Answer:
Dot Product and Angle
Given two vectors and , let be the angle between them such that . We will refer to as the included angle.
The following theorem establishes a relationship between the dot product and the included angle.
Let and be vectors in , and let be the included angle. Then \begin{equation*} \label{matlintrans} \vec{u}\dotp \vec{v}=\norm{\vec{u}}\norm{\vec{v}}\cos \theta \end{equation*}
- Proof
- Consider the triangle formed by , and .
By the Law of Cosines we have: \begin{align*} \norm{\vec{u}-\vec{v}}^2=\norm{\vec{u}}^2+\norm{\vec{v}}^2-2\norm{\vec{u}}\norm{\vec{v}}\cos \theta \end{align*}
By Theorem th:dotproductpropertiesitem:norm of Dot Product and its Properties \begin{align*} (\vec{u}-\vec{v})\dotp (\vec{u}-\vec{v})=&\vec{u}\dotp \vec{u}+\vec{v}\dotp \vec{v}-2\norm{\vec{u}}\norm{\vec{v}}\cos \theta \end{align*}
By Theorem th:dotproductpropertiesitem:distributive-again of Dot Product and its Properties \begin{align*} (\vec{u}-\vec{v})\dotp \vec{u}-(\vec{u}-\vec{v})\dotp \vec{v}=&\vec{u}\dotp \vec{u}+\vec{v}\dotp \vec{v}-2\norm{\vec{u}}\norm{\vec{v}}\cos \theta \end{align*}
By Theorem th:dotproductpropertiesitem:distributive of Dot Product and its Properties \begin{align*} \vec{u}\dotp \vec{u}-\vec{v}\dotp \vec{u}-\vec{u}\dotp \vec{v}+\vec{v}\dotp \vec{v}=&\vec{u}\dotp \vec{u}+\vec{v}\dotp \vec{v}-2\norm{\vec{u}}\norm{\vec{v}}\cos \theta \end{align*}
By Theorem th:dotproductpropertiesitem:commutative of Dot Product and its Properties \begin{align*} -2(\vec{u}\dotp \vec{v})=&-2\norm{\vec{u}}\norm{\vec{v}}\cos \theta \\ \vec{u}\dotp \vec{v}=&\norm{\vec{u}}\norm{\vec{v}}\cos \theta \end{align*}
Orthogonal Vectors
We can use Theorem th:dotproductcosine to show that two non-zero orthogonal vectors of are simply perpendicular vectors (the included angle is ). To see this, suppose that for nonzero vectors . Then from Theorem th:dotproductcosine we have Since are nonzero vectors, we have , which implies . The converse also holds. If , then the dot product is clearly 0.
The reason we prefer the term “orthogonal” to “perpendicular” in this course is because is only one example of a vector space, and the dot product is only one example of a more general product, called an inner product. For vectors in a zero dot product happens to coincide with the geometric idea of perpendicularity, but there are many vector spaces that do not possess the visual geometry of . (Later in the text, you will encounter vector spaces whose vectors are polynomial functions!) In these more abstract settings, a zero inner product still signals a special relationship between vectors. The term orthogonal captures this relationship.
By our definition, the zero vector is orthogonal to any vector. However, we will not
use the word perpendicular when the zero vector is involved, as it is not possible to
talk about an “included angle”.
Practice Problems
Problems prob:anglebetweenvectors1-prob:anglebetweenvectors4
Find the degree measure of the included angle, for each pair of vectors. Round your answers to the nearest tenth.
Find the value of for which the vector is parallel to the vector . What is the
measure of the included angle, ? Find the measure of the included angle using
Theorem ex:anglebetweenvectors. Do the two results agree?
Answer:
Imagine a clock with hands represented by vectors and , as shown below. At what
whole hour will attain its maximum value? At what whole hour will be as small as
possible?
Answer:
Let be a circle of radius . Suppose and are the endpoints of a diameter
of , and is a point on distinct from and . Show that vectors and are
orthogonal.
Assign coordinates to points , and , express vectors and in component form, then
find the dot product of and .
A rhombus is a quadrilateral with four congruent sides. Use vectors to prove that a
parallelogram is a rhombus if and only if its diagonals are perpendicular.
See section
on vector subtraction in Vector Arithmetic.
Photo Credits
The following images are courtesy of Wikimedia Commons
Hannes Grobe, Wall clock manufactured by Telefonbau & Normalzeit. CC-BY 3.0
2024-09-11 17:58:51