Tuesday, February 5

Outer product of vectors in R^3


Outer product of vectors in R^3
Cross product definition:
The outer product is also called the cross product of vectors. The formal definition is as follows:
If vector x = (x1,x2,x3) and vector y = (y1,y2,y3) are vectors in R^3, then their outer product is denoted by x X y and is defined as
x X y = (x1,x2,x3) X (y1,y2,y3)
         = (x2y3 – x3y2, -(x1y3 – x3y1), x1y2 – x2y1) <- cross="" is="" p="" product="" the="" this="" vector.="">Note that the cross product of two vectors is always another vector. Alternatively in matrix notation we can write the cross product identities as follows:
x X y = (|(x2&x3@y2&y3)|, - |(x1&x3@y1&73)|, |(x1&x2@y1&y2)|)

Properties of cross product:
(1) x X y = -y X x
That is because when we do y X x we interchange the rows of the cross product determinants. Interchange of rows results in negative value of the same determinant.

(2) x X x = 0
That is because if a determinant has two identical rows, then the value of the determinant is zero. When finding x X x, the two rows would be identical.

(3) x X ky = kx X y = k(x X y).
Here, k is a scalar. If one of the rows of a determinant has a common factor k, then the k can be taken out of the determinant and the resulting determinant value when multiplied by k gives us the value of the original determinant.

(4) x X (y + z) = x X y + x X z
This is just distributive property. The distributive property is applicable to determinants there for it also applies to the outer product.

Cross product rules:
The following rules apply to all outer products of vectors. The outer product is also called vector product of two vectors.

(1) The vector product of two vectors is always another vector.
(2) The vector product is not commutative as we saw in the first property above.
(3) The vector product is not defined for vectors in R or R^2. It is defined only for vectors in R^3 or higher order dimensions.

Sample problem:
Find the outer product of the vectors x = (1,2,3) and y = (-1,3,5).
Solution:
x X y = (|(2&3@3&5)|, - |(1&3@-1&5)|, |(1&2@-1&3)|) = (10-9, -(5+3), (3+2)) = (1,-8,5) <- answer.="" p="">

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