Vector Multiplication
Discussion
scalar-vector multiplication
Multiplication of a vector by a scalar changes the magnitude of the vector, but leaves its direction unchanged. The scalar changes the size of the vector. The scalar "scales" the vector. For example, the polar form vector…
r = r r̂ + θ θ̂
multiplied by the scalar a is…
a r = ar r̂ + θ θ̂
Multiplication of a vector by a scalar is distributive.
a(A + B) = a A + a B
Consequently, the rectangular form vector…
r = x î + y ĵ
multiplied by the scalar a is…
ar = ax î + ay ĵ
dot product
Geometrically, the dot product of two vectors is the magnitude of one times the projection of the second onto the first.
The symbol used to represent this operation is a small dot at middle height (·), which is where the name "dot product" comes from. Since this product has magnitude only, it is also known as the scalar product.
A · B = AB cos θ
The dot product is distributive…
A · (B + C) = A · B + A · C
and commutative…
A · B = B · A
Since the projection of a vector on to itself leaves its magnitude unchanged, the dot product of any vector with itself is the square of that vector's magnitude.
A · A = AA cos 0° = A^{2}
Applying this corollary to the unit vectors means that the dot product of any unit vector with itself is one. In addition, since a vector has no projection perpendicular to itself, the dot product of any unit vector with any other is zero.
î · î = ĵ · ĵ = k̂ · k̂ = (1)(1)(cos 0°) = 1
î · ĵ = ĵ · k̂ = k̂ · î = (1)(1)(cos 90°) = 0
Using this knowledge we can derive a formula for the dot product of any two vectors in rectangular form. The resulting product looks like it's going to be a terrible mess, but consists mostly of terms equal to zero.
A · B = (A_{x} î + A_{y} ĵ + A_{z} k̂) · (B_{x} î + B_{y} ĵ + B_{z} k̂)
A · B = | A_{x} î | · | B_{x} î | + | A_{x} î | · | B_{y} ĵ | + | A_{x} î | · | B_{z} k̂ | |
+ | A_{y} ĵ | · | B_{x} î | + | A_{y} ĵ | · | B_{y} ĵ | + | A_{y} ĵ | · | B_{z} k̂ | |
+ | A_{z} k̂ | · | B_{x} î | + | A_{z} k̂ | · | B_{y} ĵ | + | A_{z} k̂ | · | B_{z} k̂ |
A · B = A_{x}B_{x} + A_{y}B_{y} + A_{z}B_{z}
The dot product of two vectors is thus the sum of the products of their parallel components. From this we can derive the Pythagorean Theorem in three dimensions.
A · A = AA cos 0° = A_{x}A_{x} + A_{y}A_{y} + A_{z}A_{z}
A^{2} = A_{x}^{2} + A_{y}^{2} + A_{z}^{2}
cross product
Geometrically, the cross product of two vectors is the area of the parallelogram between them.
The symbol used to represent this operation is a large diagonal cross (×), which is where the name "cross product" comes from. Since this product has magnitude and direction, it is also known as the vector product.
A × B = AB sin θ n̂
The vector n̂ ("n hat") is a unit vector perpendicular to the plane formed by the two vectors. The direction of n̂ is determined by the right hand rule, which will be discussed shortly.
The cross product is distributive…
A × (B + C) = (A × B) + (A × C)
but not commutative…
A × B = −B × A
Reversing the order of cross multiplication reverses the direction of the product.
Since two identical vectors produce a degenerate parallelogram with no area, the cross product of any vector with itself is zero…
A × A = 0
Applying this corollary to the unit vectors means that the cross product of any unit vector with itself is zero.
î × î = ĵ × ĵ = k̂ × k̂ = (1)(1)(sin 0°) = 0
It should be apparent that the cross product of any unit vector with any other will have a magnitude of one. (The sine of 90° is one, after all.) The direction is not intuitively obvious, however. The right hand rule for cross multiplication relates the direction of the two vectors with the direction of their product. Since cross multiplication is not commutative, the order of operations is important.
- Hold your right hand flat with your thumb perpendicular to your fingers. Do not bend your thumb at anytime.
- Point your fingers in the direction of the first vector.
- Orient your palm so that when you fold your fingers they point in the direction of the second vector.
- Your thumb is now pointing in the direction of the cross product.
A right-handed coordinate system, which is the usual coordinate system used in physics and mathematics, is one in which any cyclic product of the three coordinate axes is positive and any anticyclic product is negative. Imagine a clock with the three letters x-y-z on it instead of the usual twelve numbers. Any product of these three letters that runs around the clock in the same direction as the sequence x-y-z is cyclic and positive. Any product that runs in the opposite direction is anticyclic and negative.
The cross product of a cyclic pair of unit vectors is positive. |
The cross product of an anticyclic pair of unit vectors is negative. |
Using this knowledge we can derive a formula for the cross product of any two vectors in rectangular form. The resulting product looks like it's going to be a terrible mess, and it is!
A × B = (A_{x} î + A_{y} ĵ + A_{z} k̂) × (B_{x} î + B_{y} ĵ + B_{z} k̂)
The product of two trinomials has nine terms.
A × B | = | A_{x} î | × | B_{x} î | + | A_{x} î | × | B_{y} ĵ | + | A_{x} î | × | B_{z} k̂ |
+ | A_{y} ĵ | × | B_{x} î | + | A_{y} ĵ | × | B_{y} ĵ | + | A_{y} ĵ | × | B_{z} k̂ | |
+ | A_{z} k̂ | × | B_{x} î | + | A_{z} k̂ | × | B_{y} ĵ | + | A_{z} k̂ | × | B_{z} k̂ |
Three of these are zero. Eliminate them.
A × B | = | A_{x}B_{y} k̂ | − | A_{x}B_{z} ĵ |
− | A_{y}B_{x} k̂ | + | A_{y}B_{z} î | |
+ | A_{z}B_{x} ĵ | − | A_{z}B_{y} î |
Group terms by unit vector and factor.
A × B = (A_{y}B_{z} − A_{z}B_{y}) î
There is a simpler way to write this. For those of you familiar with matrices, the cross product of two vectors is the determinant of the matrix whose first row is the unit vectors, second row is the first vector, and third row is the second vector. Symbolically…
A × B = | î | ĵ | k̂ |
A_{x} | A_{y} | A_{z} | |
B_{x} | B_{y} | B_{z} |
Expanding a 3×3 determinant by its first row is a first step. This gives us three 2×2 deteminants.
A × B = | A_{y} | A_{z} | î − | A_{x} | A_{z} | ĵ + | A_{x} | A_{y} | k̂ |
B_{y} | B_{z} | B_{x} | B_{z} | B_{x} | B_{y} |
These 2×2 deteminants can be found quickly. They also give us a solution that is presorted by unit vector, so there is no need to sort terms and factor.
A × B = (A_{y}B_{z} − A_{z}B_{y}) î