## Cross Product (made easy - 2 simple ways)

### A simple approach to compute a 3-dimensional vector cross-product

Computing the cross-product of two 3 dimensional vectors can be as easy as:

** (1 - 2) + (3 - 4) + (5 - 6)**

For each of the vector directions (I, J, K), the method is the same. You first go **(down to the right)** through the unit vector picking up 2 coefficients and return from below **(up to the right)** picking up the 2 coefficients to subtract:

#### Computing the coefficients in the [I] unit direction

(1 - 2) = (a2b3 - a3b2) [I]

#### Computing the coefficients in the [J] unit direction

(3 - 4) = (a3b1 - a1b3) [J]

#### Computing the coefficients in the [K] unit direction

(5 - 6) = (b2a1 - b1a2) [K]

### Final Solution (a x b)

a x b = (a2b3 - a3b2) [I] + (a3b1 - a1b3) [J] + (b2a1 - b1a2) [K]

### Method Advantages

The beauty of the system is it allows **the same** sweeping motion through each of the unit vectors **(first down to the right)**, then **(back up to the right)** picking up 2-coefficients each time. There is nothing you have to remember to do differently for **(I)** or **(J)** or **(K)**.

I always found this approach much simpler to remember than trying to (ignore a single direction in a column vector and then trying to remember how **(I)** was different from **(J)** and then, again, how **(K)** was different from **(I)**.

Here, just set up the matrix and simply do the same thing for each unit vector. **Try it:**

| I J K | | | | a1 a2 a3 | = a2b3-a3b2 [I] + a3b1-a1b3 [J] + b2a1-b1a2 [K] | | | b1 b2 b3 |

#### You Only Like Straight Lines? There is a Second Method

There are some that have a difficult time remembering how to go **(down to the right)** through the unit vector picking up 2 coefficients and return from below **(up to the right)** picking up the 2 coefficients to subtract. Luckily there is a second method that involves only straight lines.

The method requires that you duplicate the **(I, J)** column vectors just to the right of the **(K)** column. After doing so, your method for computing the cross product simply requires you go **(down to the right)** through the unit vector picking up 2 coefficients and return from below **(up to the right)** through the appended **(I, J)** unit vectors picking up the 2 coefficients to subtract. (note: you use the same **(K)**

#### Computing the coefficients in the [I] unit direction

(1 - 2) = (a2b3 - b2a3) [I]

#### Computing the coefficients in the [J] unit direction

(3 - 4) = (a3b1 - b3a1) [J]

#### Computing the coefficients in the [K] unit direction

(5 - 6) = (a1b2 - b1a2) [K]

### Final Solution (a x b)

a x b = (a2b3 - b2a3) [I] + (a3b1 - b3a1) [J] + (a1b2 - b1a2) [K]

(**note: only the order of multiplication has changed -- nothing else**)

Courtesy of an old Attorney and Aerospace Engineer from Texas A&M