Introducing the concept of angle is more subtle than length. To this end we need the Cauchy-Schwarz inequality:
In each inner product space #V# we have, for all #\vec{a},\vec{b}\in V#,
\[|\dotprod{\vec{a}}{\vec{b}}|\leq\norm{\vec{a}} \cdot \norm{\vec{b}}
\]
This inequality is an equality if and only if the vectors #\vec{a}# and #\vec{b}# are linearly dependent.
Let #\vec{a}# and #\vec{b}# be two vectors from #V#. If one of the two vectors is the zero vector, both statements are obviously true, since then the left-hand and right-hand side of the equation both are equal to #0#.
We will now assume that the vectors #\vec{a}# and #\vec{b}# are distinct from the zero vector. Because of the definition of the inner product, we have \[ \dotprod{(\vec{a}+\lambda \vec{b})}{(\vec{a}+\lambda \vec{b})}\geq 0\] for each scalar #\lambda#. If we expand this inner product using linearity we find \[\dotprod{\vec{a}}{\vec{a}}+2\lambda(\dotprod{\vec{a}}{\vec{b}})+\lambda^2(\dotprod{\vec{b}}{\vec{b}})\geq 0\] We can see the expression before the inequality sign as a quadratic polynomial #f(\lambda)#. Since this function has at most one zero, we know that the discriminant is less than or equal to #0#. If the function #f(\lambda)# has no roots, the discriminant is negative. The well-known discriminant formula gives us \[4(\dotprod{\vec{a}}{\vec{b}})^2-4(\dotprod{\vec{a}}{\vec{a}})(\dotprod{\vec{b}}{\vec{b}})\leq 0\] We can rewrite this to \[4(\dotprod{\vec{a}}{\vec{b}})^2\leq 4(\dotprod{\vec{a}}{\vec{a}})(\dotprod{\vec{b}}{\vec{b}})\] Dividing this by #4# and writing the right-hand side in terms of norms, this gives us \[(\dotprod{\vec{a}}{\vec{b}})^2\leq \norm{\vec{a}}^2\cdot \norm{\vec{b}}^2 \] Because both sides are positive, we can take the square root. This leads to the Cauchy-Schwarz inequality: \[|\dotprod{\vec{a}}{\vec{b}}|\leq \norm{\vec{a}}\cdot \norm{\vec{b}}\]
It remains for us to consider the case of equality. Equality occurs if and only if the discriminant is zero. If the discriminant equals #0#, there is a unique solution to the equation #f(\lambda)=0#. In other words, there is a unique #\lambda_0# satisfying \[\dotprod{(\vec{a}+\lambda_0 \vec{b})}{(\vec{a}+\lambda_0\vec{b})}=0\] In view of the definition of the inner product this means that #\vec{a}+\lambda_0\vec{b}=\vec{0}#, which we can rewrite to #\vec{a}=-\lambda_0\vec{b}#. This expresses linear dependence of #\vec{a}# and #\vec{b}#.
In the inner product space #\mathbb{R}^3# with standard inner product the inequality for #\vec{a}=\rv{1,1,2}# and #\vec{b} =\rv{x,y,z}# reads
\[
(x+y+2z)^2 \leq (1^2 + 1^2 + 2^2)\cdot (x^2+y^2+z^2)=6(x^2+y^2+z^2)
\] Note that here both sides have been squared.
In the inner product space of all real continuous functions on #\ivcc{a}{b}# with the inner product of functions #\dotprod{f}{g} = \int_a^bf(x)\cdot g(x)\dd x# we find, for each pair of real continuous functions #f# and #g#, on #\ivcc{a}{b}#
\[
\left|\int_a^b f(x){g(x)}\,\dd x\right|^2\ \leq \ \int_a^b\left|f(x)\right|^2\,\dd x\int_a^b \left|g(x)\right|^2\,\dd x
\]These formulas are often used for estimates in Analysis of functions. For the functions #\ee^{x^2}# and # \ee^{-x^2}# on the interval #\ivcc{0}{1}# the Cauchy-Schwarz inequality gives
\[1= \left|\int_0^1 \ee^{x^2}\cdot\ee^{-x^2}\, \dd x \right|^2 \leq \int_0^1 \ee^{2x^2}\, \dd x \, \int_0^1\ee^{-2x^2}\, \dd x \] Here too, both sides of the Cauchy-Schwarz inequality have been squared.
The following characteristics of the norm have been announced previously.
Let #V# be an inner product space. For all vectors #\vec{a}# and #\vec{b}# and all scalars #\lambda# we have
- Positivity: #\norm{\vec{a}}\geq 0#, with equality if and only if #\vec{a} =\vec{0}#
- Triangle inequality: \(\norm{\vec{a} + \vec{b}} \leq \norm{\vec{a}} + \norm{\vec{b}}\)
- Multiplicativity: #\norm{\lambda\cdot\vec{a}} = | \lambda | \cdot \norm{\vec{a}}#
The triangle inequality reflects the inequality known from the space of arrows:
The diagonal is the segment from #\vec{0}# to #\vec{a}+\vec{b}#. This is the shortest path from #\vec{0}# to #\vec{a}+\vec{b}#.
The first part follows immediately from the third property of the definition of inner product.
For the triangle inequality, we need the Cauchy-Schwarz inequality: \[ \begin{array}{rcl} \norm{\vec{a}+\vec{b}}^2&=& \dotprod{(\vec{a}+\vec{b})}{(\vec{a}+\vec{b})}\\ &&\phantom{xx}\color{blue}{\text{definition norm}}\\& = & \dotprod{\vec{a}}{\vec{a}}+\dotprod{\vec{b}}{\vec{b}}+\dotprod{\vec{a}}{\vec{b}}+\dotprod{\vec{b}}{\vec{a}}\\&&\phantom{xx}\color{blue}{\text{bilinearity}}\\&=&\norm{\vec{a}}^2+ \norm{\vec{b}}^2+2\dotprod{\vec{a}}{\vec{b}}\\&&\phantom{xx}\color{blue}{\text{definition norm and symmetry}}\\&\leq&\norm{\vec{a}}^2+ \norm{\vec{b}}^2+2|\dotprod{\vec{a}}{\vec{b}}|\\&&\phantom{xx}\color{blue}{{x}\leq \abs{x}}\\& \leq& \norm{\vec{a}}^2+ \norm{\vec{b}}^2+2 \norm{\vec{a}}\cdot \norm{\vec{b}}\\&&\phantom{xx}\color{blue}{\text{Cauchy-Schwarz}}\\&=&(\norm{\vec{a}}+ \norm{\vec{b}})^2\\&&\phantom{xx}\color{blue}{\text{factored}}\\ \end{array}\] The triangle inequality follows by taking square roots at both sides.
The third part follows from the linearity and symmetry of the inner product: \[(\lambda \vec{a},\lambda \vec{a}) = \lambda (\vec{a},\lambda\vec{a})= \lambda
\cdot\lambda(\vec{a},\vec{a})= \lambda^2\cdot(\dotprod{\vec{a}}{\vec{a}})=\abs{\lambda}^2\cdot(\dotprod{\vec{a}}{\vec{a}})\] If we take the square root of the rightmost and the leftmost side, we get the equality \[\norm{\lambda\cdot \vec{a}} = |\lambda |\cdot \norm{\vec{a}}
\]
The triangle inequality gives us an upper bound on the length of a sum of two vectors. However, from the triangle inequality we can also find a lower bound. The inequality holds for all #\vec{a}# and #\vec{b}#, and so also for the vectors #\vec{a}-\vec{b}# and #\vec{b}#. We then find \[\norm{\vec{a}} \,=\, \norm{(\vec{a} -\vec{b})+\vec{b}}\leq\norm{\vec{a} -\vec{b}} + \norm{\vec{b}}
\]so\[ \norm{\vec{a} -\vec{b}}\ \geq\ \norm{\vec{a}} - \norm{\vec{b}}
\]This statement applies to all vectors. In particular for the vectors #\vec{a}# and #-\vec{b}#, it gives us \[\norm{\vec{a} + \vec{b}} \,\geq \, \norm{\vec{a}} -\norm{-\vec{b}} \,=\, \norm{\vec{a}} -\norm{\vec{b}}\] Interchanging #\vec{a} # and #\vec{b}# now produces \[
\norm{\vec{b} +\vec{a}} \geq \norm{\vec{b}} - \norm{\vec{a}}\] so, for all vectors #\vec{a}# and #\vec{b}#, we find \[
\norm{\vec{a} +\vec{b}}\ \geq \abs{\norm{\vec{a}} - \norm{\vec{b}}}
\]We conclude that \[\abs{\norm{\vec{a}} - \norm{\vec{b}}}\, \leq\, \norm{\vec{a}+\vec{b}}\, \leq \,\norm{\vec{a}}+\norm{\vec{b}}\]
A vector space having a length concept that meets the requirements of the theorem is also referred to as a normed vector space.
The triangle inequality for distance, #d(\vec{a},\vec{c})\le d(\vec{a},\vec{b}) + d(\vec{b},\vec{c}) #, is a direct consequence of the triangle inequality for length:\[\begin{array}{rcl}d(\vec{a},\vec{c})&=&\norm{\vec{a}-\vec{c}}\\ &&\phantom{xx}\color{blue}{\text{definition distance}}\\ &=&\norm{(\vec{a}-\vec{b})+(\vec{b}-\vec{c})}\\ &&\phantom{xx}\color{blue}{\text{sum rewritten with }\vec{b}}\\ &\le &\norm{\vec{a}-\vec{b}}+\norm{\vec{b}-\vec{c}}\\ &&\phantom{xx}\color{blue}{\text{triangle inequality for length}}\\ &=&d(\vec{a},\vec{b}) + d(\vec{b},\vec{c}) \\ &&\phantom{xx}\color{blue}{\text{definition distance}}\end{array}\]
Thanks to the Cauchy-Schwarz inequality we can also use the inner product on a real vector space to define the angle between two vectors:
Let #V# be an inner product space and let #\vec{a}# and #\vec{b}# be two vectors distinct from the zero vector. Then there exists a real number #\varphi\in\ivcc{0}{\pi}# such that \[\cos(\varphi) = \frac{\dotprod{\vec{a}}{\vec{b}}}{\norm{\vec{a}}\cdot \norm{\vec{b}}}\] The number #\varphi# is called the angle between the two vectors #\vec{a}# and #\vec{b}#. The angle does not depend on the length of #\vec{a}# or the length of #\vec{b}#.
We recall the Cauchy-Schwarz inequality \[|\dotprod{\vec{a}}{\vec{b}}|\ \leq\ \norm{\vec{a}} \cdot \norm{\vec{b}}\] We divide both sides by #\norm{\vec{a}}\cdot\norm{\vec{b}}# to get \[ \frac{|\dotprod{\vec{a}}{\vec{b}}|}{\norm{\vec{a}} \cdot \norm{\vec{b}}}\leq 1\] Application of the definition of absolute value gives the following inequalities: \[-1\ \leq \ \frac{\dotprod{\vec{a}}{\vec{b}}}{\norm{\vec{a}} \cdot \norm{\vec{b}}} \leq \ 1\]Thus there exists an angle #\varphi# such that
\[\cos(\varphi) = \frac{\dotprod{\vec{a}}{\vec{b}}}{\norm{\vec{a}}\cdot \norm{\vec{b}}}\]
The angle does not depend on the length: \[\cos(\varphi) = \frac{\dotprod{\vec{a}}{\vec{b}}}{\norm{\vec{a}}\cdot \norm{\vec{b}}}\ = \dotprod{\left( \frac{1}{\norm{\vec{a}}} \vec{a}\right)}{\left( \frac{1}{\norm{\vec{b}}} \vec{b}\right)}\]
In the figure below we see how the angle and the inner product are related to each other. If we rewrite the equation we get
\[\dotprod{\vec{a}}{\vec{b}}=\norm{\vec{a}} \cdot \norm{\vec{b}} \cdot \cos (\varphi)
\]
We keep the lengths #\vec{a}# and #\vec{b}# equal (this does not matter for the inner product). You can drag the non-horizontal vector in order to see what happens to the angle. We can see from the equation that the absolute value of the inner product is at a maximum if the angle equals #0# or #180# degrees.
What angle do the vectors #\vec{a} = \rv{2,1,-1}# and #\vec{b}=\rv{-2,2,4}# make in the inner product space #\mathbb{R}^3# with standard inner product?]
Give your answer in radians.
#\varphi = # # {{\pi}\over{3}}#
First we calculate the inner product of the vectors #\vec{a}# and #\vec{b}# and their lengths
\[\begin{array}{rclcl}
\dotprod{\vec{a}}{\vec{b}} &=& (2)\cdot (-2) + (1)\cdot (2) + (-1)\cdot (4) &=&-6 \\
\norm{\vec{a}} &=& \sqrt{(2)^2+ (1)^2 + (-1)^2} &=&\sqrt{6} \\
\norm{\vec{b}} &=& \sqrt{(-2)^2+ (2)^2 + (4)^2} &=&2\sqrt{6} \end{array}
\] It follows that the angle #\varphi# between #\vec{a}# and #\vec{b}# satisfies
\[\cos(\varphi) = \frac{\dotprod{\vec{a}}{\vec{b}}}{\norm{\vec{a}}\cdot\norm{\vec{b}}}=\frac{-6}{\sqrt{6} \cdot 2\sqrt{6}} = {{1}\over{2}}\] We conclude that the answer is #\varphi = {{\pi}\over{3}}#.
Angle
