Dot products and Cross products
Dot Products
Given any two vectors and in or , we define their dot product as follows:
In , and ,
In , and ,
Remark: The dot product of any two vectors is a scalar i.e. a real number.
We can easily derive the following properties of dot product from its definition:
Let be vectors in or and be a real number.
1.
2.
3.
4.
5.
It turns out that for any two non-zero vectors and in or , it can be shown that their dot product can also be expressed as follows:
when both vectors are non-zero and is the angle in the range from to formed by translating two non-zero vectors such that their tails meet at a point.
In the applet below, bring the two vectors together in the plane to find out their dot product.
Question: Describe what happens to two non-zero vectors and and their dot product when (a) (b) (c) (d) (e)
Definition: Two vectors are orthogonal if their dot product is zero i.e. if two vectors are orthogonal, they are either both non-zero and perpendicular to each other, or at least one of the vectors is a zero vector.
Remark: Zero vector is orthogonal to any vector.
Exercise: (a) Let and . Show that they are orthogonal. (b). Let and . Find the angle between and .
Theorem: Given any two vectors and in or ,
The proof is as follows:
Orthogonal Projections
Given non-zero vectors and , let be the line through the tail of in the direction of . We define the (orthogonal) projection of onto , denoted by , to be the vector pointing from the tail of such that its head is the foot of the perpendicular from the head of to the the line .
In the applet below, the red vector is the projection of onto .
If , its magnitude is and it is in the same direction as . Hence, we have
()
If , its magnitude is and it is in opposite direction to . Hence, we have
(Note: When , the tail of is the foot of the perpendicular and hence .)
Combining above, we have the formula for the projection:
Using the dot product, we can also rewrite the formula as follows:
.
In (), the factor multiplying to the unit vector in the direction of is called the scalar projection of onto , denoted by i.e.
It is also called the scalar component of in the direction of .
Exercise: Let and . Compute .
Exercise: Given any non-zero vectors and , let i.e. . Show that and are orthogonal. (Decomposition of a vector into its orthogonal components)
An Application - Work
A constant force is applied to an object. We define the work done (W) by the force over the displacement of the object as follows:
Therefore, we have .
Example: Given a force that moves an object from to . Find the work done by .
. Therefore, we have
(J stands for "joule", the unit for energy i.e. work done).
Cross Product
Given two vectors and in , we are going to define the cross product of these two vectors. But first of all, we need to recall the definition of the determinant of a matrix:
Suppose is a matrix. Its determinant
Then we can define the determinant of a matrix as follows:
Example:
Write and . The cross product, is defined as follows:
Remark:
- Cross product of two vectors in is a vector in .
- The order of taking cross product is important: In general,
Properties of Cross Product
Recall that if we interchange any two rows of a matrix, its determinant will be changed as follows:
Therefore, by the definition of cross product, we have the following properties:
- and in particular, .
- Suppose . (Note: In some textbooks, this is called the scalar triple product). In particular, i.e. is orthogonal to both and .
- Let and be real numbers, then .
- For any vectors , and . (Distributive laws).
In the applet below, you can do the following:
- First construct two vectors and by typing "u=vector((1,2,3))" and "v=vector((4,5,6))".
- Type "cross(u,v)" to get the cross product .
- Draw the plane containing the two vectors through the origin by typing "plane((0,0,0),u,v)". Then you will see that the cross product is perpendicular to the plane.
- Verify the right-hand rule using some other vectors.
- You can also find , and .
We have the following nice theorem about the norm of a cross product:
Theorem: , where is the angle between the two vectors.
Proof:
(Note: positive square root here because for ).
Remark: is exactly the area of the parallelogram spanned by and .
Exercise: Find the area of the triangle with vertices , and using the above theorem.
The Volume of a Parallelepiped
Given three vectors in . We can compute its scalar triple product . It turns out that it is related to the volume of the parallelepiped formed by these three vectors, as shown in the applet below.
In general, the volume of a parallelepiped is the product of the area of its base parallelogram and its height (the distance between the two planes that contain the base and top respectively). By the previous theorem, the area of the base parallelogram is . Moreover, the height of the parallelepiped is . Therefore, we have
Volume =
In the applet below, the dotted line contains , which is perpendicular to both planes containing the top and base of the parallelepiped. The red vector is and its length is exactly the height of the parallelepiped.
In this example, . By the right-hand rule, the red vector is in the same direction as . However, if you interchange and , the red vector will point to the opposite direction. Its length is still the height of the parallelepiped though.
Moreover, you can check that the sign of to see whether and are in the same side of the plane containing the base.
Question: