Line in $\mathbb{R}^2$:

$y = mx + b$

Line in $\mathbb{R}^n$:

$L = { \vec{x} + t \vec{v}   |   t \in \mathbb{R} }$

Given two point vectors $\vec{P_1}$ and $\vec{P_2}$, what is the line that passes through them ?

Solution:

$L = \vec{P_1} + t(\vec{P_1} - \vec{P_2})   |   t \in \mathbb{R}$

or

$L = \vec{P_2} + t(\vec{P_1} - \vec{P_2})   |   t \in \mathbb{R}$

See video.

1. $\| \vec{x} \cdot \vec{y} \| \le \|\vec{x}\|\|\vec{y}\|$, aka Cauchy-Schwartz Inequality
2. $\| \vec{x} + \vec{y} \| \le \|\vec{x}\| + \|\vec{y}\|$, aka Triangular Inequality
3. $\vec{a} \cdot \vec{a} = \|\vec{a}\|^2$, which can be easily verified.
4. $\vec{a} \cdot \vec{b} = \|\vec{a}\| \|\vec{b}\| \cos{\theta}$, where $\theta$ is the angle between $\vec{a}$ and $\vec{b}$. Can be verified based on (3).
5. $\displaystyle \vec{a} \times \vec{b} = \|\vec{a}\| \|\vec{b}\| \sin{\theta} \, n$, where $n$ is the unit vector perpendicular to the plane containing $\vec{a}$ and $\vec{b}$. Note the cross product between two vectors is applicable only in $\mathbb{R}^3$. Based on (4), there is a nice proof at Khan Academy.
6. $\vec{a} \times (\vec{b} \times \vec{c}) = \vec{b} (\vec{a} \cdot \vec{c}) - \vec{c} (\vec{a} \cdot \vec{b})$. Very optional with nice proof at Khan Academy.
7. How to determine the linear equation, $Ax + By + Cz = D$, of a plane in $\mathbb{R}^3$ if given a point on the plane, and a normal vector, which is the vector perpendicular to the plane ?
8. Conversely, how to determine the normal vector if given the linear equation of a plane in $\mathbb{R}^3$ ? Turns out if the plan is $Ax + By + Cz = D$, the normal vector $\vec{n}$ is simply $Ai + Bj + Ck$, where $i, j, k$ are the unit vectors of the respective $x, y, z$ coordinates.
9. Given a plane, $Ax + By + Cz = D$, and a point vector $x_0i + y_0j + z_0k$ outside of the plane, how to determine the shortest distance $d$ from the point vector to the plane ? Turns out: $\displaystyle d = \frac{Ax_0 + By_0 + Cz_0 - D}{\sqrt{A^2 + B^2 + C^2}}$

Footnotes:

• When is $\| \vec{x} \cdot \vec{y} \| = \|\vec{x}\|\|\vec{y}\|$ ?
• When is $\| \vec{x} + \vec{y} \| = \|\vec{x}\| + \|\vec{y}\|$ ?
• What is the angle $\theta$ between two vectors when one of them is a zero vector ?
• The word "perpendicular" is meaningful only when the angle is defined.
• When the dot product of two vectors is zero, are they always perpendicular ? Or, are they always orthogonal ? Why ?
• Intuition of a dot product: how much are the two vectors moving in the same direction ?
• Intuition of a cross product: how much is the area (as a parallelogram) between the two vectors ?
• The dot product is a scalar, but the cross product is a vector.
• The cross product of two vectors is always orthogonal to each of the two original vectors. This can be used to find the normal vector of a plane.
• Video on finding the distance between planes as an interesting application.

$$\begin{aligned} \sin \frac{u}{2} &= \pm \sqrt{ \frac{1 - \cos u}{2} } \\ \cos \frac{u}{2} &= \pm \sqrt{ \frac{1 + \cos u}{2} } \\ \tan \frac{u}{2} &= \pm \sqrt{ \frac{1 - \cos u}{1 + \cos u} } = \frac{\sin u}{1 + \cos u} = \frac{1 - \cos u}{\sin u} \\ \end{aligned}$$

Useful and not hard to verify.

$$\begin{aligned} \sin(A+B) &= \sin A\cos B + \cos A\sin B \\ \sin(A-B) &= \sin A\cos B - \cos A\sin B \\ \cos(A+B) &= \cos A\cos B - \sin A\sin B \\ \cos(A-B) &= \cos A\cos B + \sin A\sin B \\ \tan(A+B) &= \frac{\tan A + \tan B}{1 - \tan A \tan B} \\ \tan(A-B) &= \frac{\tan A - \tan B}{1 + \tan A \tan B} \\ 1 + \tan^2 x &= \sec^2 x \\ \end{aligned}$$

Note for $\sin(A+B)$ and $\cos(A+B)$, Khan Academy has some nice proofs. Notice $\cos -\theta = \cos \theta$ and $\sin -\theta = - \sin \theta$. The other identities are not hard to verify.

Euler's Formula

On the other hand, you may never have to memorize these formulas. Here is why. By Euler's formula:

$$\begin{aligned} e^{i(x+y)} &= \cos(x+y) + i \sin(x+y) \\ &= e^{ix} e^{iy} = (\cos x + i \sin x) \cdot (\cos y + i \sin y) \\ &= (\cos x \cdot \cos y - \sin x \cdot \sin y) + i(\sin x \cdot \cos y + \cos x \cdot \sin y) \\ \end{aligned}$$

and therefore

$$\begin{aligned} \cos(x+y) &= \cos x \cdot \cos y - \sin x \cdot \sin y \\ \sin(x+y) &= \sin x \cdot \cos y + \cos x \cdot \sin y \\ \end{aligned}$$