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--- lecture_18 [2015/04/02 10:25] – rupert
+++ lecture_18 [2017/04/06 10:04] (current) – rupert
@@ Line 1: / Line 1: @@
-==== Remark ====
+====== The dot product ======
-$\|\vec{AB}\|$ is the distance from point $A$ to point $B$, since this is the length of vector which takes point $A$ to point $B$.
+==== Definition of the dot product ====
-=== Examples ===
+{{page>dot product}}
-  * The distance from $A=(1,2)$ to $B=(-3,4)$ is $\|\def\m#1{\begin{bmatrix}#1\end{bmatrix}}\vec{AB}\|=\left\|\m{-3\\4}-\m{1,2}\right\|=\left\|\m{-4\\2}\right\|=\sqrt{(-4)^2+2^2}=\sqrt{20}=2\sqrt{5}$.
+Note that while $\vec v$ and $\vec w$ are vectors, their dot product $\vec v\cdot \vec w$ is a scalar.
-  * The length of the main diagonal of the unit cube in $\mathbb{R}^3$ is the distance between $0=(0,0,0)$ and $A=(1,1,1)$, which is $\|\vec{0A}\|=\left\|\m{1\\1\\1}\right\|=\sqrt{1^2+1^2+1^2}=\sqrt3$.
-==== Scalar multiplication and direction ====
+=== Example ===
-Multiplying a vector by a scalar changes its length, but doesn't change its direction.
+If $\vec v=\m{3\\5}$ and $\vec w=\m{4\\-7}$, then $\vec v\cdot \vec w=\m{3\\5}\cdot \m{4\\-7} = 3(4)+5(-7)=-23$.
-==== Definition: unit vectors ====
+===== Properties of the dot product =====
-{{page>unit vector}}
+For any vectors $\vec v$, $\vec w$ and $\vec u$ in $\mathbb{R}^n$, and any scalar $c\in \mathbb{R}$:
-==== Proposition: finding a unit vector in the same direction as a given vector ====
+  - $\def\dp#1#2{\vec #1\cdot \vec #2}\dp vw=\dp wv$ (the dot product is commutative)
+  - $\vec u\cdot(\vec v+\vec w)=\dp uv+\dp uw$
+  - $(c\vec v)\cdot \vec w=c(\dp vw)$
+  - $\dp vv=\|\vec v\|^2\ge 0$, and $\dp vv=0 \iff \vec v=0_{n\times 1}$
-If $\vec v$ is a non-zero vector, then $\vec w=\frac1{\|\vec v\|}\vec v$ is a unit vector (in the same direction as $\vec v$).
+The proofs of these properties are exercises.
-=== Proof ===
+===== Angles and the dot product =====
-Using the formula $\|c\vec v\|=|c|\,\|\vec v\|$ and the fact that $\|\vec v\|>0$, we have
+==== Theorem: the relationship between angle and the dot product ====
-\[ \|\vec w\|=\left\|\frac1{\|\vec v\|}\vec v\right\|=\left|\frac1{\|\vec v\|}\right|\,\|\vec v\|=\frac1{\|\vec v\|}\,\|\vec v\| = 1.\]
-So $\vec w$ is a unit vector, and since it's scalar multiple of $\vec v$, it's in the same direction as $\vec v$. ■
-==== Example ====
+If $\vec v$ and $\vec w$ are non-zero vectors in $\mathbb{R}^n$, then
+\[ \dp vw=\|\vec v\|\,\|\vec w\|\,\cos\theta\]
+where $\theta$ is the angle between $\vec v$ and $\vec w$.
-What is unit vector in the same direction as $\vec v=\m{1\\2}$?
+The proof will be given soon, but for now here is an example.
-We have $\|\vec v\|=\sqrt{1^2+2^2}=\sqrt5$, so the proposition tells us that is $\vec w=\frac1{\|\vec v\|}\vec v = \frac1{\sqrt 5}\vec v=\frac1{\sqrt5}\m{1\\2}=\m{1/\sqrt{5}\\2/\sqrt5}$ is a unit vector in the same direction as $\vec v$.
+=== Example ===
-===== Addition of vectors =====
+If $\vec v=\m{1\\2}$ and $\vec w=\m{-2\\1}$, then $\dp vw=1(-2)+2(1)=-2+2=0$. On the other hand, we have $\|\vec v\|=\sqrt5=\|\vec w\|$, so the angle $\theta$ between $\vec v$ and $\vec w$ satisfies
+\[ 0=\dp vw=\sqrt 5\times \sqrt 5 \times \cos\theta\]
+so $5\cos\theta=0$, so $\cos\theta=0$, so $\theta=\pi/2$ or $\theta=3\pi/2$ (measuring angles in radians). This tells us that the angle between $\vec v$ and $\vec w$ is a right angle. We say that these vectors are **orthogonal**. We can draw a convincing picture which indicates that these vectors are indeed at right angles:
-If $\vec v=\vec{AB}$, then $\vec v$ moves $A$ to $B$, so $A+\vec v=B$.
+{{ :orth1.png?nolink&300 |}}
-If $\vec w=\vec {BC}$, then $\vec w$ moves $B$ to $C$, so $B+\vec w=C$.
+=== Proof of the Theorem ===
-What about $\vec v+\vec w$? We have $A+\vec v+\vec w=B+\vec w=C$. So $\vec v+\vec w=\vec{AC}$.
+We wish to show that $\def\vv{\vec v}
+\def\ww{\vec w}\vv\cdot\ww=\|\vv\|\,\|\ww\|\cos\theta$ where $\theta$ is the angle between $\vv$ and $\ww$.
-This gives us the triangle law for vector addition: $\vec v$, $\vec w$ and $\vec v+\vec w$ may be arranged to form a triangle:
+Recall the [[wp>cosine rule]]:{{ :t1.png?nolink&300 |}}
-{{ :tri.png?nolink&300 |}}
+Consider a triangle with two sides $\vv$ and $\ww$. By the triangle rule for vector addition, the third side $\vec x$ has $\ww+\vec x=\vv$, so $\vec x=\vv-\ww$:{{ :t2.png?nolink&300 |}}
-We get another triangle by starting at $A$ and translating first by $\vec w$ and then by $\vec v$; the other side of this triangle is $\vec w+\vec v$ But we know that $\vec v+\vec w=\vec w+\vec v$, so we can put these two triangles together to get the parallelogram law for vector addition:
+Applying the cosine rule gives \[ \|\vv-\ww\|^2=\|\vv\|^2+\|\ww\|^2-2\|\vv\|\,\|\ww\|\,\cos\theta.\]
+On the other hand, we know that $\|\vec x\|^2=\vec x\cdot\vec x$, so
+\begin{align*}\|\vv-\ww\|^2&=(\vv-\ww)\cdot(\vv-\ww)\\&=\vv\cdot\vv+\ww\cdot\ww-\ww\cdot\vv-\vv\cdot\ww\\&=\|\vv\|^2+\|\ww\|^2-2\vv\cdot\ww.\end{align*}
+So \[\|\vv\|^2+\|\ww\|^2-2\|\vv\|\,\|\ww\|\,\cos\theta=\|\vv\|^2+\|\ww\|^2-2\vv\cdot\ww\cos\theta.\] Subtracting $\|\vv\|^2+\|\ww\|^2$ from both sides and dividing by $-2$ gives $\vv\cdot\ww=\|\vv\|\,\|\ww\|\cos\theta$. ■
+==== Corollary ====
+If $\vv$ and $\ww$ are non-zero vectors and $\theta$ is the angle between them, then $\cos\theta=\displaystyle\frac{\vv\cdot\ww}{\|\vv\|\,\|\ww\|}$.
+==== Corollary ====
+If $\vv$ and $\ww$ are non-zero vectors with $\vv\cdot\ww=0$, then $\vv$ and $\ww$ are **orthogonal**: they are at right-angles.
+==== Examples ====
+  - The angle $\theta$ between $\def\c#1#2{\begin{bmatrix}#1\\#2\end{bmatrix}}\c12$ and $\c3{-4}$ has \[ \cos\theta=\frac{\c12\cdot\c3{-4}}{\left\|\c12\right\|\,\left\|\c3{-4}\right\|} =\frac{3-8}{\sqrt5\sqrt{25}}=-\frac1{\sqrt5},\] so $\theta=\cos^{-1}(-1/\sqrt5) \approx 2.03\,\text{radians}\approx 116.57^\circ$.
+  - The points $A=(2,3)$, $B=(3,6)$ and $C=(-4,5)$ are the vertices of a right-angled triangle. Indeed, we have $\vec{AB}=\c36-\c23=\c13$ and $\vec{AC}=\c{-4}5-\c23=\c{-6}2$, so $\vec{AB}\cdot\vec{AC}=\c13\cdot\c{-6}2=1(-6)+3(2)=0$, so the sides $AB$ and $AC$ are at right-angles.
+  - To find a unit vector orthogonal to the vector $\vv=\c12$, we can first observe that $\ww=\c{-2}1$ has $\vv\cdot\ww=0$, so $\vv$ and $\ww$ are orthogonal; and then consider the vector $\vec u=\frac1{\|\ww\|}\ww$, which is a unit vector in the same direction as $\ww$, so is also orthogonal to $\vv$. Hence $\vec u=\frac1{\sqrt5}\c{-2}1=\c{-2/\sqrt5}{1/\sqrt5}$ is a unit vector orthogonal to $\vv=\c12$.
-{{ :par.png?nolink&300 |}}