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

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