1. X as the right-left axis pointing right, Z as the forward-backward axis pointing forward, and Y as the up-down axis pointing up is (+x +z +y). Right hand.
2. Swap Z and Y, keeping their direction.
3. Swap Y and Z, keeping their direction. What…?

1. (+y -z +x) is right-handed.
2. Swap X and Y, keeping their direction. Flip Z's direction. Swap Z and Y. I don't know if you want me to use rotations, and even with rotations, we still need to mirror the space against the plane of the other two axes.
3. Same but from the last step of b backwards.

1. CW.
2. CCW.
3. CCW.
4. CW.

1. \(15\).

2. with distributive property we can take out the \(2\):

   \(\sum_{i=1}^{5}(2i) = 2 (\sum_{i=1}^{5} i) = 30.\)

3. \( \prod_{i=1}^{5}(2i) = 2^5 (2 * 3 * 4 * 5) = 2^5 * 2^3 * 3 * 5 = 2^8 * 15 = 256 * 15 = 1000 + 250 + 30 + 2000 + 500 + 60 = 3000 + 750 + 90 = 3840 \)
4. \(\prod_{i=0}^{4}(7 (i + 1)) = 7^5 (1 * 2 * 3 * 4 * 5) = 2^3 * 3^1 * 5^1 * 7^5 = blah. \)
5. I've seen this trick before. It's a square-ish divided by two as in: \( 100 + 101 / 2 = 5050 \).

1. \( 30 \tau / 360 = 3 \tau / 36 = \tau / 12 \)
2. There are 8 slices of \(45^{\circ}\) in a circle, therefore \(-45^{\circ} = -(\tau / 8) rad\).
3. There are 6 slices of \(60^{\circ}\) in a circle, therefore \(60^{\circ} = (\tau / 6) rad\).
4. There are four slices of \(90^{\circ}\) in a circle, therefore \(90^{\circ} = (\tau / 4) rad\).
5. There are two slices of \(180^{\circ}\) in a circle, therefore \(-180^{\circ} = -(\tau / 2) rad\).
6. \( 225\tau/360 = (40 + 5)\tau/(60 + 12) = 45\tau/72 = (10 + 5)\tau/(20 + 4) = 15\tau/24 = 5\tau/8 \)

… blah.

blah!

The square of the length of the hippopotamus, in a right-angled triangle, is equal to the sum of the squares of the lengths of the other two sides.

1. Show that:

   \( ( \sin(\alpha) / \csc(\alpha) ) + ( \cos(\alpha) / \sec(\alpha) ) = 1 \)

   We use the identities \( \sec(\alpha) = 1 / \cos(\alpha) \) and \( \csc(\alpha) = 1 / \sin(\alpha) \) to rewrite:

   \( ( \sin(\alpha) / (1 / \sin(\alpha)) ) + ( \cos(\alpha) / (1 / \cos(\alpha)) ) = 1 \)

   Algebraisting:

   \( \sin(\alpha)^2 + \cos(\alpha)^2 = 1 \)

   Which is one of the Pythagorean identities.

2. Show that:

   \( ( \sec^2(\theta) - 1 ) / ( \sec^2(\theta) ) = \sin^2(\theta) \)

   Multiply both sides by \( \sec^2(\theta) \):

   \( \sec^2(\theta) - 1 = \sin^2(\theta) * \sec^2(\theta) \)

   Replace \( \sec(\theta) = 1 / \cos(\theta) \), that is, \( \sec^2(\theta) = 1 / \cos^2(\theta) \):

   \( \sec^2(\theta) -1 = \sin^2(\theta) / \cos^2(\theta) \)

   Replace \( \sin(\theta) / \cos(\theta) = \tan(\theta) \):

   \( \sec^2(\theta) - 1 = \tan^2(\theta) \)

   Add \( 1 \) to both sides:

   \( 1 + \tan^2(\theta) = \sec^2(\theta) \)

   We got one of the Pythagorean identities.

3. Show that \( 1 + \cot^2(t) = csc^2(t) \).

   ?! It is already one of the known identities.

   In the unit circle, this is rewriten as \( 1 + (x/y)^2 = (r/y)^2 \).

   Power distribution law of \((a / b)² = a² / b²):

   \( 1 + x^2/y^2 = r^2 / y^2 \)

   It's the unit circle, so replace \(r = 1\):

   \( 1 + x^2/y^2 = 1 / y^2 \)

   Substract \(x^2/y^2\) on both side:

   \( 1 = 1 / y^2 - x^2/y^2 \)

   Division manipulation:

   \( 1 = (1 - x^2) / y^2 \)

   Pythagoras said \(r^2 = x^2 + y^2\) and we are on a unit circle, therefore \(1 = x^2 + y^2\), and after subtracting \(x^2\) from both sides we get \(1 - x^2 = y^2\). Replace \(1-x^2\) with \(y^2\):

   \( 1 = y^2 / y^2 \)

   Same divided with same equals \( 1 \):

   \( 1 = 1 \)

4. Show that \( \cos(\phi) ( \tan(\phi) + \cot(\phi) ) = \csc(\phi) \).

   Distribute \(\cos(\phi)\) onto \(\tan(\phi) + \cot(\phi)\) and replace \(\tan = \sin / \cos\) and \(\cot = \cos / \sin\).

   \( \cos(\phi)\sin(\phi)/\cos(\phi) + \cos(\phi)\cos(\phi)/\sin(\phi) = 1/\sin(\phi) \)

   Same divided with same equals one and some multiplying same with same is same with the power of two.

   \( \sin(\phi) + \cos^2(\phi)/\sin(\phi) = 1/\sin(\phi) \)

   Multiply both sides by \(\sin(\phi)\).

   \( \sin^2(\phi) + \cos^2(\phi) = 1 \)

   Which is a Pythagorean identity.

Let \( a = [-3 \ 8], b = \begin{bmatrix} 4 \\ 0 \\ 5 \end{bmatrix}, c = \begin{bmatrix} 16 \\ -1 \\ 4 \\ 6 \end{bmatrix}. \)

1. • \(a\) is a row vector of dimension 2.
   • \(b\) is a column vector of dimension 3.
   • \(c\) is a column vector of dimension 4.
2. \(b_y + c_w + a_x + b_z = 8 + 6 + (-3) + 5 = 16\).

1. Weight is scalar.
2. Speed is scalar.
3. Two blocks north is a vector. Two blocks magnitude north is the direction.
4. 600 mph is the magnitude and Los Angeles to New York is the direction. 33k feet is a 33k feet magnitude vector directed at the zenith.

1. \(

a = [0 \ 2], b = -a, c = [0.5 \ 2], d = c, e = [0.5 \ -3], f = [-2 \ 0], g = [-2 \ 1], h = [2.5 \ 2], i = [6 \ 1]. \)

1. False. Vectors are defined by magnitude and direction. They have no location.
2. True. \([a_1 \ a_2 \ ... \ a_n] = [a_1 \ 0 \ ... \ 0] + [0 \ a_2 \ ... \ 0] + ... + [0 \ 0 \ ... \ a_n]\).
3. False, due to the commutativity of vector addition.
4. False, it gives the displacement from the origin to the point. Multiply the vector by the scalar \(-1\) to get the vector from the point to the origin.

1. \(-[3 \ 7] = [-3 \ -7]\).
2. \(\|[-12 \ 5]\| = \sqrt((-12) * (-12) + 5 * 5) = \sqrt(144 + 25) = \sqrt(169 \textnormal{nice}) = 13\).
3. \(\|[8 \ -3 \ 1/2]\| = \sqrt(8 * 8 + (-3) * (-3) + (1/2) * (1/2)) = \sqrt(64 + 9 + 1/4) = \sqrt(73.25)\).
4. \(3[4 \ -7 \ 0] = [12 \ -21 \ 0]\).
5. \([4 \ 5]/2 = [2 \ 2.5]\).

1. \([12 \ 5] / \|[12 \ 5]\| = [12 / 13 \ 5 / 13]\).
2. \([0 \ 743.632] / \|[0 \ 743.632]\| = [0 \ 1]\).
3. \([8 \ -3 \ 1/2] / \|[8 \ -3 \ 1/2]\| = [8 \ -3 \ 1/2] / \sqrt(73.25)\).
4. \([-12 \ 3 \ -4] / \|[-12 \ 3 \ -4]\| = [-12 \ 3 \ -4] / 13 = \).
5. \([1 \ 1 \ 1 \ 1] / \|[1 \ 1 \ 1 \ 1]\| = [1 \ 1 \ 1 \ 1] / sqrt(4) = [0.5 \ 0.5 \ 0.5 \ 0.5]\).

1. \( [7 \ -2 \ -3] + [6 \ 6 \ -4] = [7 + 6 \ (-2) + 6 \ (-3) + (-4)] = [13 \ 4 \ -7] \).
2. \( [2 \ 9 \ -1] + [-2 \ -9 \ 1] = [2 + (-2) \ 9 + (-9) \ (-1) + 1] = [0 \ 0 \ 0] \).
3. …

Call the left vector \(a\) and the right vector \(b\).

1. \( \left\|\begin{bmatrix} 10 - (-14) \\ 6 - 30 \\ \end{bmatrix}\right\| = \sqrt(24 * 24 + 24 * 24) = \sqrt(1152) \).
2. \(\sqrt(12 * 12 + 25) = 13\).
3. \( \sqrt((3 - 8)^2 + (10 - (-7))^2 + (7 - 4)^2) = \sqrt(25 + 289 + 9) = \sqrt(323) \).
4. \(\sqrt((-2 - 6)^2 + ((-4) - (-7))^2 + (9 - 9.5)^2)\)…

1. \(\begin{bmatrix} 2 \\ 6 \end{bmatrix} \cdot \begin{bmatrix} -3 \\ 8 \end{bmatrix} = 2*(-3) + 6*8 = -6 + 48 = 42\).
2. \(-7[1 \ 2] \cdot [11 \ -4] = (-7)*1*11 + (-7)*11*(-4) = -77 + 28*11 = -77 + 8 + 80 + 20 + 200 = 11 + 220 = 231\)
3. \( 10 + (-5) * 4 + 1 * (-13) + 3 * 9 = 10 + (-20) + (-13) + 27 = 4 \)
4. \(3 \begin{bmatrix} -2 \\ 0 \\ 4 \end{bmatrix} \cdot \left ( \begin{bmatrix} 8 \\ -2 \\ 3/2 \end{bmatrix} + \begin{bmatrix} 0 \\ 9 \\ 7 \end{bmatrix} \right ) = 3 \begin{bmatrix} -2 \\ 0 \\ 4 \end{bmatrix} \cdot \left ( \begin{bmatrix} 8 \\ 7 \\ 8.5 \end{bmatrix} \right ) = 3 * (-2) * 8 + 3 * 0 * 7 + 3 * 4 * 8.5 = -48 + 0 + 12 * 8.5 = -48 + 6 + 16 + 80 = -48 + 22 + 80 = -26 + 80 = 54 \)

\(v = \begin{bmatrix} 4 \\ 3 \\ -1 \end{bmatrix}, \hat{n} = \begin{bmatrix} \sqrt(2)/2 \\ \sqrt(2)/2 \\ 0 \end{bmatrix}\).

Parallel to \(\hat{n}\):

\(\hat{n} (\hat{n} \cdot v) = \hat{n} (4 * \sqrt(2)/2 + 3 * \sqrt(2)/2) = \hat{n} \sqrt(2) (2 + 3/2) = \hat{n} \sqrt(2) 7/2 = \begin{bmatrix} 7 \sqrt(2)^2 / 4 \\ 7 \sqrt(2)^2 / 4 \\ 0 \end{bmatrix}\)

Perpendicular to \(\hat{n}\):

\(\hat{n} \times v = \begin{bmatrix} 3*0 - (-1)*(\sqrt(2)/2) \\ (-1)*(\sqrt(2)/2) - 4*0 \\ 4*(\sqrt(2)/2) - 3*(\sqrt(2)/2) \end{bmatrix} = \begin{bmatrix} \sqrt(2)/2 \\ -\sqrt(2)/2) \\ \sqrt(2)/2 \end{bmatrix}\)

Show the law of cosines using \(a \cdot b = \|a\| \|b\| \cos(\theta)\).

Law of cosines:

For triangle sides \(a, b, c\), and their opposite angles \(\alpha, \beta, \gamma\) (of corners \(A, B, C\)).

\(a^2 = b^2 + c^2 - 2bc\cos(\alpha)\)

\(b^2 = a^2 + c^2 - 2ac\cos(\beta)\)

\(c^2 = a^2 + b^2 - 2ab\cos(\gamma)\)

\(\vec{a}\) is the vector pointing from \(B\) to \(C\) or \(C\) to \(B\), \(\vec{b}\) from \(C\) to \(A\) or \(A\) to \(C\), and \(vec{c}\) from \(B\) to \(A\) or \(A\) to \(B\). Their magnitude are each equal to the sides \(a, b, c\).

We know that \(\|\vec{c}\| = \|\vec{a} + \vec{b}\|\)

The LHS is equal to:

\(\sqrt(\vec{c} \cdot \vec{c})\)

The RHS is equal to:

\(\sqrt((\vec{a} + \vec{b}) \cdot (\vec{a} + \vec{b})) =\)

\(\sqrt((\vec{a} \cdot (\vec{a} + \vec{b}) + \vec{b} \cdot (\vec{a} + \vec{b}))) =\)

\(\sqrt(\vec{a} \cdot \vec{a} + \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{a} + \vec{b} \cdot \vec{b}) =\)

\(\sqrt(\|\vec{a}\|^2 + \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{a} + \|\vec{b}\|^2) =\)

\(\sqrt(\|\vec{a}\|^2 + 2 \vec{a} \cdot \vec{b} + \|\vec{b}\|^2) =\)

\(\sqrt(\|\vec{a}\|^2 + 2 \vec{a} \cdot \vec{b} + \|\vec{b}\|^2)\)

Raise by the power of two.

\(\|\vec{c}\|^2 =\|\vec{a}\|^2 + 2 \vec{a} \cdot \vec{b} + \|\vec{b}\|^2\)

Reorder.

\(\|\vec{c}\|^2 =\|\vec{a}\|^2 + \|\vec{b}\|^2 + 2 \vec{a} \cdot \vec{b}\)

The angle \(\gamma\) is the one at the intersection of side \(a\) and side \(b\), so we can use the definition of dot product, \(\vec{a} \cdot \vec{b} = \|\vec{a}\| \|\vec{b}\| \cos{\gamma}\).

\(\|\vec{c}\|^2 =\|\vec{a}\|^2 + \|\vec{b}\|^2 + 2 \|\vec{a}\| \|\vec{b}\| \cos(\gamma)\)

Replace normals with the name of the sides.

\(c^2 = a^2 + b^2 + 2 a b \cos(\theta)\)

Show the law of cosines using \(a \cdot b = \|a\| \|b\| \cos(\theta)\).

Law of cosines says that for triangle sides \(a, b, c\), and their opposite angles \(\alpha, \beta, \gamma\) (of corners \(A, B, C\)).

\(a^2 = b^2 + c^2 - 2bc\cos(\alpha)\)

\(b^2 = a^2 + c^2 - 2ac\cos(\beta)\)

\(c^2 = a^2 + b^2 - 2ab\cos(\gamma)\)

\(\vec{a}\) is the vector pointing from \(B\) to \(C\) or \(C\) to \(B\) and has the magnitude of the side \(a\), \(\vec{b}\) points from \(C\) to \(A\) or \(A\) to \(C\) and has the magnitude of side \(b\), \(vec{c}\) points from \(B\) to \(A\) or \(A\) to \(B\) and has the magnitude of side \(c\).

Take \(c^2 = a^2 + b^2 - 2ab\cos(\gamma)\).

Consider the LHS.

\(c^2\)

Replace length of side \(c\) with the distance of \(\vec{c}\).

\(\|\vec{c}\|^2\)

Deconstructing it to its axis aligned parts.

\(c_x^2 + c_y^2\)

Consider the RHS.

\(a^2 + b^2 - 2ab\cos(\gamma)\)

Replace lengths of sides with their vector magnitudes.

\(\|\vec{a}\|^2 + \|\vec{b}\|^2 - 2\|\vec{a}\|\|\vec{b}\|\cos(\gamma)\)

Use the vector magnitude defined as dot product.

\(\vec{a} \cdot \vec{a} + \vec{b} \cdot \vec{b} - 2\|\vec{a}\|\|\vec{b}\|\cos(\gamma)\)

Deconstruct vectors to their axis aligned parts.

\(a_x^2 + a_y^2 + b_x^2 + b_y^2 - 2(a_x^2 + a_y^2)(b_x^2 + b_y^2)\cos(\gamma)\)

Distribute multiplication over addition.

\(a_x^2 + a_y^2 + b_x^2 + b_y^2 - 2(a_x^2(b_x^2 + b_y^2) + a_y^2(b_x^2 + b_y^2))\cos(\gamma)\)

And \(a_x^2 + a_y^2 + b_x^2 + b_y^2 - 2(a_x^2b_x^2 + a_x^2b_y^2 + a_y^2b_x^2 + a_y^2b_y^2)\cos(\gamma)\).

And \(a_x^2 + a_y^2 + b_x^2 + b_y^2 - 2a_x^2b_x^2 \cos(\gamma)- 2a_x^2b_y^2\cos(\gamma) - 2a_y^2b_x^2\cos(\gamma) + 2a_y^2b_y^2\cos(\gamma)\).

Replace \(\|\vec{a}\|\|\vec{b}\|\cos(\gamma)\) with \(\vec{a}\cdot\vec{b}\).

\(\vec{a} \cdot \vec{a} + \vec{b} \cdot \vec{b} - 2\vec{a}\cdot\vec{b}\)

\(\vec{c}\cdot\vec{c} = \vec{a} \cdot \vec{a} + \vec{b} \cdot \vec{b} - 2\vec{a}\cdot\vec{b}\)

\(\vec{c}\cdot\vec{c} - \vec{a} \cdot \vec{a} - \vec{b} \cdot \vec{b} = 2\vec{a}\cdot\vec{b}\)

\((\vec{c}\cdot\vec{c} - \vec{a} \cdot \vec{a} - \vec{b} \cdot \vec{b}) / 2 = \vec{a}\cdot\vec{b}\)

Show the law of cosines using \(a \cdot b = \|a\| \|b\| \cos(\theta)\).

Law of cosines says that for triangle sides \(a, b, c\), and their opposite angles \(\alpha, \beta, \gamma\) (of corners \(A, B, C\)).

\(a^2 = b^2 + c^2 - 2bc\cos(\alpha)\)

\(b^2 = a^2 + c^2 - 2ac\cos(\beta)\)

\(c^2 = a^2 + b^2 - 2ab\cos(\gamma)\)

\(\vec{a}\) is the vector pointing from \(B\) to \(C\) or \(C\) to \(B\) and has the magnitude of the side \(a\), \(\vec{b}\) points from \(C\) to \(A\) or \(A\) to \(C\) and has the magnitude of side \(b\), \(vec{c}\) points from \(B\) to \(A\) or \(A\) to \(B\) and has the magnitude of side \(c\).

Take \(c^2 = a^2 + b^2 - 2ab\cos(\gamma)\).

Consider the RHS.

\(a^2 + b^2 - 2ab\cos(\gamma)\)

Replace lengths of sides with their vector magnitudes.

\(\|\vec{a}\|^2 + \|\vec{b}\|^2 - 2\|\vec{a}\|\|\vec{b}\|\cos(\gamma)\)

Using \(\vec{a}\cdot\vec{b} = \|\vec{a}\|\|\vec{b}\|\cos{\gamma}\), replace \(\|\vec{a}\|\|\vec{b}\|\cos{\gamma}\).

\(\vec{a} \cdot \vec{a} + \vec{b} \cdot \vec{b} - 2\vec{a}\cdot\vec{b}\)

Deconstruct vectors to their axis aligned parts.

\(a_x^2 + a_y^2 + b_x^2 + b_y^2 - 2(a_x^2 + a_y^2)(b_x^2 + b_y^2)\)

Distribute multiplication over addition.

\(a_x^2 + a_y^2 + b_x^2 + b_y^2 - 2(a_x^2(b_x^2 + b_y^2) + a_y^2(b_x^2 + b_y^2))\)

And \(a_x^2 + a_y^2 + b_x^2 + b_y^2 - 2(a_x^2b_x^2 + a_x^2b_y^2 + a_y^2b_x^2 + a_y^2b_y^2)\).

And \(a_x^2 - 2a_x^2b_x^2 + b_x^2 + a_y^2 + 2a_y^2b_y^2 + b_y^2 - 2a_x^2b_y^2 - 2a_y^2b_x^2\).

And \(a_x^2 - 2a_x^2b_x^2 + b_x^2 + a_y^2 + 2a_y^2b_y^2 + b_y^2 - 2a_x^2b_y^2 - 2a_y^2b_x^2\).

c² = a² + b² -2abcos(gamma)

= a_x² + a_y² + b_x² -2*a_x*b_x

= a_x² -2*a_x*b_x + b_x² + a_y²

= (a_x - b_x)² + a_y²

-—

=|c|² = |a+b|² = =(sqrt(a+b))² =

Via: https://math.mit.edu/~djk/18_01/chapter05/proof05.html.

This is Professor Kleitman's solution which I will try and reproduce it from memory – I've read it a few tens of minutes ago – reducing rigor in the process.

Let's say A, B, C are points on corners of a triangle.

We can represent the sides AB, AC, and BC as the vectors \((A_x - B_x, A_y - B_y)\), \((A_x - C_x, A_y - C_y)\), and \((B_x - C_x, B_y - C_y)\).

Because \vec{AB}, \vec{AC}, and \vec{BC} express sides of a triangle, we know that \(\vec{AB} = \vec{AC} - \vec{BC}\).

The square of the magnitude of \vec{AB}, \(\|\vec{AB}\|^2\), is equal to the dot product of \vec{AB} with itself.

\(\|\vec{AB}\|^2 = \vec{AB} \cdot \vec{AB}\)

In the RHS, we replace \(\vec{AB} = \vec{AC} - \vec{BC}\).

\((\vec{AC} - \vec{BC}) \cdot (\vec{AC} - \vec{BC})\)

We then replace with the axis aligned vector component representation of the vector.

\((\vec{AC} - \vec{BC})_x^2 + (\vec{AC} - \vec{BC})_y^2\)

Algebraisting.

\(\vec{AC}_x^2 - 2\vec{AC}_x\vec{BC}_x + \vec{BC}_x^2 + \vec{AC}_y^2 - 2\vec{AC}_y\vec{BC}_y + \vec{BC}_x^2\)

\(\vec{AC}_x^2 + \vec{AC}_y^2 + \vec{BC}_x^2 + \vec{BC}_y^2 - 2\vec{AC}_x\vec{BC}_x - 2\vec{AC}_y\vec{BC}_y\)

\(\vec{AC}_x^2 + \vec{AC}_y^2 + \vec{BC}_x^2 + \vec{BC}_y^2 - 2(\vec{AC}_x\vec{BC}_x + \vec{AC}_y\vec{BC}_y)\)

Use vector dot products.

\(\vec{AC} \cdot \vec{AC} + \vec{BC} \cdot \vec{BC} - 2(\vec{AC} \cdot \vec{BC})\)

Dot product of a vector with itself is the square of the vector's magnitude.

\(\|\vec{AC}\|^2 + \|\vec{BC}\|^2 - 2(\vec{AC} \cdot \vec{BC})\)

Dot product of a vector with another is the product of their magnitude and the cosine of the angle between them.

\(\|\vec{AC}\|^2 + \|\vec{BC}\|^2 - 2\|\vec{AC}\|\|\vec{BC}\|\cos(\gamma)\)

What I like about this thing is the duality of point and vector, as mentioned in the chapter. \((A_x, A_y\) can be seen as a vector from the origin to the point \(A\) and \(\vec{AB\}\) is the vector from point \(A\) to point \(B\).

Use \( a \cdot b = a_x b_x + a_y b_y \) and trigonometric identities to prove the geometric interpretation of the dot product in 2D.

Let there be \(\vec{a}, \vec{b}\) two 2d vectors and \(\theta_1, \theta_2\) their angles as measured counter clockwise from the positive side of the X axis. Additionally, let \(\theta\) be the small angle between \(\vec{a}\) and \(\vec{b}\).

The axis aligned values of the two vectors are \(\vec{a} = (\vec{a}_x, \vec{a}_y) = (\|\vec{a}\|\cos\theta_1, \|\vec{a}\|\sin\theta_1)\) and \(\vec{b} = (\vec{b}_x, \vec{b}_y) = (\|\vec{b}\|\cos\theta_2, \|\vec{b}\|\sin\theta_2)\).

Replace \(a_x, b_x, a_y, b_y\) with their axis aligned values in the dot product equation.

\(a \cdot b = (\|\vec{a}\|\cos\theta_1) (\|\vec{b}\|\cos\theta_2) + (\|\vec{a}\|\sin\theta_1) (\|\vec{b}\|\sin\theta_2)\)

Take out \(\|\vec{a}\|\|\vec{b}\|\).

\(a \cdot b = \|\vec{a}\|\|\vec{b}\| ((\cos\theta_1)(\cos\theta_2) + (\sin\theta_1)(\sin\theta_2))\)

By \(\cos(a-b) = \cos a \cos b + \sin a \sin b\).

\(a \cdot b = \|\vec{a}\|\|\vec{b}\|\cos(\theta_2 - \theta_1)\)

We know that \(\theta = \theta_2 - \theta_1\).

\(a \cdot b = \|\vec{a}\|\|\vec{b}\|\cos(\theta)\)

There you go.

Show \(\|a \times b\| = \|a\|\|b\|sin(\theta)\).

First multiply each side by itself.

\( \|a \times b\|^2 = (\|a\|\|b\|sin(\theta))^2 \)

\(\|\begin{bmatrix} ... \end{bmatrix}\|_1\)

\(\|\begin{bmatrix} ... \end{bmatrix}\|_2\)

\(\|\begin{bmatrix} ... \end{bmatrix}\|_3\)

\(\|\begin{bmatrix} ... \end{bmatrix}\|_\infty\)

1. 1. \(\begin{bmatrix} 3 & 4 \end{bmatrix}\)

      \(\|\begin{bmatrix} 3 & 4 \end{bmatrix}\|_1 = |3| + |4| = 3 + 4 = 7\)

      \(\|\begin{bmatrix} 3 & 4 \end{bmatrix}\|_2 = \sqrt(3^2 + 4^2) = \sqrt(9 + 16) = \sqrt(25) = 5\)

      \(\|\begin{bmatrix} 3 & 4 \end{bmatrix}\|_3 = (3^3 + 4^3)^\frac{1}{3} = (27 + 64)^\frac{1}{3} = (91)^\frac{1}{3} \approx 4.5 \)

      \(\|\begin{bmatrix} 3 & 4 \end{bmatrix}\|_\infty = 4\)

   2. \(\begin{bmatrix} -2 & 10 & -7 \end{bmatrix}\)

      \(\|\begin{bmatrix} -2 & 10 & -7 \end{bmatrix}\|_1 = |-2| + |10| + |-7| = 2 + 10 + 7 = 19\)

      \(\|\begin{bmatrix} -2 & 10 & -7 \end{bmatrix}\|_2 = \sqrt((-2)^2 + 10^2 + (-7)^2) = \sqrt(4 + 100 + 49) = 153\)

      \(\|\begin{bmatrix} -2 & 10 & -7 \end{bmatrix}\|_3 = (|-2|^3 + 10^3 + |-7|^3)^\frac{1}{3} = (8 + 1000 + 343)^\frac{1}{3} = (1351)^\frac{1}{3} \approx 11.05\)

      \(\|\begin{bmatrix} -2 & 10 & -7 \end{bmatrix}\|_\infty = 10\)

   3. \(\begin{bmatrix} 6 & 1 & -9 \end{bmatrix}\)

      \(\|\begin{bmatrix} 6 & 1 & -9 \end{bmatrix}\|_1 = |6| + |1| + |-9| = 6 + 1 + 9 = 16\)

      \( \|\begin{bmatrix} 6 & 1 & -9 \end{bmatrix}\|_2 = \sqrt((6)^2 + (1)^2 + (-9)^2) = \sqrt(36 + 1 + 81) \approx 4.904868131524016 \)

      \( \|\begin{bmatrix} 6 & 1 & -9 \end{bmatrix}\|_3 = (|6|^3 + |1|^3 + |-9|^3)^\frac{1}{3} = (216 + 1 + 729)^\frac{1}{3} = 946^\frac{1}{3} \approx 9.81 \)

      \( \|\begin{bmatrix} 6 & 1 & -9 \end{bmatrix}\|_\infty = 9 \)

   4. \(\begin{bmatrix} 5 & -12 \end{bmatrix}\)

      \( \|\begin{bmatrix} 5 & -12 \end{bmatrix}\|_1 = 5 + 12 = 17 \)

      \(\|\begin{bmatrix} 5 & -12 \end{bmatrix}\|_2 = \sqrt(25 + 144) = \sqrt(169 \textnormal{nice}) = 13 \)

      \(\|\begin{bmatrix} 5 & -12 \end{bmatrix}\|_3 = \)

      \(\|\begin{bmatrix} 5 & -12 \end{bmatrix}\|_\infty\)

   5. \(\begin{bmatrix} -2 & -2 & -2 & -2 \end{bmatrix}\)

(cl-labels ((L (v p)
              (expt (cl-loop for x in v
                             sum (expt (abs x) p))
                    (/ 1.0 p))))
  (let ((vectors '((3 4)
                   (-2 10 -7)
                   (6 1 -9)
                   (5 -12)
                   (-2 -2 -2 -2)))
        (ps '(1 2 3 100 10000)))
    (cl-loop
     for v in vectors
     collect (cl-loop
              for p in ps
              collect (L v p)))))

1. • For \(L_1\), the points for which \(\|x\|_1 = 1\) would be:

     \(\{(x, y) \mid (y = x + 1 \land x \in [-1, 0]) \lor (y = x - 1 \land x \in [0, 1]) \lor (y = -x + 1 \land x \in [0, 1]) \lor (y = -x - 1 \land x \in [-1, 0]) \}\)

   • For \(L_2\), the points for which \(\|x\|_2 = 1\) would be:

     \(x^2 + y^2 = 1\)

   • For \(L_\infty\), the points for which \(\|x\|_\infty = 1\) would be:

     \(\{(x, y) \mid (|x| = 1 \land |y| \leq 1) \lor (|y| = 1 \land |x| \leq 1)\}\)

In a box shaped 2 feet wide, 2 feet long, and 2 feet tall.

\(\|\begin{bmatrix} 2 & 2 & 2 \end{bmatrix}\| = \sqrt(2^2 + 2^2 + 2^2) = \sqrt(12) \approx 3.5\)

\(a = \begin{bmatrix} -1 & 3 \end{bmatrix}\)

\(b = \begin{bmatrix} 1 & 3 \end{bmatrix}\)

\(c = \begin{bmatrix} 3 & -2 \end{bmatrix}\)

\(d = \begin{bmatrix} -1 & -2 \end{bmatrix}\)

\(e = \begin{bmatrix} -6 & 4 \end{bmatrix}\)

\(f = \begin{bmatrix} -1 & -3 \end{bmatrix}\)

\(a + b + c + d + e + f = \begin{bmatrix} -5 & 3 \end{bmatrix}\)

\(a + b + c + d + e + f = ((a + b + c + d + e + f)_x, (a + b + c + d + e + f)_y)\)

\( (a + b + c + d + e + f)_x = a_x + b_x + c_x + d_x + e_x + f_x = -1+1+3-1-6-1 = -5\)

\( (a + b + c + d + e + f)_y = a_y + b_y + c_y + d_y + e_y + f_y = 3+3-2-2+4-3 = 3\)

Seems to be a left-handed coordinate system. If I place my left pointing finger on the vector \(a\) and left middle finger on the vector \(b\) the thumb points at the direction of the displayed \(\|a \times b\|\).

Oh, oops. You are supposed to make a thumbs up gesture with your hands and see which one's fingers point from the vector on the left side of the cross product operation to the one on the right while the thumb points at the direction of the cross product's vector result. Anyway, all the same, it is the left hand.

1. \(P_\textnormal{UpperLeft} = c + r \cdot \begin{bmatrix} -1 & 1 \end{bmatrix}\)

   \(P_\textnormal{UpperRight} = c + r \cdot \begin{bmatrix} 1 & 1 \end{bmatrix}\)

   \(P_\textnormal{LowerLeft} = c + r \cdot \begin{bmatrix} -1 & -1 \end{bmatrix}\)

   \(P_\textnormal{LowerRight} = c + r \cdot \begin{bmatrix} 1 & -1 \end{bmatrix}\)

2. \(P_\textnormal{UpperLeftNear} = c + r \cdot \begin{bmatrix} -1 & 1 & -1 \end{bmatrix}\)

   \(P_\textnormal{UpperRightNear} = c + r \cdot \begin{bmatrix} 1 & 1 & -1 \end{bmatrix}\)

   \(P_\textnormal{LowerLeftNear} = c + r \cdot \begin{bmatrix} -1 & -1 & -1 \end{bmatrix}\)

   \(P_\textnormal{LowerRightNear} = c + r \cdot \begin{bmatrix} 1 & -1 & -1 \end{bmatrix}\)

   \(P_\textnormal{UpperLeftFar} = c + r \cdot \begin{bmatrix} -1 & 1 & 1 \end{bmatrix}\)

   \(P_\textnormal{UpperRightFar} = c + r \cdot \begin{bmatrix} 1 & 1 & 1 \end{bmatrix}\)

   \(P_\textnormal{LowerLeftFar} = c + r \cdot \begin{bmatrix} -1 & -1 & 1\end{bmatrix}\)

   \(P_\textnormal{LowerRightFar} = c + r \cdot \begin{bmatrix} 1 & -1 & 1\end{bmatrix}\)

1. \(\textnormal{Point position in relation to NPC} = \begin{cases} (x - p) \cdot v > 0, & \textnormal{Point in front of the NPC} \\ (x - p) \cdot v < 0, & \textnormal{Point is behind of the NPC} \end{cases}\)
2. \(p = \begin{bmatrix} -3 & 4 \end{bmatrix}, v = \begin{bmatrix} 5 & -2 \end{bmatrix}\).
   1. \(\begin{aligned} x & = \begin{bmatrix} 0 & 0 \end{bmatrix}, \\ (x - p) \cdot v & = (\begin{bmatrix} 0 & 0 \end{bmatrix} - \begin{bmatrix} -3 & 4 \end{bmatrix}) \cdot \begin{bmatrix} 5 & -2 \end{bmatrix} \\ & = \begin{bmatrix} 3 & -4 \end{bmatrix} \cdot \begin{bmatrix} 5 & -2 \end{bmatrix} \\ & = (3) \cdot (5) + (-4) \cdot (-2) \\ & = (15) + (8) \\ & = 23 > 0 \implies \textnormal{Point x is in front of the NPC} \end{aligned}\)
   2. \(\begin{aligned} x & = \begin{bmatrix} 1 & 6 \end{bmatrix}, \\ (x - p) \cdot v & = (\begin{bmatrix} 1 & 6 \end{bmatrix} - \begin{bmatrix} -3 & 4 \end{bmatrix}) \cdot \begin{bmatrix} 5 & -2 \end{bmatrix} \\ & = \begin{bmatrix} 4 & 2 \end{bmatrix} \cdot \begin{bmatrix} 5 & -2 \end{bmatrix} \\ & = (4) \cdot (5) + (2) \cdot (-2) \\ & = (20) + (-4) \\ & = 16 > 0 \implies \textnormal{Point x is in front of the NPC} \end{aligned}\)
   3. \(\begin{aligned} x & = \begin{bmatrix} -6 & 0 \end{bmatrix}, \\ (x - p) \cdot v & = (\begin{bmatrix} -6 & 0 \end{bmatrix} - \begin{bmatrix} -3 & 4 \end{bmatrix}) \cdot \begin{bmatrix} 5 & -2 \end{bmatrix} \\ & = \begin{bmatrix} -3 & -4 \end{bmatrix} \cdot \begin{bmatrix} 5 & -2 \end{bmatrix} \\ & = (-3) \cdot (5) + (-4) \cdot (-2) \\ & = (-15) + (8) \\ & = -7 < 0 \implies \textnormal{Point x is behind the NPC} \end{aligned}\)
   4. \(\begin{aligned} x & = \begin{bmatrix} -4 & 7 \end{bmatrix}, \\ (x - p) \cdot v & = (\begin{bmatrix} -4 & 7 \end{bmatrix} - \begin{bmatrix} -3 & 4 \end{bmatrix}) \cdot \begin{bmatrix} 5 & -2 \end{bmatrix} \\ & = \begin{bmatrix} -1 & 3 \end{bmatrix} \cdot \begin{bmatrix} 5 & -2 \end{bmatrix} \\ & = (-1) \cdot (5) + (3) \cdot (-2) \\ & = (-5) + (-6) \\ & = -11 < 0 \implies \textnormal{Point x is behind the NPC} \end{aligned}\)
   5. \(\begin{aligned} x & = \begin{bmatrix} 5 & 5 \end{bmatrix}, \\ (x - p) \cdot v & = (\begin{bmatrix} 5 & 5 \end{bmatrix} - \begin{bmatrix} -3 & 4 \end{bmatrix}) \cdot \begin{bmatrix} 5 & -2 \end{bmatrix} \\ & = \begin{bmatrix} 8 & 1 \end{bmatrix} \cdot \begin{bmatrix} 5 & -2 \end{bmatrix} \\ & = (8) \cdot (5) + (1) \cdot (-2) \\ & = (40) + (-2) \\ & = 38 > 0 \implies \textnormal{Point x is in front of the NPC} \end{aligned}\)
   6. \(\begin{aligned} x & = \begin{bmatrix} -3 & 0 \end{bmatrix}, \\ (x - p) \cdot v & = (\begin{bmatrix} -3 & 0 \end{bmatrix} - \begin{bmatrix} -3 & 4 \end{bmatrix}) \cdot \begin{bmatrix} 5 & -2 \end{bmatrix} \\ & = \begin{bmatrix} 0 & -4 \end{bmatrix} \cdot \begin{bmatrix} 5 & -2 \end{bmatrix} \\ & = (0) \cdot (5) + (-4) \cdot (-2) \\ & = (0) + (8) \\ & = 8 > 0 \implies \textnormal{Point x is in front of the NPC} \end{aligned}\)
   7. \(\begin{aligned} x & = \begin{bmatrix} -6 & -3.5 \end{bmatrix}, \\ (x - p) \cdot v & = (\begin{bmatrix} -6 & -3.5 \end{bmatrix} - \begin{bmatrix} -3 & 4 \end{bmatrix}) \cdot \begin{bmatrix} 5 & -2 \end{bmatrix} \\ & = \begin{bmatrix} -3 & -7.5 \end{bmatrix} \cdot \begin{bmatrix} 5 & -2 \end{bmatrix} \\ & = (-3) \cdot (5) + (-7.5) \cdot (-2) \\ & = (-15) + (15.0) \\ & = 0.0 = 0 \implies \textnormal{Point x is to the side of the NPC} \end{aligned}\)

(let* ((p '(-3 4))
       (p-x (car p))
       (p-y (cadr p))
       (v '(5 -2))
       (v-x (car v))
       (v-y (cadr v)))
  (cl-loop for i from 1
           for x in '((0 0)
                      (1 6)
                      (-6 0)
                      (-4 7)
                      (5 5)
                      (-3 0)
                      (-6 -3.5))
           concat
           (let* ((x-x (car x))
                  (x-y (cadr x))
                  (at-point-x (- x-x p-x))
                  (at-point-y (- x-y p-y))
                  (mult-x (* v-x at-point-x))
                  (mult-y (* v-y at-point-y))
                  (dot-product (+ mult-x mult-y)))
             (concat
              "   " (number-to-string i) ") \\(\\begin{aligned}\n"
              "      x & = \\begin{bmatrix} " (number-to-string x-x) " & " (number-to-string x-y) " \\end{bmatrix}, \\\\\n"
              "        (x - p) \\cdot v & = (\\begin{bmatrix} " (number-to-string x-x) " & " (number-to-string x-y) " \\end{bmatrix} - \\begin{bmatrix} "
              (number-to-string p-x) " & " (number-to-string p-y)
              " \\end{bmatrix})\n"
              "          \\cdot\n"
              "          \\begin{bmatrix} " (number-to-string v-x) " & " (number-to-string v-y) " \\end{bmatrix} \\\\\n"
              "        & = \\begin{bmatrix} " (number-to-string at-point-x) " & " (number-to-string at-point-y) " \\end{bmatrix} \\cdot \\begin{bmatrix} " (number-to-string v-x) " & " (number-to-string v-y) " \\end{bmatrix} \\\\\n"
              "        & = (" (number-to-string at-point-x) ") \\cdot (" (number-to-string v-x) ") + (" (number-to-string at-point-y) ") \\cdot (" (number-to-string v-y) ") \\\\\n"
              "        & = (" (number-to-string mult-x) ") + (" (number-to-string mult-y) ") \\\\\n"
              "        & = " (number-to-string dot-product) " "
              (cond ((< dot-product 0) "<") ((> dot-product 0) ">") (t "="))
              " 0 \\implies\n"
              "      \\textnormal{"
              (cond ((< dot-product 0) "Point x is behind the NPC")
                    ((> dot-product 0) "Point x is in front of the NPC")
                    ((and (= p-x x-x) (= p-y x-y)) "Point x is at the same position as the NPC")
                    (t "Point x is to the side of the NPC"))
              "}\n"
              "      \\end{aligned}\\)\n"))))

p' = p + (k - 1) ( p . n ) n.

[1 0 0] + (k - 1) ( [1 0 0] . [n_x n_y n_z] ) [n_x n_y n_z].

[1 0 0] + (k - 1) n_x [n_x n_y n_z].

[1 0 0] + [(k - 1) n_x n_x, (k - 1) n_x n_y, (k - 1) n_x n_z].

p' = [1 + (k - 1) n_x n_x, (k - 1) n_x n_y, (k - 1) n_x n_z].

\(p' = cos\theta(p - (p \cdot n)n) + \sin(\theta)(n \times p) + (p \cdot n)n\)

p = [1, 1, 1];

\(\cos\theta([1, 0, 0] - ([1, 0, 0] \cdot [n_x, n_y, n_z])[n_x, n_y, n_z]) + \sin(\theta)([n_x, n_y, n_z] \times [1, 0, 0]) + ([1, 0, 0] \cdot [n_x, n_y, n_z])[n_x, n_y, n_z]\)

\(\cos\theta([1, 0, 0] - (n_x)[n_x, n_y, n_z]) + \sin(\theta)([n_x, n_y, n_z] \times [1, 0, 0]) + ([1, 0, 0] \cdot [n_x, n_y, n_z])[n_x, n_y, n_z]\)

\(\cos\theta([1, 0, 0] - [n_x^2, n_x n_y, n_x n_z]) + \sin(\theta)([n_x, n_y, n_z] \times [1, 0, 0]) + ([1, 0, 0] \cdot [n_x, n_y, n_z])[n_x, n_y, n_z]\)

\(\cos\theta([1- n_x^2, -n_x n_y, -n_x n_z]) + \sin(\theta)([n_x, n_y, n_z] \times [1, 0, 0]) + ([1, 0, 0] \cdot [n_x, n_y, n_z])[n_x, n_y, n_z]\)

\(\cos\theta([1- n_x^2, -n_x n_y, -n_x n_z]) + \sin(\theta)([n_y 0 - n_z 0, n_z 1 - n_x 0, n_x 0 - n_y 1]) + ([1, 0, 0] \cdot [n_x, n_y, n_z])[n_x, n_y, n_z]\)

\(\cos\theta([1- n_x^2, -n_x n_y, -n_x n_z]) + \sin(\theta)([0, n_z 1, - n_y 1]) + ([1, 0, 0] \cdot [n_x, n_y, n_z])[n_x, n_y, n_z]\)

\(\cos\theta([1- n_x^2, -n_x n_y, -n_x n_z]) + \sin(\theta)([0, n_z 1, - n_y 1]) + (n_x)[n_x, n_y, n_z]\)

\(\cos\theta([1- n_x^2, -n_x n_y, -n_x n_z]) + \sin(\theta)([0, n_z 1, - n_y 1]) + [n_x n_x, n_x n_y, n_x n_z]\)

\([\cos\theta(1- n_x^2), -\cos\theta n_x n_y, -\cos\theta n_x n_z] + [0, \sin(\theta) n_z, - \sin(\theta) n_y] + [n_x n_x, n_x n_y, n_x n_z]\)

\([\cos\theta(1- n_x^2), -\cos\theta n_x n_y + \sin(\theta) n_z, -\cos\theta n_x n_z - \sin(\theta) n_y] + [n_x n_x, n_x n_y, n_x n_z]\)

\([\cos\theta(1- n_x^2) + n_x n_x, -\cos\theta n_x n_y + \sin(\theta) n_z + n_x n_y, -\cos\theta n_x n_z - \sin(\theta) n_y + n_x n_z]\)

Sure hope I've got it right ~;~.

1. Upright space.
2. World peace.
3. Object space.
4. Camera space.

The point would be move counterclockwise by 42 degrees, then translated along the world z axis by -6 units and -12 units along the world x axis.

Or am I supposed to translate along the x and z axes of the vectors after their rotation? I am not sure about that.

1. The zero vector \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} can be expressed as a linear combination of any other 3d vectors.

   In the linear combination \begin{aligned}a \begin{bmatrix} 1 \\ 0 \\ 0\end{bmatrix} + b \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}\end{aligned} set their linear contributions to zero, i.e. \(a = 0, b = 0\).

2. We can attempt to find the values of two of these vector's linear contributions in a linear combination such that the linear combination would equal to the third.

   \begin{aligned} \begin{bmatrix} 1 \\ 0 \\ 2 \end{bmatrix} & = a \begin{bmatrix} -1 \\ 1 \\ 2 \end{bmatrix} + b \begin{bmatrix} 0 \\ 1 \\ 2 \end{bmatrix} \\ \end{aligned}

   We can express this combination as a system of equations.

   \begin{aligned} 1 & = a (-1) + b (0) \\ 0 & = a ( 1) + b (1) \\ 2 & = a ( 2) + b (2) \\ \end{aligned}

   Take the second equation and substract \(-b\) from both sides and you get \(a = -b\).

   Assign the newly discovered value of \(a\) into the first equation and you get \(1 = (-b)(-1) = b\).

   We now know \(a = -1\) and \(b = 1\), so assign their values in the third equation and you get \(2 = (-1) 2 + 2 = 0\). That is a contradiction, which means that the system of equation has no solution and that the vectors are linearly independent.

3. We can see that we can get one of the vectors from the other three in the following linear combination, which means this set of vectors is linearly dependent.

   \begin{aligned}\begin{bmatrix} -1 \\ 2 \\ 3 \end{bmatrix} + \begin{bmatrix} 1 \\ -2 \\ 3 \end{bmatrix} + \begin{bmatrix} 1 \\ 2 \\ -3 \end{bmatrix} = \begin{bmatrix} -1 + 1 + 1 \\ 2 -2 + 2 \\ 3 + 3 - 3 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} \end{aligned}

4. Let us solve the corresponding system of equations.

   \begin{aligned} 1 = a 0 + b (-2) \\ 2 = a 1 + b (-4) \\ 3 = a 5 + b (-6) \end{aligned}

   From the first equation we find \( 1 = -2b \) and then \( b = -\frac{1}{2} \). We set \(b\)'s value in the second equation as \( 2 = a -\frac{1}{2}(-4) = a + 2 \). Subtract two from the two sides in \( 2 - 2 = a + 2 - 2 \) and we get \( a = 0 \).

   We set both currently known values of the unknowns into the third equation and get \( 3 = (0) 5 -\frac{1}{2}(-6) = 3 \). Oh crap. \(\begin{bmatrix} 1 \\ 2 \\ 3\end{bmatrix}\) is obviously \( 0 \begin{bmatrix} 0 \\ 1 \\ -5\end{bmatrix} + (\frac{-1}{2})\begin{bmatrix} -2 \\ -4 \\ -6 \end{bmatrix}\). I am so dumb.

5. Following our previous exercise we try and look at the number to see an obvious pattern and we see that \(0 = 1 - 1\), \(-5 = -4 -1\), and \(4 = 9 - 5\). Therefore the linear combination \(\begin{bmatrix} 0 \\ -5 \\ 4 \end{bmatrix} = \begin{bmatrix} 1 \\ -4 \\ 9 \end{bmatrix} - \begin{bmatrix} 1 \\ 1 \\ 5 \end{bmatrix}\) implies that this here set of vectors is linearly dependent.

6. \(1 = a(-1) + b\) → \(1 + a - 1 = -a + b + a - 1\) → \(a = b - 1\).

   We set \(a\) into the second equation as \(2 = 2(b-1) -2b\) → \(2 = 2b -2 -2b\) → \(2 = -2\), a contradiction implying the vectors are linearly independent.

Reading on the Wikipedia I see that two vectors \(a\) and \(b\) are orthogonal if and only if \(a \cdot b = 0\). I didn't see it stated clearly and I am too dumb to figure it out from the text. It makes sense. The 90 degrees projection / shadow of one vector onto another when these two vectors are orthogonal would be of zero length.

Also, LibreTexts™ Mathematics's text on "Orthogonal Vectors and Matrices" is gosh darn good.

1. We immediately see that the first vector is equal to the \(\hat{x}\) unit vector, the second is equal to \(4\hat{z}\), and the third to \(2\hat{y}\). Multiplying each vector by a scalar does not change the angles between the vectors, so these stay orthogonal.
2. \(\begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} \cdot \begin{bmatrix} -1 \\ 2 \\ 3 \end{bmatrix} = 1 (-1) + 2 (2) + 3 (3) = -1 + 4 + 9 = 12 \neq 0\). These two aren't orthogonal, therefore this set of vectors is not orthogonal.

3. \(\begin{bmatrix} 0 \\ 4 \\ 1 \end{bmatrix} \cdot \begin{bmatrix} 0 \\ -1 \\ 4 \end{bmatrix} = 0 - 4 + 4 = 0\). This pair is orthogonal.

   \(\begin{bmatrix} 0 \\ 4 \\ 1 \end{bmatrix} \cdot \begin{bmatrix} 8 \\ 0 \\ 0 \end{bmatrix} = 0(8) + 4(0) + 1(0) = 0\). This pair is orthogonal.

   \(\begin{bmatrix} 0 \\ 4 \\ 1 \end{bmatrix} \cdot \begin{bmatrix} 0 \\ -1 \\ 4 \end{bmatrix} = 0(0) + 4(-1) + 1(4) = 0\). This pair is orthogonal.

   All three pairs of vectors are orthogonal to each other which means this is an orthogonal set of vectors.

…

Vectors are orthonormal if and only if they are both orthogonal and of unit length, right? So we test that all vector pairs of vectors are \(a \cdot b = 0\) and that every vector \(\|v\| = 1\).

1. We know from problem 4 that these three are orthogonal, but the second and third vectors' lengths are 4 and 2, not 1, therefore this set of vectors are not orthonormal.
2. We already know they are not orthogonal, so they are not orthonormal.
3. These three obviously each has the length of 1 are obviously orthogonal, therefore they are orthonormal.
4. \(\begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} \cdot \begin{bmatrix} 0 \\ 0.707 \\ 0.707 \end{bmatrix} = 0.707 \neq 0\) therefore they are not orthogonal and therefore are not orthonormal.
5. [0, 0.707, -0.707] ⋅ [0, 0.707, 0.707] = 0.707² - 0.707² = 0. The pairing the third and the first and the third and the second in a dot product operation would obviously yield zero, therefore they are orthogonal vectors. [0, 0.707, -0.707] ⋅ [0, 0.707, -0.707] = 0.707² + 0.707² = 2 (0.707²) ≈ 1, but not quite, so these two 0.707 vectors are almost unit vectors, but a bit shorter by a few ten thousandth of a unit. Adapting this book's First Law of Computer Graphics – if the vectors look like they're orthonormal, they are orthonormal.
6. God almighty, I don't want to make any more of these calculations.

Looking at the book's solution, I see that what I have here is the transpose of R_x(θ). I would multiply an horizontal vector \(v\) and the rotation matrix like \(v R_x(-22^\circ)\), but I see that the book also multiplies it this way, even with its rotation matrix made of horizontal basis vectors \(R_x(\theta) = \begin{bmatrix} - p - \\ - q' - \\ - r' - \end{bmatrix}\).

Let's just try it with a vertical basis rotation matrix and an horizontal basis rotation matrix.

Their rotation matrix \(R_x(90^\circ)\):

My badly constructed rotation(?!) matrix:

A vector pointing at \(\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}^T\) turned \(90^\circ\) around the X axis would indeed point at \(\begin{bmatrix} 1 \\ -1 \\ 1 \end{bmatrix}^T\), and I was very very wrong about how one constructs matrices from basis vectors.

The source of the confusion was probably looking at

and seeing all those tall vectors and confusing it with

Anyway, inner dot product between x component of vector with x components of the basis vectors, v_y with p_y q_y and r_y, and v_z with p_z q_z and r_z.

Okay, so now that we know lets feed it:

That's about that.

Using the left hand rule for a left hand space, we get basis vectors \(p'=\begin{bmatrix} cos(\theta) \\ 0 \\ -sin(\theta) \end{bmatrix}^T\), \(q'=\begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}^T\), \(r'=\begin{bmatrix} sin(\theta) \\ 0 \\ cos(\theta) \end{bmatrix}^T\).

Stackin' 'em up in a transformation matrix we get:

That's a pretty pretty pretty good looking matrix, innit?

Feed it with 30^ˆ:

Construct for \(\hat{n} = \begin{bmatrix} 0.267 & -0.535 & 0.802 \end{bmatrix}, \theta = -15^\circ, R(\hat{n}, \theta)\).

After reading the derivation, I want to write it again here from memory.

We split v into its parallel and perpendicular components \(v = v_\parallel + v_\perp\).

We find the parallel component with the inner dot product \(v_\parallel = (\hat{n} \cdot v) \hat{n}\).

Setting the value of the parallel componet to the inner dot product we get the perpendicular component \(v_\perp = v - (\hat{n} \cdot v) \hat{n}\).

A vector perpendicular to both \(\hat{n}\) and v_⊥ is found using the cross product \(w = \hat{n} \times v_\perp\).

Because the parallel component of v does not change after a rotation, i.e. \(v'_\parallel = v_\parallel\), we can write \(v' = v'_\perp + v'_\parallel\) as \(v' = v'_\perp + v_\parallel\).

The vector v_⊥ and w form the mutually perpendicular basis vectors of the plane on which we rotate \(v'_\perp\), so \(v'_\perp = cos(\theta) v_\perp + sin(\theta) w\).

Setting the value of \(v'_\perp\) into \(v' = v'_\perp + v_\parallel\) we get \(v' = cos(\theta) v_\perp + sin(\theta) w\).

All we now need to do is to set v with our concrete basis vectors p, q, and r, and algebraise.

Now let's skip all this boring stuff and use the existing book solution.

Mind you, SVG's vertical Y axis points downwards on the screen.

1, 45°

cos 45°, sin 45°

cos 45 tau / 360, sin 45 tau / 360

(cl-loop for v in '((1 (deg 45))
                    (3 (deg 0))
                    (4 (deg 90))
                    (10 (deg -30))
                    (5.5 (rad 0.5)))
         collect (let* ((radius (car v))
                        (type (caadr v))
                        (angle (cadadr v)))
                   (cond ((eq type 'rad)
                          (list (* radius (cos angle))
                                (* radius (sin angle))))
                         ((eq type 'deg)
                          (list (* radius (cos (* float-pi angle (/ 180.0))))
                                (* radius (sin (* float-pi angle (/ 180.0)))))))))

AAAAA.

• 10, 20

  sqrt(10 * 10 + 20 * 20), atan2(20, 10)

  (sqrt (+ (* 10 10) (* 20 20))), (atan (/ 20.0 10.0))

  22.360679774997898, 1.1071487177940904 rad

• -12, -5

  sqrt((-12) * (-12) + (-5) * (-5)), atan2(-5, -12)

  (sqrt (+ (* -12 -12) (* -5 -5))), (- (atan (/ -5.0 -12.0)) float-pi)

  (sqrt (+ 144 25)), -2.7468015338900313 rad

  (sqrt 169 nice)), -2.7468015338900313 rad

  13, -2.7468015338900313 rad

• 0, 4.5

  sqrt(4.5 * 4.5), atan2(4.5, 0)

  4.5, 90 °

• -3, 4

  (sqrt (+ (* -3 -3) (* 4 4))), atan2(4, -3)

  (sqrt (+ (* -3 -3) (* 4 4))), atan2(4, -3)

  (sqrt (+ 9 16)), atan2(4, -3)

  (sqrt 25), atan2(4, -3)

  5, atan2(4, -3)

• 0, 0

  0, 0

• -5280, 0

  5280, (atan2 0 -5280)

  5280, (+ (atan (/ 0 -5280)) float-pi)

  5280, (+ (atan 0) float-pi)

  5280, (+ 0 float-pi)

  5280, \(\frac{\tau}{2}\)

(defun atan2 (y x)
  (cond ((= x 0.0) (cond ((> y 0.0) float-pi)
                         ((< y 0.0) float-pi)
                         ((= y 0.0) 0.0)))
        ((> x 0.0) (atan (/ y x)))
        ((< x 0.0) (cond ((>= y 0.0) (+ (atan (/ y x)) float-pi))
                         ((< y 0.0) (- (atan (/ y x)) float-pi))))))

(cl-loop for v in '((10 20)
                    (-12 -5)
                    (0 4.5)
                    (-3 4)
                    (0 0)
                    (-5280 0))
         collect (let ((x (float (car v)))
                       (y (float (cadr v))))
                   (list (sqrt (+ (* x x) (* y y)))
                         (atan2 y x))))

• 4, 120°, 5.

  4 cos 120 °, 4 sin 120 °, 5.

  (* 4 (cos (* (float 120) float-pi (/ (float 180))))), (* 4 (sin (* (float 120) float-pi (/ (float 180))))), 5.

  -2, 3.464101615137755, 5.

• 2, 45°, -1.

  (* 2.0 (cos (* 45.0 float-pi (/ 180.0)))), (* 2.0 (sin (* 45.0 float-pi (/ 180.0)))), -1.

  1.4142135623730951, 1.414213562373095, -1.

• 6, -π/6 rad, -3.

  (* 6 (cos (- (/ float-pi 6.0)))), (* 6 (sin (- (/ float-pi 6)))), -3.

  5.196152422706632, -2.9999999999999996, -3.

• 3, 3/π, 1

  (* 3 (cos (* 3 float-pi)), (* 3 (sin (* 3 float-pi)), 1

  -3.0, 1.102182119232618e-15, 1

(cl-loop for v in `((4 (deg 120) 5)
                    (2 (deg 45) -1)
                    (6 (rad ,(- (/ float-pi 6))) -3)
                    (3 (rad ,(* 3 float-pi)) 1))
         collect (let* ((radius (float (car v)))
                        (type (car (cadr v)))
                        (angle (cadr (cadr v)))
                        (theta (cond ((eq type 'deg) (* angle float-pi (/ 180.0)))
                                     ((eq type 'rad) angle)))
                        (z (caddr v)))
                   (list (* radius (cos theta))
                         (* radius (sin theta))
                         z)))

• 1, 1, 1.

  (sqrt 2.0), (atan2 1.0 1.0), 1.

  1.4142135623730951, (atan (/ 1.0 1.0)), 1.

  1.4142135623730951, (atan 1), 1.

  1.4142135623730951, 0.7853981633974483, 1.

• 0, -5, 2.

  5, (atan2 -5.0 0.0), 2.

  5, -90 °, 2.

• -3, 4, -7.

  (sqrt (+ (* -3 -3) (* 4 4))), (atan2 4 -3), -7.

  5, (+ (atan 4 -3) float-pi), -7.

  5, 5.355890089177974, -7.

• 0, 0, -3.

  0, 0, -3.

• 4, π / 3, 3 π / 4.

  4 cos (3 π / 4) cos (π / 3), 4 cos (3 π / 4) sin (π / 3), 4 cos 3 π / 4.

  (* 4 (cos (* 3.0 float-pi (/ 4.0))) (cos (* float-pi (/ 3.0)))), (* 4.0 (cos (* 3.0 float-pi (/ 4.0))) (sin (/ float-pi 3.0))), (* 4.0 (cos (* 3.0 float-pi (/ 4.0)))).

  -1.4142135623730951, -2.449489742783178, -2.82842712474619.

• 5, -5 π / 6, π / 3.

  5 cos (π / 3) cos (-5 π / 6), 5 cos (π / 3) sin (-5 π / 6), 5 cos (π / 3).

  (* 5.0 (cos (/ float-pi 3.0)) (cos (* -5.0 float-pi (/ 6.0)))), (* 5.0 (cos (/ float-pi 3.0)) (sin (* -5.0 float-pi (/ 6.0)))), (* 5.0 (cos (/ float-pi 3.0))).

  -2.1650635094610964, -1.250000000000001, 2.5000000000000004.

• 2, -π / 6, π.

  2 cos(π) cos(-π / 6), 2 cos(π) sin(-π / 6), 2 cos(π).

  (* 2.0 (cos float-pi) (cos (/ float-pi -6.0))), (* 2.0 (cos float-pi) (sin (/ float-pi -6.0))), (* 2.0 (cos float-pi).

  (-1.7320508075688774 0.9999999999999999 -2.0).

• 8, 9 π / 4, π / 6.

  (list (* 8.0 (cos (/ float-pi 6.0)) (cos (* 9.0 float-pi (/ 4.0)))) (* 8.0 (cos (/ float-pi 6.0)) (sin (* 9.0 float-pi (/ 4.0)))) (* 8.0 (cos (/ float-pi 6.0))))

  (4.8989794855663575 4.898979485566356 6.92820323027551)

• • x = r cos p sin h
  • y = -r sin p
  • z = r cos p cos h

• 4, π / 3, 3 π / 4.

  r = 4.

  h = π / 3.

  p = 3 π / 4.

  d = r cos p = 4 cos (3π/4).

  y = -r sin p = -4 sin (3π/4).

  x = d sin h = r cos p sin h = 4 cos (3π4) sin (π/3).

  z = d cos h = r cos p cos h = 4 cos (3π4) cos (π/3).

  `((,(* 4.0
         (cos (* 3.0 float-pi (/ 4.0)))
         (sin (/ float-pi 3.0))))
    (,(* -4.0
         (sin (* 3.0 float-pi (/ 4.0)))))
    (,(* 4.0
         (cos (* 3.0 float-pi (/ 4.0)))
         (cos (/ float-pi 3.0)))))
  

• 5, -5 π / 6, π / 3.

  r = 5.

  h = -5π/6.

  p = π/3.

  d = r cos p = 5 cos (π/3)

  y = -5 sin (π/3)

  x = r cos p sin h = 5 cos (π/3) sin (-5π/6)

  z = r cos p cos h = 5 cos (π/3) cos (-5π/6)

  `((,(* 5.0 (cos (/ float-pi 3.0)) (sin (* -5.0 float-pi (/ 6.0)))))
    (,(* -5.0 (sin (/ float-pi 3.0))))
    (,(* 5.0 (cos (/ float-pi 3.0)) (cos (* -5.0 float-pi (/ 6.0))))))
  

• 2, -π / 6, π.

  r = 2.

  h = -π/6.

  p = π.

  d = r cos p = 2 cos π.

  y = -r sin p = -2 cos π.

  x = r cos p sin h = 2 cos (π) sin (-π/6).

  z = r cos p cos h = 2 cos (π) cos (-π/6).

  `((,(* 2.0 (cos float-pi) (sin (/ float-pi -6.0))))
    (,(* -2.0 (sin float-pi)))
    (,(* 2.0 (cos float-pi) (cos (/ float-pi -6.0)))))
  

• 8, 9 π / 4, π / 6.

  r = 8.

  h = 9π/4.

  p = π/6.

  d = r cos p = 8 cos (π/6).

  y = -r sin p = -8 cos (π/6).

  x = r cos p sin h = 8 cos (π/6) sin (9π/4).

  z = r cos p cos h = 8 cos (π/6) cos (9π/4).

  `((,(* 8.0 (cos (/ float-pi 6.0)) (sin (* 9.0 float-pi (/ 4.0)))))
    (,(* 8.0 (sin (/ float-pi 6.0))))
    (,(* 8.0 (cos (/ float-pi 6.0)) (sin (* 9.0 float-pi (/ 4.0))))))
  

Using the above diagrams we can find the Gamemath polar coordinates of some \(v = \begin{bmatrix} x(v) & y(v) & z(v) \end{bmatrix}\) whose three Cartesian coordinates are known in a left handed 3D space.

We will use:

• Radius using Pythagoras's thingamajig!

  \begin{aligned} r = sqrt(x(v)^2 + y(v)^2 + z(v)^2) \end{aligned}

• Heading using Eytan Mark Bet!

  \begin{aligned} h = \textnormal{atan2}(x(v), z(v)) \end{aligned}

• And pitch using algebra! (not \(p = \textnormal{atan2}(y(v), d) = \textnormal{atan2}(y(v), \sqrt{x(v)^2 + z(v)^2})\), which is close but no cigar, alas.)

  \begin{aligned} y = & - r \sin p & & \textnormal{divide by} \ (-r) \\ -y / r = & \sin p & & \textnormal{strain through} \arcsin \\ \arcsin(-y / r) = & p \\ p = & \arcsin(-y / r) \end{aligned}

With that we can define the GaMaPoCoSy, or the GameMath Polar Coordinate System:

• \(x = \sqrt{2}; y = 2 \sqrt{3}; z = -\sqrt{2}\).

  \begin{aligned} x = & \sqrt{2} \\ y = & 2 \sqrt{3} \\ z = & -\sqrt{2} \\ r = & \sqrt{x^2 + y^2 + z^2} \\ = & \sqrt{\sqrt{2}^2 + (2 \sqrt{3})^2 + (-\sqrt{2})^2} \\ = & \sqrt{2 + 4 * 3 + 2} \\ = & \sqrt{4 + 12} \\ = & \sqrt{16} \\ = & 4 \\ h = & \textnormal{atan2}(x, z) \\ = & \textnormal{atan2}(\sqrt{2}, -\sqrt{2}) = \\ = & \arctan(\sqrt{2} / (-\sqrt{2})) + 180^\circ = \\ = & \arctan(-1) + 180^\circ = \\ = & \textnormal{(+ (atan -1) float-pi)} \\ = & 2.356194490192345 \\ p = & \arcsin(-y/r) \\ = & \arcsin(-2 \sqrt{3} / r) \\ = & -1.0471975511965976 \end{aligned}

• \(x = 2 \sqrt{3}; y = 6; z = -4\).

  \begin{aligned} x = & 2 \sqrt{3} \\ y = & 6 \\ z = & -4 \\ r = & \sqrt{x^2 + y^2 + z^2} \\ = & \sqrt{(2 \sqrt{3})^2 + 6^2 + (-4)^2} \\ = & \sqrt{4 \times 3 + 36 + 16} \\ = & \sqrt{12 + 36 + 16} \\ = & \sqrt{64} \\ = & 8 \\ h = & \textnormal{atan2}(x, z) \\ = & \textnormal{atan2}(2 \sqrt{3}, -4) = \\ = & \arctan(2 \sqrt{3} / (-4)) + 180^\circ = \\ = & 2.4278682746450277 \\ p = & \arcsin(-y/r) \\ = & \arcsin(-6 / 8) \\ = & \arcsin(-3 / 4) \\ = & -0.848062078981481 \end{aligned}

• \(x = -1; y = -1; z = -1\).

  \begin{aligned} x = & -1 \\ y = & -1 \\ z = & -1 \\ r = & \sqrt{x^2 + y^2 + z^2} \\ = & \sqrt{3(-1)^2} \\ = & \sqrt{3} \\ = & 1.7320508075688772 \\ h = & \textnormal{atan2}(x, z) \\ = & \textnormal{atan2}(-1, -1) = \\ = & \arctan(-1 / -1) - 180^\circ = \\ = & \arctan(1) - 180^\circ = \\ = & \textnormal{(- (atan 1.0) float-pi)} \\ = & -2.356194490192345 \\ p = & \arcsin(-y/r) \\ = & \arcsin(1 / 1.7320508075688772) \\ = & \textnormal{(asin (/ 1.7320508075688772))} \\ = & 0.6154797086703875 \end{aligned}

• \(x = 2; y = -2 \sqrt{3}; z = 4\).

  \begin{aligned} x = & 2 \\ y = & -2 \sqrt{3} \\ z = & 4 \\ r = & \sqrt{x^2 + y^2 + z^2} \\ = & \sqrt{4 + 4 \times 3 + 16} \\ = & \sqrt{4 + 12 + 16} \\ = & \sqrt{16 + 16} \\ = & \sqrt{32} \\ = & \sqrt{(2 \times 2) \times (2 \times 2) \times 2} \\ = & \sqrt{4^2 \times 2} \\ = & \sqrt{4^2} \sqrt{2} \\ = & 4 \times \sqrt{2} \\ = & \textnormal{(* 4 (sqrt 2))} \\ = & 5.656854249492381 \\ h = & \textnormal{atan2}(x, z) \\ = & \textnormal{atan2}(2, 4) \\ = & \arctan(2 / 4) \\ = & \arctan(1 / 2) \\ = & \textnormal{(atan (/ 2.0))} \\ = & 0.4636476090008061 \\ p = & \arcsin(-y/r) \\ = & \arcsin(2 \sqrt{3} / 5.656854249492381) \\ = & \textnormal{(asin (* 2 (sqrt 3) (/ 5.656854249492381)))} \\ = & 0.6590580358264089 \end{aligned}

• \(x = -\sqrt{3}; y = -\sqrt{3}; z = 2 \sqrt{2}\).

  \begin{aligned} x = & -\sqrt{3} \\ y = & -\sqrt{3} \\ z = & 2 \sqrt{2} \\ r = & \sqrt{x^2 + y^2 + z^2} \\ = & \sqrt{3 + 3 + 4 * 2} \\ = & \sqrt{6 + 8} \\ = & \sqrt{14} \\ = & \textnormal{(sqrt 14)} \\ = & 3.7416573867739413 \\ h = & \textnormal{atan2}(x, z) \\ = & \textnormal{atan2}(-\sqrt{3}, 2 \sqrt{2}) \\ = & \arctan(-\sqrt{3} / (2 \sqrt{2})) \\ = & \textnormal{(atan (/ (- (sqrt 3)) (* 2.0 (sqrt 2))))} \\ = & -0.5494672447576273 \\ p = & \arcsin(-y/r) \\ = & \arcsin(\sqrt{3} / 3.7416573867739413) \\ = & \textnormal{(asin (/ (sqrt 3) 3.7416573867739413))} \\ = & 0.4812753739423435 \end{aligned}

• \(x = 3; y = 4; z = 12\).

  \begin{aligned} x = & 3 \\ y = & 4 \\ z = & 12 \\ r = & \sqrt{x^2 + y^2 + z^2} \\ = & \sqrt{9 + 16 + 12 * 12} \\ = & \sqrt{9 + 16 + 100 + 20 + 20 + 4} \\ = & \sqrt{9 + 16 + 144} \\ = & \sqrt{25 + 144} \\ = & \sqrt{169 \ \textnormal{nice}} \\ = & \textnormal{(sqrt 169) ;; nice} \\ & \textnormal{Hey, now!} \\ & \textnormal{I'm rote learning and it's extremely boring,} \\ & \textnormal{so I'm trying to entertain myself a wee bit} \\ & \textnormal{by adding a lame joke here and there between} \\ & \textnormal{the atan2s and sqrts and asins, it's no sin.} \\ & \textnormal{So give me a rest now why won'tcha!} \\ = & 13.0 \\ h = & \textnormal{atan2}(x, z) \\ = & \textnormal{atan2}(3, 12) \\ = & \arctan(3 / 12) \\ = & \textnormal{(atan (/ 3.0 12.0))} \\ = & 0.24497866312686414 \\ p = & \arcsin(-y/r) \\ = & \arcsin(-4 / 13) \\ = & \textnormal{(asin (/ -4.0 13.0))} \\ = & -0.312766721941545 \end{aligned}

(defun kakafarm/vec3-cartesian-to-polar (v)
  (let* ((x (float (car v)))
         (y (float (cadr v)))
         (z (float (caddr v)))
         (radius (sqrt (+ (* x x) (* y y) (* z z)))))
    (if (> radius 0.0)
        (let ((pitch (asin (/ (- y) radius))))
          (if (>= (abs pitch)
                  (* 2.0 float-pi 0.9999))
              (list radius 0.0 pitch)
            (list radius (kakafarm/atan2 x z) pitch)))
      (list 0.0 0.0 0.0))))


(cons (list 'radius 'heading 'pitch)
      (cl-loop for v in (list (list (sqrt 2) (* 2 (sqrt 3)) (- (sqrt 2)))
                              (list (* 2 (sqrt 3)) 6 -4)
                              (list -1 -1 -1)
                              (list 2 (* -2 (sqrt 3)) 4)
                              (list (- (sqrt 3)) (- (sqrt 3)) (* 2 (sqrt 2)))
                              (list 3 4 12))
               collect (kakafarm/vec3-cartesian-to-polar v)))

Nice!

• A fixed radius is the surface of a sphere.
• A fixed heading would be a plane that includes the Y axis.
• A fixed pitch would be the surface of two mirror cones with their tips touching at the origin. Looks like light cones.

Har har… at least you didn't make a penguin eaten by a polar bear joke.

I suppose the authors were also losing their minds writing this book the same way I'm losing my mind reading it.

• State "DONE" from "TODO" [2026-10-20 Tue 16:16]
• a -> 5.
• b -> 3.
• c -> 6.
• d -> 1.
• e -> 2.
• f -> 4.

• State "DONE" from "TODO" [2026-10-20 Tue 17:42]
• a -> 3. Canonical.
• b -> 4. Canonical.
• c -> 5. Non-canonical. In a canonical trio, when pitch is 1/4 or -1/4 a turn, bank must be zero, which isn't the case here.
• d -> 1. Canonical.
• e -> 2. Canonical.
• f -> 3. Non-canonical. In a canonical trio, pitch must be within \([-\frac{\tau}{4} \textnormal{rad}, +\frac{\tau}{4} \textnormal{rad}]\).
• g -> 5. Canonical.
• h -> 2. Non-canonical. Pitch cannot be smaller than \(-90^\circ\).
• i -> 6. Cnonical.

• State "DONE" from "TODO" [2026-10-21 Wed 01:17]

• a - [-1 0 0 0].

  this is the identity quaternion.

  It fits 2.

• b - [0.653 -0.653 -0.271 -0.271].

  ;; p = arcsin(-2((-0.271)(-0.271) - 0.653(-0.653)))
  
  (* (/ 180.0 float-pi)
     (asin (* -2.0
              (- (* -0.271 -0.271)
                 (* 0.653 -0.653)))))
  

• c - [0.364 -0.].