2 - Vectors

Informally, a vector is resented as an arrow, or points in ${\mathbb{R}}^n$ or ${\mathbb{C}}^n$, which are written as column vectors $$\vec{x} = \begin{bmatrix}x_{1}\x_{2}\ \vdots \x_{n}\end{bmatrix}$$ Where $x_i\in \mathbb{R}$ or $x_i\in \mathbb{C}$. for short hand, we may write $\overrightarrow{x}=[x_i{]}_n$.
If we have two vectors $\overrightarrow{x}=[x_i{]}_n,\overrightarrow{y}=[y_i{]}_n$, then $\overrightarrow{x}+\overrightarrow{y}=[x_i+y_i{]}_n$. Geometrically, if we have two arrows, for $\overrightarrow{x},\overrightarrow{y}$. Then, one can equivalently either put a copy of $\overrightarrow{x}$ with the tail at the head of $\overrightarrow{y}$ or put a copy of $\overrightarrow{y}$ with the tail at the head of $\overrightarrow{x}$.

One can also change the length of the vector using a positive scalar, shortening it if the scalar is smaller than one, otherwise making it longer. if the scalar is negative, the scaled vector points opposite to the original.
$a\overrightarrow{x}=[a{x}_i{]}_n$ where $a\in \mathbb{R}$.

Dot product:

if $\overrightarrow{x},\overrightarrow{y}\in {\mathbb{R}}^n$, then the dot product $\overrightarrow{x}\cdot \overrightarrow{y}:=\sum _{i=1}^n{x}_i{y}_i\in \mathbb{R}$. The same thing holds for dot products of vectors in ${\mathbb{C}}^n$.
Geometrically, think law of cosines:

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but, $\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{a}$, equivalently $$|\vec{a}- \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 -2|\vec{a}||\vec{b}|\cos(\theta)$$
Then, assuming $a,b\in {\mathbb{R}}^n$ or ${\mathbb{C}}^n$, and doing a little algebra, we have that:

$\overrightarrow{a}\cdot \overrightarrow{b}$ is the length of the projection of $\overrightarrow{a}$ onto $\overrightarrow{b}$ or vice versa.
Moreover, $\overrightarrow{a}\perp \overrightarrow{b}$ if and only if $\overrightarrow{a}\cdot \overrightarrow{b}=0$.

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Now, the length of the projection of $\overrightarrow{a}$ on $\overrightarrow{u}$ is $|\overrightarrow{a}|\cos \theta$. We can scale the unit vector $\hat{u}$ (along $\overrightarrow{u}$) by this amount to get $$\vec{a}{||} = \hat{u}(|\vec{a}|\cos \theta) $$
But, $\overrightarrow{a}\cdot \overrightarrow{u}=|\overrightarrow{a}||\overrightarrow{u}|\cos \theta$, and $\hat{u}=\frac{\overrightarrow{u}}{|\overrightarrow{u}|}$. Puting it all together, we get $$\vec{a} = \frac{{\vec{a} \cdot \vec{u}}}{|\vec{u}|^2}$$

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for the paralellogram formed by $\overrightarrow{a}$ and $\overrightarrow{b}$ the area is $|\overrightarrow{a}||\overrightarrow{b}|\sin \theta$. we know that $cos\theta =\frac{\overrightarrow{a}\cdot \overrightarrow{b}}{|\overrightarrow{a}||\overrightarrow{b}|}$. Putting in together, the area of the paralellogram $A$ is given by: $$ A^2 = |\vec{a}|^2|\vec{b}|^2 - (\vec{a} \cdot \vec{b})^2$$
Allowing $a,b\in {\mathbb{R}}^2$, $\overrightarrow{a}=\left[\begin{array}{c}{a}_1\\ a_2\end{array}\right]$, $\overrightarrow{b}=\left[\begin{array}{c}{b}_1\\ b_2\end{array}\right]$.
we have $$ A = a_{1}b_{2} - a_{2}b_{1}$$
This expression is commonly given as a determinant $$A = \begin{vmatrix}
a_{1} \ a_{2} \
b_{1} \ b_{2}
\end{vmatrix} $$

Similarly, in ${\mathbb{R}}^3$, if $\overrightarrow{a}=\left[\begin{array}{c}{a}_1\\ a_2\\ a_3\end{array}\right]$, $\overrightarrow{b}=\left[\begin{array}{c}{b}_1\\ b_2\\ b_3\end{array}\right]$, $\overrightarrow{c}=\left[\begin{array}{c}{c}_1\\ c_2\\ c_3\end{array}\right]$

Then, the area of the parallelopiped is given by $$\begin{vmatrix}
a_{1} \ a_{2} \ a_{3} \
b_{1} \ b_{2} \ b_{3} \
c_{1} \ c_{2 } \ c_{3}
\end{vmatrix} = a_{1}\begin{vmatrix}
b_{2} \ b_{3} \
c_{2} \ c_{3}
\end{vmatrix} - a_{2}\begin{vmatrix}
b_{1} \ b_{3} \
c_{1} \ c_{3}
\end{vmatrix} + a_{3}\begin{vmatrix}
b_{1} \ b_{2} \
c_{1} \ c_{2}
\end{vmatrix}$$

we will not PROVE THIS FACT.

Cross product in ${\mathbb{R}}^3$:

again revisiting the discussion of the area of a parallelogram, only this time in 3d space:
for the parallelogram formed by $\overrightarrow{a}$ and $\overrightarrow{b}$ the area is $|\overrightarrow{a}||\overrightarrow{b}|\sin \theta$. we know that $cos\theta =\frac{\overrightarrow{a}\cdot \overrightarrow{b}}{|\overrightarrow{a}||\overrightarrow{b}|}$. Putting in together, the area of the parallelogram $A$ is given by: $$ A^2 = |\vec{a}|^2|\vec{b}|^2 - (\vec{a} \cdot \vec{b})^2$$
letting $\overrightarrow{a},\overrightarrow{b}\in {\mathbb{R}}^3$ we obtain that $$A = \sqrt{ (a_{1}b_{2} - a_{2}b_{1})^2 + (a_{2}b_{3} -a_{3}b_{2})^2 +(a_{1}b_{3}-a_{3}b_{1})^2 }$$
Now, notice A = $|\overrightarrow{a}\times \overrightarrow{b}|$ as computing the determinant gives us $$\vec{a} \times \vec{b} = \begin{bmatrix}
a_{1}\a_{2}\a_{3}
\end{bmatrix}\times \begin{bmatrix}
b_{1} \ b_{2} \ b_{3}
\end{bmatrix}= \begin{vmatrix}
i \ \ \ j \ \ \ k \
a_{1} \ a_{2} \ a_{3} \
b_{1} \ b_{2} \ b_{3}
\end{vmatrix} = (a_{1}b_{2} - a_{2}b_{1})i + (a_{2}b_{3} -a_{3}b_{2})j +(a_{1}b_{3}-a_{3}b_{1})k $$

Now, here is the the cool thing, intuitively, we know that two vectors in ${\mathbb{R}}^3$ specify a plane.

Notice that if $\overrightarrow{x}=[x_1,x_2,x_3{]}^T$ (I don't like writing column vectors anymore haha so I will do row transpose)
is on the plane of the vectors $\overrightarrow{a},\overrightarrow{b}$ (both tails attached to the origin) then from the picture above, it must be that there exists scalars $\lambda ,\mu$ such that $\lambda \overrightarrow{a}+\mu \overrightarrow{b}=\overrightarrow{x}$ that is, for every point $\overrightarrow{x}$ on this plane, we can scale $\overrightarrow{a},\overrightarrow{b}$ to complete the parallelogram. Then we have the set of equations as follows:

Using any two equations, and solving for the scalars, and using the third, we get

but notice the coefficients are actually the co-ordinates of $\overrightarrow{a}\times \overrightarrow{b}$.
In particular, The plane formed by the vectors $\overrightarrow{a},\overrightarrow{b}$ is given by $(\overrightarrow{a}\times \overrightarrow{b})\cdot \overrightarrow{x}=\overrightarrow{0}$. That is that the cross product of two vectors in ${\mathbb{R}}^3$ is perpendicular or orthogonal to the plane formed by those vectors.