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Home , Cross product

University of Babylon Subject : III College of nd Class : 2 year - first semester Mechanical Engineering Dept. Date: / 10 / 2016

2016 \ 2017 : Understanding the Cross

The cross product accumulates interactions between different dimensions. Taking two vectors, we can write every combination of components in a grid:

This completed grid is the , which can be separated into the:

,

the interactions between similar dimensions (x*x, y*y, z*z)

 Cross product,

the interactions between different dimensions (x*y, y*z, z*x, etc.)

The dot product (a⃗ ⋅b⃗ ) measures similarity because it only accumulates interactions in matching dimensions. It’s a simple calculation with 3 components.

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The cross product (written a⃗ ×b⃗) has to measure a half-dozen “cross interactions”. The calculation looks complex but the concept is simple: accumulate 6 individual differences for the total.

Instead of thinking “When do I need the cross product?” think “When do I need interactions between different dimensions?”.

Area, for example, is formed by vectors pointing in different directions (the more orthogonal, the better). Indeed, the cross product measures the spanned by two 3d vectors:

(The “cross product” assumes 3d vectors, but the concept extends to higher dimensions.)

Defining the Cross Product

The dot product represents vector similarity with a single number:

(Remember that trig functions are percentages.) Should the cross product (difference between interacting vectors) be a single number too?

Let’s try. Sine is the percentage difference, so we could use:

Instead, let’s express these unique differences as a vector:

 The size of the cross product is the numeric “amount of difference” (with sin(θ) as the percentage)  The direction of the cross product is based on both inputs: it’s the direction orthogonal to both (i.e., favoring neither)

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A vector result represents the x*y and x*z separately, even though y and z are both “100% different” from x.

(Should the dot product be turned into a vector too? Well, we have the inputs and a similarity percentage. There’s no new direction that isn’t available from either input.)

Geometric Interpretation

Two vectors determine a , and the cross product points in a direction different from both (source):

Here’s the problem: there’s two directions. By convention, we assume a “right-handed system” (source):

If you hold your first two fingers like the diagram shows, your thumb will point in the direction of the cross product. I make sure the orientation is correct by sweeping my

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first finger from a⃗ to b⃗ . With the direction figured out, the magnitude of the cross product is ∥a∥∥b∥ sin(θ)

, which is proportional to the magnitude of each vector and the “difference percentage” (sine).

The Cross Product For Orthogonal Vectors

To remember the right hand rule, write the xyz order twice: xyzxyz. Next, find the pattern you’re looking for:

 xy => z (x cross y is z)  yz => x (y cross z is x; we looped around: y to z to x)  zx => y

Now, xy and yx have opposite signs because they are forward and backward in our xyzxyz setup.

So, without a formula, you should be able to calculate:

Again, this is because x cross y is positive z in a right-handed . I used unit vectors, but we could scale the terms:

Calculating The Cross Product

A single vector can be decomposed into its 3 orthogonal parts:

When the vectors are crossed, each pair of orthogonal components (like ax ×by ) casts a vote for where the orthogonal vector should point. 6 components, 6 votes, and their total is the cross product. (Similar to the , where each axis casts a vote for the direction of greatest increase.)

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 xy => z and yx => -z (assume a⃗ is first, so xy means ax by)  yz => x and zy => -x  zx => y and xz => -y xy and yx fight it out in the z direction. If those terms are equal, such as in (2,1,0)×(2,1,1)

, there is no cross product component in the z direction (2 – 2 = 0).

The final combination is:

where n⃗ is the to a⃗ and b⃗

.

Don’t let this scare you:

 There’s 6 terms, 3 positive and 3 negative  Two dimensions vote on the third (so the z term must only have y and x components)  The positive/negative order is based on the xyzxyz pattern

If you like, there is an algebraic proof, that the formula is both orthogonal and of size ∥a∥∥b∥ sin(θ)

Example Time

Again, we should do simple cross products in our head:

Why? We crossed the x and y axes, giving us z (or i⃗ ×j⃗ =k⃗, using those unit vectors). Crossing the other way gives −k⃗

Here’s how I walk through more complex examples:

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 Let’s do the last term, the z-component. That’s (1)(5) minus (4)(2), or 5 – 8 = -3. I did z first because it uses x and y, the first two terms. Try seeing (1)(5) as “forward” as you scan from the first vector to the second, and (4)(2) as backwards as you move from the second vector to the first.  Now the y component: (3)(4) – (6)(1) = 12 – 6 = 6  Now the x component: (2)(6) – (5)(3) = 12 – 15 = -3

So, the total is (−3,6,−3)

So, let’s start with the two vectors

and then the cross product is given by the formula,

This is not an easy formula to remember. There are two ways to derive this formula. Both of them use the fact that the cross product is really the of a 3x3 . If you don’t know what this is that is don’t worry about it. You don’t need to know anything about matrices or to use either of the methods. The notation for the determinant is as follows,

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The first row is the standard vectors and must appear in the order given here. The second row is the components of and the third row is the components of . Now, let’s take a look at the different methods for getting the formula.

The first method uses the Method of Cofactors. If you don’t know the method of cofactors that is fine, the result is all that we need. Here is the formula.

where,

This formula is not as difficult to remember as it might at first appear to be. First, the terms alternate in and notice that the 2x2 is missing the column below the vector that multiplies it as well as the row of standard basis vectors.

The second method is slightly easier; however, many textbooks don’t cover this method as it will only on 3x3 determinants. This method says to take the determinant as listed above and then copy the first two columns onto the end as shown below.

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We now have three diagonals that move from left to right and three diagonals that move from right to left. We multiply along each diagonal and add those that move from left to right and subtract those that move from right to left.

This is best seen in an example. We’ll also use this example to illustrate a fact about cross products.

Example If and compute each of the following.

(a)

(b)

Solution

(a) Here is the computation for this one.

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(b) And here is the computation for this one.

Notice that switching the order of the vectors in the cross product simply changed all the signs in the result. Note as well that this means that the two cross products will point in exactly opposite directions since they only differ by a sign. We’ll formalize up this fact shortly when we list several facts.

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There is also a geometric interpretation of the cross product. First we will let θ be the between the two vectors and and assume that , then we have the following fact,

and the following figure.

There should be a natural question at this point. How did we know that the cross product pointed in the direction that we’ve given it here?

First, as this figure, implies the cross product is orthogonal to both of the original vectors. This will always be the case with one exception that we’ll get to in a second.

Second, we knew that it pointed in the upward direction (in this case) by the “right hand rule”. This says that if we take our right hand, start at and rotate our fingers towards our thumb will point in the direction of the cross product. Therefore, if we’d sketched in above we would have gotten a vector in the downward direction.

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Example

A plane is defined by any three points that are in the plane. If a plane contains the points , and find a vector that is orthogonal to the plane.

Solution

The one way that we know to get an orthogonal vector is to take a cross product. So, if we could find two vectors that we knew were in the plane and took the cross product of these two vectors we know that the cross product would be orthogonal to both the vectors. However, since both the vectors are in the plane the cross product would then also be orthogonal to the plane.

So, we need two vectors that are in the plane. This is where the points come into the problem. Since all three points lie in the plane any vector between them must also be in the plane. There are many ways to get two vectors between these points. We will use the following two,

The cross product of these two vectors will be orthogonal to the plane. So, let’s find the cross product.

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So, the vector will be orthogonal to the plane containing the three points.

Now, let’s address the one time where the cross product will not be orthogonal to the original vectors. If the two vectors, and , are parallel then the angle between them is either 0 or 180 degrees. From (1) this implies that,

From a fact about the magnitude we saw in the first section we know that this implies

In other words, it won’t be orthogonal to the original vectors since we have the zero vector. This does give us another test for parallel vectors however.

Fact

If then and will be parallel vectors.

Let’s also formalize up the fact about the cross product being orthogonal to the original vectors.

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Fact

Provided then is orthogonal to both and .

Here are some nice properties about the cross product.

Properties

If , and are vectors and c is a number then,

The determinant in the last fact is computed in the same way that the cross product is computed. We will see an example of this computation shortly.

There are a of geometric applications to the cross product as well. Suppose we have three vectors , and and we form the three dimensional figure shown below.

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The area of the parallelogram (two dimensional front of this object) is given by,

and the volume of the (the whole three dimensional object) is given by,

Note that the absolute value bars are required since the quantity could be negative and volume isn’t negative.

We can use this volume fact to determine if three vectors lie in the same plane or not. If three vectors lie in the same plane then the volume of the parallelepiped will be zero.

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Example Determine if the three vectors , and lie in the same plane or not.

Solution

So, as we noted prior to this example all we need to do is compute the volume of the parallelepiped formed by these three vectors. If the volume is zero they lie in the same plane and if the volume isn’t zero they don’t lie in the same plane.

So, the volume is zero and so they lie in the same plane.

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Appendix

Connection with the Determinant

You can calculate the cross product using the determinant of this matrix:

There’s a neat connection here, as the determinant (“signed area/volume”) tracks the contributions from orthogonal components.

There are theoretical reasons why the cross product (as an orthogonal vector) is only available in 0, 1, 3 or 7 dimensions. However, the cross product as a single number is essentially the determinant (a signed area, volume, or hypervolume as a ).

Connection with

Curl measures the twisting a vector field applies to a point, and is measured with a vector perpendicular to the surface. Whenever you hear “perpendicular vector” start thinking “cross product”.

We take the “determinant” of this matrix:

Instead of , the interaction is taking a partial derivative. As before, the i⃗ component of curl is based on the vectors and derivatives in the j⃗ and k⃗

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directions.

Relation to the

The cross and dot product are like the orthogonal sides of a :

For unit vectors, where ∥a∥=∥b∥=1

, we have:

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