A bit of Trigonometry

Right angle triangle

  • sin(θ) = Opposite / Hypotenuse
  • cos(θ) = Adjacent / Hypotenuse
  • tan(θ) = Opposite / Adjacent
  • Pythagoras Theorem: \(a^2 + b^2 = c^2\)

Other triangle

  • The Law of Cosines: \(c^2=a^2+b^2\)
  • The Law of Sines: \(\frac{a}{sinA}=\frac{b}{sinB}=\frac{c}{sinC}\)

Vector Magnitude and Direction

  • Magnitude of u (distance): \(\|\vec{u}\|=\sqrt{\|u_1\|^2+\|u_2\|^2}\) (ie. Pythagorean Theorem)
  • Direction vector of u(u1, u2) is a unit vector with the same direction: \(\vec{w}=(\frac{u1}{\|u\|},\frac{u2}{\|u\|})\)
  • For right triangle, you also have the following properties: \(cos(\theta)=\frac{u_1}{\|u\|}; cos(\alpha)=\frac{u_2}{\|u\|}\)

Dot product of a and b = scalar

  • Dot product of x and y:
    • \(\vec{x}\cdot\vec{y}=\|x\|\|y\|cos\theta=x_1y_1+x_2y_2\)
  • It involves multiplying two vectors together to get a scalar, not another vector (for this reason, the dot product is often referred to as a scalar product).
  • Two vectors a and b are orthogonal if and only if \(\vec{a}\cdot\vec{b} = 0 \)

Cross product of a and b = another vector

  • it multiplies two vectors together to obtain a third vector rather than a scalar.
  • a×b is a vector that is perpendicular to both a and b.

Orthogonal projection of a vector onto another

  • By definition: \(Proj_y \vec{x} = \vec{z} = (\frac{\vec{x}\cdot\vec{y}}{\|y\|^2})\vec{y}\) (ie. a scalar times vector y)
  • For better understand the orthogonal projection, you can watch this video
  • Lets prove the formula above:
  1. \(cos\theta = \frac{\|\vec{z}\|}{\|\vec{x}\|}\) (for right angle trigonometry)
  2. \(cos\theta = \frac{\vec{x}\cdot\vec{y}}{\|\vec{x}\|\|\vec{y}\|}\) (for dot product)
  3. Then, \(\frac{\|\vec{z}\|}{\|\vec{x}\|} = \frac{\vec{x}\cdot\vec{y}}{\|\vec{x}\|\|\vec{y}\|}\)
  4. \(\|z\| = \frac{\vec{x}\cdot\vec{y}}{\|\vec{y}\|}\) (Multiply \(\|\vec{x}\|\) on each side)
  5. Since vector z and y are at the same direction, they will have the same direction vector. \(\frac{\vec{z}}{\|\vec{z}\|}=\frac{\vec{y}}{\|\vec{y}\|}\).
  6. Then, we will have \(\vec{z} = (\frac{\vec{x}\cdot\vec{y}}{\|\vec{y}\|^2}) \vec{y}\)

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