1. Vectors
1.1. Notation
1.1.1. Vector "v" would can be written as a v with a line over it, an arrow over it, or in bold
1.1.2. A vector starting at the point (2,3) and ending at the point (7,6) on the x/y plane would be written as <(7-2),(6-3)>, or <5,3>. Similarly a vector on the x/y/z plane traveling 1 unit in the x direction, 2 units y, and 3 units z would be written <1,2,3>.
1.2. Definition
1.2.1. A vector is a mathematical object that represents a direction and a length (magnitude)
1.3. Unit/Basis Vectors
1.3.1. On the x/y plane, the two unit vectors are i hat and j hat (unable to insert proper vector notation in this program). These represent <1,0> and <0,1> on the x/y plane, respectively. Similarly, on the x/y/z plane there are three "basis vectors" i hat, j hat, and k hat. These represent <1,0,0>, <0,1,0>, and <0,0,1> on the x/y/z plane, respectively.
1.3.2. If given a problem with A*i hat plus B*j hat, where A and B are constants, the vector is said to travel A units in the x direction and B units in the y direction, or <A,B>
1.4. Magnitude
1.4.1. Magnitude refers to the length of a vector. The notation for magnitude is to surround the original vector notation with double bars on each side (i.e. ||v|| but with a line above the v).
1.4.2. To find the magnitude, take the vector and create a right triangle using the x and y axis. Then, use the Pythagorean Theorem to find the length of the hypotenuse (which is the vector line). For example, a vector <2,5> would create a right triangle with legs 2 units and 5 units. Then, to find the magnitude you would use the theorem to solve for the hypotenuse (sqrt(2^2+5^2)).
1.5. Direction
1.5.1. The direction of a vector is the angle it travels from the x axis in the first quadrant. Therefore, you would create a right triangle using the x and y axis, then utilize tangent to find the angle that the hypotenuse is from the x axis. For example, with vector <2,5> you would create the triangle and find the inverse tangent of 5/2 to solve for the angle desired. The direction angle of the vector would then be 68.20 degrees from the x axis in the 1st quadrant.
1.6. Finding a Unit Vector
1.6.1. a unit vector is a vector with the same direction as your initial vector, but with a magnitude of 1.
1.6.2. a unit vector is found by dividing the initial vector by the magnitude of the vector ((v hat)/(||v hat||))
1.6.3. For example, a vector <3,4> would have magnitude sqrt(3^2+4^2)=5. <3,4>/5 would be the same as scalar multiplication of the vector by (1/5), or (1/5)*<3,4>. Therefore the unit vector for <3,4> would be <(3/5),(4/5)>.
1.7. Scalar Multiplication
1.7.1. Any real number multiplied by a vector simply creates a new vector where each coordinate is multiplied by that real number constant. Therefore, 3*<1,2> would simply equal <3,6>.
1.8. Addition/Subtraction
1.8.1. To add vectors, simply add the coordinates respectively. For example <2,3>+<0,5> = <(2+0),(3+5)> or <2,8>
1.8.2. To subtract vectors, simply subtract the coordinates respectively. For example, <2,3> - <0,5> = <(2-0),(3-5)> or <2,-2>
1.9. Dot Product
1.9.1. If you have two vectors a hat = <a^1,a^2,a^3> and b hat = <b^1,b^2,b^3>, then the dot product is (a hat) * (b hat) = (a^1*b^1)+(a^2*b^2)+(a^3*b^3). Note that this gives a numerical answer and NOT a new vector.
1.9.2. The dot product can be used to find the angle between two vectors. For example if theta is the angle between a hat and b hat, then ||a hat|| * ||b hat|| * cos(theta) = a hat ⋅ b hat. Therefore by doing (a hat ⋅ b hat)/(||a hat||*||b hat||) you get cos(theta). Then, using inverse cos you can find the angle theta between the two vectors.
1.9.3. a hat and b hat are orthogonal (perpendicular) if and only if the dot product of the two is equal to 0.
1.10. Cross Product
1.10.1. If you have two vectors, (u hat) × (v hat) is a new vector with length; ||(u hat) × (v hat)|| = ||(u hat)||*||(v hat)||*sin(theta), and direction follows the "right hand rule." This new vector will be orthogonal (perpendicular) to both the original vectors.
1.10.2. To determine the direction of the new vector, graph the original two vectors. Then, place your thumb on the first vector, and your pointer finger on the second vector. Then, whichever way your pointer finger points when stuck straight out is the direction of the new vector. In the case of (u hat) × (v hat), lay your thumb on (u hat) and pointer finger on (v hat).
1.10.2.1. Gyazo For example, in the image linked, if the red vector is the first in the cross product and blue is the second, your thumb would lay on the red vector and your pointer on the blue. Then, you would stick out your middle finger, which would be pointed back into the image, meaning the new vector direction would go away from you on the z axis.
1.10.3. (u hat) × (v hat) is only defined when u hat and v hat are 3-dimensional vectors (R^3).
1.10.4. (u hat) × (v hat) is not equivalent to (v hat) × (u hat).
1.11. Lines and Planes
1.11.1. Torque
1.11.1.1. Torque is the rotational force. Think a wrench on a bolt; the wrench is the vector of rotation, being pivoted at the bolt, which is the axis of rotation. The force being applied upward on the wrench is then the force vector, which causes the rotation vector to spin around the axis.
1.11.1.2. Torque is the cross product of the rotational vector and the force vector.
1.11.1.3. https://i.imgur.com/l6hR2Mw.jpg
1.11.2. Planes in R3
1.11.2.1. A plane can be found from vectors which lie on the plane, often given by three points, like so https://i.imgur.com/e1GJX3f.jpg
1.11.2.2. In order to find a vector perpendicular to the plane, simply use two given vectors that lie on the plane and find their cross product: https://i.imgur.com/bVhDqUl.jpg
1.11.2.3. A plane containing the point (x,y,z) and the nominal vector <a,b,c> is written as a(x-x1) + b(y-y1) + c(z-z1) = 0. https://i.imgur.com/DWc9aFA.jpg
1.11.2.4. To find the intersection of two planes, set their equations equal and solve, letting one variable equal to some constant and finding a point on the line. https://i.imgur.com/ZynNxT2.jpg
1.11.2.4.1. Next, subtract out the values of x y and z from the <x,y,z> vector, creating your <x-x1,y-y1,z-z1> vector. Set equal to t times the direction vector and solve for your parametric representation. https://i.imgur.com/Wpad8Ca.jpg
1.11.3. Lines in R3
1.11.3.1. Lines in R3 will be represented by a point on the line and a direction vector: https://i.imgur.com/Q2XFWj2.jpg
1.11.3.2. If the point (x,y,z) is on the line, any vector given by <x-1,y-1,z-1> or any constant will be parallel to the line. By using this parallel vector and setting equal to a constant scalar t multiplied by the direction vector, we can solve x y and z for t and write a parametric representation of the line. https://i.imgur.com/ijLvBnz.jpg