English | Español

Try our Free Online Math Solver!

Online Math Solver

 

 

 

 

 

 

 

 
 
 
 
 
 
 
 
 

 

 

 
 
 
 
 
 
 
 
 

Please use this form if you would like
to have this math solver on your website,
free of charge.


Vectors

44.3 Example. Let and compute the cross product of these vectors.

Solution:

In terms of the standard basis vectors you can check the multiplication table . An easy way to remember
the multiplication table is to put the vectors clockwise in a circle. Given two of the three
vectors their product is either plus or minus the remaining vector. To determine the sign you step from
the first vector to the second, to the third: if this makes you go clockwise you have a plus sign, if you
have to go counterclockwise, you get a minus.

The products of and are all you need to know to compute the cross product. Given two vectors
and write them as and, and multiply as follows

This is a useful way of remembering how to compute the cross product, particularly when many of the
components and are zero.

44.4 Example. Compute

There is another way of remembering how to find ×. It involves the “triple product” and
determinants. See § 44.3.

44.2. Algebraic properties of the cross product

Unlike the dot product, the cross product of two vectors behaves much less like ordinary multiplication.
To begin with, the product is not commutative – instead one has

for all vectors and.

This property is sometimes called “anti-commutative.”

Since the crossproduct of two vectors is again a vector you can compute the cross product of three
vectors You now have a choice: do you first multiply and , or and , or and ? With
numbers it makes no difference (e.g. 2× (3 × 5) = 2 ×15 = 30 and (2 × 3) ×5 = 6 ×5 = also 30) but
with the cross product of vectors it does matter: the cross product is not associative, i.e.

for most vectors

so"×” is not associative

The distributive law does hold, i.e.



is true for all vectors

Also, an associative law, where one of the factors is a number and the other two are vectors, does
hold. I.e

holds for all vectors, and any number t. We were already using these properties when we multiplied
in the previous section .

Finally, the cross product is only defined for space vectors, not for plane vectors.

44.3. The triple product and determinants

Definition 44.5. The triple product of three given vectors,, and is defined to be

In terms of the components of ,, and one has

This quantity is called a determinant, and is written as follows

There’s a useful shortcut for computing such a determinant:
after writing the determinant, append a fourth and a fifth column
which are just copies of the first two columns of the determinant.
The determinant then is the sum of six products,
one for each dotted line in the drawing . Each term has a sign:
if the factors are read from top-left to bottom-right, the term is
positive
, if they are read from top-right to bottom left the term
is negative .

This shortcut is also very useful for computing the cross product. To compute the cross product of two
given vectors and you arrange their components in the following determinant

This is not a normal determinant since some of its entries are vectors, but if you ignore that odd circumstance
and simply compute the determinant according to the definition (67), you get (68).

An important property of the triple product is that it is much more symmetric in the factors ,,
than the notation suggests.

Theorem 44.6. For any triple of vectors one has

and

In other words, if you exchange two factors in the product it changes its sign. If you
“rotate the factors,” i.e. if you replace by, by and by , the product doesn’t change at all.

44.4. Geometric description of the cross product

 

Theorem 44.7.

Proof. We use the triple product:

since for any vector . It follows that × is perpendicular to.
Similarly, • (×) = • (×) =!0 shows that is perpendicular to.

Theorem 44.8.

Proof. Bruce just slipped us a piece of paper with the following formula on it :

After setting andand diligently computing both sides we find that this formula
actually holds for any pair of vectors ,! The (long) computation which implies this identity will be
presented in class (maybe).

If we assume that Bruce’s identity holds then we get

since . The theorem is proved.

These two theorems almost allow you to construct the cross product of two vectors geometrically. If
and are two vectors, then their cross product satisfies the following description:

(1) If and are parallel, then the angle θ between them vanishes, and so their cross product is
the zero vector . Assume from here on that and are not parallel.

(2) × is perpendicular to both and . In other words, since and are not parallel, they
determine a plane, and their cross product is a vector perpendicular to this plane.

(3) the length of the cross product× is sin θ.

There are only two vectors that satisfy conditions 2 and 3: to determine which one of these is
the cross product you must apply the Right Hand Rule (screwdriver rule, corkscrew rule, etc.) for
,,×: if you turn a screw whose axis is perpendicular to and in the direction from to , the
screw moves in the direction of ×.


Alternatively, without seriously injuring yourself, you should be able to make a fist with your right
hand, and then stick out your thumb, index and middle fingers so that your thumb is , your index
finger is and your middle finger is ×. Only people with the most flexible joints can do this with
their left hand.

45. A few applications of the cross product

45.1. Area of a parallelogram

Let ABCD be a parallelogram. Its area is given by “height times base ,” a formula which should be
familiar from high school geometry.

If the angle between the sides AB and AD is θ, then the height of the parallelogram is ,
so that the area of ABCD is

The area of the triangle ABD is of course half as much,


These formulae are valid even when the points A, B, C, and D are points in space. Of course they
must lie in one plane for otherwise ABCD couldn’t be a parallelogram.

 

Prev Next