot + \hat k) + \cdots\\ \amp = A_x B_x (1) + A_x B_y (0) + A_x B_z (0) + \cdots\\ \amp = A_x B_x + A_y B_y + A_z B_z. \end{align*}
Of course this will equal $A B \cos\theta$ with angle $\theta$ between the two vectors.
$$\vec A \cdot \vec B = A B \cos\theta = A_x B_x + A_y B_y + A_z B_z.\tag{3.8.5}$$
Cross Product of Two Vectors Using Cartesian Unit Vectors: I have already given formulas for this above.
\begin{equation*} \vec A \times \vec B = \left(A_y B_z - A_z B_y\right) \hat i + \left(A_z B_x - A_x B_z\right) \hat j + \left(A_x B_y - A_y B_x\right) \hat k. \end{equation*}
We can show this to be the case now by expicitly working out the cross product.
\begin{align*} \vec A \times \vec B \amp = (A_x \hat i + A_y \hat j + A_z \hat k) \times (B_x \hat i + B_y \hat j + B_z \hat k) \\ \amp = A_x B_x (\hat i \times \hat i) + A_x B_y (\hat i \times \hat j) + A_x B_z (\hat i \times + \hat k) + \cdots\\ \amp = A_x B_x (0) + A_x B_y (\hat k) + A_x B_z (- \hat j) + \cdots\\ \amp = [A_x B_y (\hat k) + A_x B_z (- \hat j)] + [A_y B_x (-\hat k) + A_y B_z (\hat i)] \\ \amp \ \ \ \ \ \ + [A_z B_x (\hat j) + A_z B_y (-\hat i)] \\ \amp = \left(A_y B_z - A_z B_y\right) \hat i + \left(A_z B_x - A_x B_z\right) \hat j + \left(A_x B_y - A_y B_x\right) \hat k. \end{align*}
Often, we express this result in the form of a determinant.
$$\vec A \times \vec B = \begin{vmatrix} \hat i \amp \hat j \amp \hat k \\ A_x \amp A_y \amp A_z \\ B_x \amp B_y \amp B_z \\ \end{vmatrix} \tag{3.8.6}$$

Using $\vec A = A_x \hat i + A_y \hat j + A_z \hat k\text{,}$ prove that $\vec A \cdot \vec A = A_x^2 + A_y^2 + A_z^2\text{.}$

Hint

Expand

N/A

Solution

We display the calculation for $x$ in detail and infer those for $y$ and $z$ from the symmetry.

\begin{align*} \vec A \cdot \vec A \amp =(A_x \hat i + A_y \hat j + A_z \hat k)\cdot (A_x \hat i + A_y \hat j + A_z \hat k) \\ \amp = A_x \left( A_x \hat i\cdot \hat i + A_y \hat i \cdot \hat j + A_z \hat i \cdot \hat k \right) + (x \rightarrow y) + (x \rightarrow z) \\ \amp = A_x \left( A_x \times 1 + A_y \times 0 + A_z \times 0 \right) + (x \rightarrow y) + (x \rightarrow z) \\ \amp = A_x^2 + (x \rightarrow y) + (x \rightarrow z) \\ \amp = A_x^2 + A_y^2 + A_z^2. \end{align*}

Consider two vectors expressed in terms of Cartesian unit vectors.

\begin{equation*} \vec A = 3\hat i + 4\hat j - 12 \hat k,\ \ \vec A = -3\hat i + 4\hat j + 12 \hat k \end{equation*}

Find (a) magnitudes $A$ and $B\text{,}$ (b) sum $\vec A + \vec B\text{,}$ (c) difference $\vec A - \vec B\text{,}$ (d) dot product $\vec A \cdot \vec B\text{,}$ (e) cross product $\vec A \times \vec B\text{,}$ (f) angle between the two vectors, and (g) direction cosines of $\vec A\text{.}$

(a) $13 \text{,}$ $13\text{,}$ (b) $8 \hat j \text{,}$ (c) $6 \hat i -24 \hat k \text{,}$ (d) $-137 \text{,}$ (e) $96 \hat i + 24 \hat k \text{,}$ (f) $\cos^{-1}(-137/169)\text{,}$ (g) $\cos^{-1}(3/13)$