Finding the cross product of two vectors is a useful way to find a new vector from another vector. The cross product creates a new vector that is orthogonal (perpendicular) to both input vectors.
You can use the cross product to find the perpendicular vector for one rotation axis to another rotation axis. This is very useful in physics applications, such as in angular displacement and velocity calculations.
However, if you are trying to find the cross product of two arbitrary vectors, you need to be careful. There is no formula for finding the cross product of two arbitrary vectors!
You can only find the cross product of two orthogonal (perpendicular) vectors. So, how do you find the cross product of any arbitrary pair of vectors then? Well, this article will discuss that and other helpful tips regarding the cross product.
Example using cross products
Now let’s look at an example. Suppose you are given the following vector:
a = A × B × C × D
Where A is the length of the first axis (in this case, it’s a unit vector, since the first axis is the “x” axis, so its length is 1), B is the length of the second axis (in this case, it’s “y” so its length is 2), C is the third axis (in this case it’s “z” so its length is 3), and D is a direction (say, counter-clockwise).
Now suppose we are given vectors I = I × J × K and J = J × K. How do we find vector K? First let us find out what cross products are:
Example using determinants
A great example of how to find the vector using determinants is given by the blog post above. Given a 3×3 matrix and a 2×2 matrix, the determinant of the first matrix multiplied by the second matrix can give you the vector.
Let’s use an easy example: A=|1| and B=|1 1|. Then (A B) = |1 1| × |1 1| = |−1| × |0| = −1||0=the vector.
This is because: AB=B(A)B=I=(−1)|0I=(−1)|0×(0)|0=(−1)(0)=−1||0.
This was done using properties of matrices, not determinants!
So, although this seems hard, you do not need to know how to use determinants to find the vector from two other matrices.
However, there are some cases where you need to use determinants to find the vector. These cases will be discussed later in this article.\r
…Continued from above.
So now that we know how to find the vector using only cross products and properties of cross products, let’s get back to answering our original question: Can we find the oriented edge segment from only two points?
The answer is…it depends! More specifically, it depends on whether or not one of the points is on the line segment. If one point is on the line segment, then we can!
“On Plane” Points
Consider two points A(x′A,y′A) and B(x′B,y′B). Assume that point A(x′A,y′A) lies on plane Π and point B(x′B,y′B) does not lie on plane Π.
- Point B(x″b , y″b ) does not lie on plane Π.
- Therefore there exists a perpendicular bisector \({{l}}_{{{\rm{T}}}_{{{\rm{X}}}}}\) of line segment {{{\rm{T}}}{{\rm{X}}}} connecting points \({{X}}\) and \({{X}}\) such that \({{{l}}}_{{{{\rm {T}}}_{{{{\rm {X}}} }}}}\) does not intersect plane Π.
Understanding the relationship between the two methods
A very interesting observation is that if you can find the vector v using only the dot product, then you can find the same vector using only the cross product and vice versa.
If you are given the vector v as the result of a dot product, then you can use Cramer’s theorem to show that any other vector w can be used to produce the same scalar triple product.
w = (I × J) × K where I, J, and K are the fundamental bases.
This is because if you take any vector w and perform the cross product with K, then you will get (I × J) × K which is equal to I × (J × K). Since I = v, we get that v = (I × J) × K which shows that v can be obtained by taking any other vector w and performing a cross product with K.
When should I use each method?
When solving determinant problems, you should always check if there is a vector solution. If there is, then you can solve the problem by finding the corresponding vector.
If there is not a vector solution, then you must solve for the scalar solution by finding the magnitude of each factor. You can then solve for the vector using one of the properties of cross products.
If no scalar or vector solutions exist, then you must find another solution. Unfortunately, there are some problems that have no solutions except for scalars. In these cases, try to find a simpler problem that has a scalar solution and expand it to find the solution to your original problem.
When solving linear systems using Gaussian elimination, always begin with solving the first column. Then check to see if the first column is solved correctly by putting it back into the original equation.
What are the properties of the cross product?
There are three properties of the cross product, or ways in which you can use it. The first is called the parallelogram rule.
When you compute the area of a parallelogram (a quadrilateral with opposite sides of equal length), you add the areas of the two sides.
That is, you compute the area of one side times the length of the other side. This area is then multiplied by a certain ratio to get the area of the opposite side.
You can do this with any parallelogram, such as a box. You would take the length, width, and height, and multiply them by each other to get the area.
Cross products behave like sides in this sense: You take two vectors and multiply them by each other to get a third vector that has an area equal to that of one vector times its reverse (the other vector).
I × J = k, where k is a vector with magnitude equal to the area of the parallelogram determined by I and J and direction counterclockwise when looking along J toward i
Imagine you are given three vectors and you are asked to find the magnitude of the vector that is the sum of these three vectors.
You could calculate the determinant of the three given vectors and use that to find the magnitude of the fourth vector, but there is an easier way!
First, calculate the area of a parallelogram with sides I and J. Then, take a fourth vector K with magnitude equal to this area and direction counterclockwise when looking along J toward I. The sum of these two vectors is your desired vector.
To better understand this, look at some examples. Given vectors i,
This method can be applied to any set of four length-dimensional (or higher) vectors.
(I × J) × K = 0 iff I × J = k where k is a length-dimensional (or higher) unit vector in the direction of I×J.
I × (J × K) = 0 iff I×K = J where K is a length-dimensional (or higher) unit vector in the direction of I×J.
(I × J) × K = 0 iff I × J = k where k is a length-dimensional (or higher) unit vector in the direction opposite that of I×J.
(I′′ − 2I′ + 3I + 2)2 − 6(2I′ − 1)(3I + 1)=0iff 2(1−1)=0iff 2=0iff .5=0iff .5^2=1 iff .5^2=(1−1)(1−1)=0if(A=(B=(C=(D=(E=(F())))))))))). Of course you could also just check all values between [−6;6] for each variable but this would take much longer!
This method works for finding any scalar value between two given values by using only determinants! Try it out yourself! …
I × J = -J × I
Another way to find the vector is by using the property that the cross product of two vectors is a vector perpendicular to both. You can use this property to find the vector of a determinant, not just evaluate one.
Parallel vectors do not interact, so the property fails for those cases. For example, if you had two vectors, one horizontal and one vertical, then their cross product would be a third horizontal vector.
You can evaluate any single element (called scalar) of a determinant by using basic algebra. For example, if you had the 2 × 3 determinant + + – , then you could subtract 4 from each |4| to get an equivalent 2 × 2 determinant . Then you could evaluate (1 − 4) × (2 − 3) = −6.
(I + J) × K = I × K + J × K + K × J
The last property of cross products that we will discuss is how to find the vector K when you have the scalar K and the vectors I, J, and I + J.
Once again, as in the above case, you can find K by using the Cartesian theorem:
K = (I + J) × K
Then you can find the vector K by taking the scalar K and crossing it with I × J. The order does not matter, since scalars are just a single value with no direction.
This is a very useful way to find vectors! You do not have to calculate all of the components of the vector K, you can just find one vector and cross it with another to get a new vector.
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