vectors notes

uni/mmme/1026_maths_for_engineering/vectors.md

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+---
+author: Akbar Rahman
+date: \today
+title: Vectors
+tags: [ uni, nottingham, mechanical, engineering, mmme1026, maths, vectors ]
+---
+
+Vectors have a *magnitude* (size) and *direction*.
+Examples of vectors include force, velocity, and acceleration.
+
+In type, vectors are notated in **bold**: $\pmb{a}$.
+In handwriting is is \underline{underlined}.
+
+# Vector Algebra
+
+## Equality
+
+Two vectors are said to be equal if their magnitudes and directions are equal.
+
+You can also do this by checking if their vertical and horizontal components are equal.
+
+## Addition
+
+Two vectors, $\pmb{a}$ and $\pmb{b}$, can be summed together by summing their components.
+
+You can also do this graphically by drawing $\pmb{a}$ and then $\pmb{b}$ by putting its tail on
+the tip of $\pmb{a}$.
+The sum of the vectors is from the tail of $\pmb{a}$ to the tip of $\pmb{b}$:
+
+![](./images/vimscrot-2022-02-18T16:34:17,290235454+00:00.png)
+
+Vector addition is associative[^d_associative] and commutative[^d_commutative].
+
+## Zero Vector
+
+The *zero vector* is denoted by $\pmb{0}$ and has zero magnitude and arbitrary direction.
+
+$$\pmb{a} + \pmb 0 = \pmb a$$
+
+If $\pmb a + \pmb b = 0$ then it is normal to write
+
+
+$$\pmb b = -\pmb a$$
+
+$-\pmb a$ is a vector with the same magnitude to $\pmb a$ but opposite direction.
+
+## Multiplication
+
+### Multiplication by a Scalar
+
+Let $k$, an arbitrary scalar and $\pmb a$, an arbitrary vector.
+
+- $k\pmb a$ is a vector of magnitude $|k|$ times that of $\pmb a$ and is parralel to it
+- $0\pmb a = 0$
+- $1\pmb a = a$
+- $(-k)\pmb a = -(k\pmb a)$
+- $(-1)\pmb a = = -\pmb a$
+- $k(\pmb a + \pmb b) = k\pmb a + k\pmb b$
+- $(k_1 + k_2)\pmb a = k_1\pmb a + k_2\pmb a$
+- $(k_1k_2)\pmb a = k_1(k_2\pmb a)$
+-
+### The Scalar Product (Inner Product, Dot Product)
+
+The scalar product of two vectors $\pmb a$ and $\pmb b$ is a scalar defined by
+
+$$\pmb a \cdot \pmb b = |\pmb a||\pmb b|\cos\theta$$
+
+where $\theta$ is the angle between the two vectors (note that $\cos\theta = \cos(2\pi - \theta)$).
+This definition does not depend on a coordinate system.
+
+- The dot product is commutative[^d_commutative]
+- The dot product is distributive[^d_distributive]
+- If $\pmb a$ is perpendicular to $\pmb b$, then $\pmb a \cdot \pmb b = 0$ and they are said to be
+  orthogonal
+- If $\pmb a \cdot \pmb b = 0$ then either
+
+    i. The vectors are orthogonal
+    ii. One or both of the vectorse are zero vectors
+
+- $\pmb a \cdot \pmb a = |\pmb a|^2 = a^2$
+- If $\pmb a = (a_1, a_2, a_3)$ and $\pmb b = (b_1, b_2, b_3)$ then
+
+    $$\pmb a \cdot \pmb b = a_1b_1 + a_2b_2 + a_3b_3$$
+
+The base vectors are said to be *orthonormal* when $\pmb i^2 = \pmb j^2 = \pmb k^2 = 1$ and
+$i\cdot j = i\cdot k = j\cdot k = 0$.
+
+### The Vector Product (Cross Product)
+
+The vector product between two vectors is defined by:
+
+$$\pmb a \times \pmb b = |\pmb a||\pmb b|\sin\theta \pmb n$$
+
+where $0 \le \theta \le \pi$ is the angle between $\pmb a$ and $\pmb b$ and $\pmb n$ is a unit
+vector such that the three vectors from a right handed system:
+
+![](./images/vimscrot-2022-02-18T20:11:12,072203286+00:00.png)
+
+- $\pmb a \times \pmb b = -\pmb b \times \pmb a$ (the vector product is non-commutative[^d_commutative])
+- If $\pmb a \times \pmb b = 0$ then either
+
+    i. The vectors are parralel
+    ii. One or both of the vectors are a zero vector
+- $(k_1\pmb a)\times(k_2\pmb b) = (k_1k_2)(\pmb a \times \pmb b)$ where $k_1$, $k_2$ are scalars
+- If $\pmb a = (a_1, a_2, a_3)$ and $\pmb b = (b_1, b_2, b_3)$ then
+
+    $$\pmb a \times \pmb b = (a_2b_3 - a_3b_2, a_1b_3-a_3b_1, a_1b_2-a_2b_1)$$
+
+- In the notation of determinants, provided we **expand by row 1**:
+
+    $$\pmb a \times \pmb b = \begin{vmatrix} \pmb i & \pmb j & \pmb k \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix}$$
+
+    This is technically not a determinant because not all the elements are numbers but shhhhhh...
+
+
+### Scalar Triple Product
+
+\begin{align*}
+        [ \pmb a, \pmb b, \pmb c ] &= \pmb a \cdot (\pmb b \times \pmb c) \\
+                                   &= \pmb b \cdot (\pmb c \times \pmb a) \\
+                                   &= \pmb c \cdot (\pmb a \times \pmb b) \\
+                                   &= (\pmb b \times \pmb c) \cdot \pmb a \\
+                                   &= (\pmb c \times \pmb a) \cdot \pmb b \\
+                                   &= (\pmb a \times \pmb b) \cdot \pmb c
+\end{align*}
+
+In terms of determinants:
+
+$$
+[ \pmb a, \pmb b, \pmb c ] = \begin{vmatrix} 
+                a_1 & a_2 & a_3 \\ 
+                b_1 & b_2 & b_3 \\ 
+                c_1 & c_2 & c_3 
+        \end{vmatrix}
+$$
+
+If $[\pmb a, \pmb b, \pmb c] = 0$ then the vectors are coplanar.
+
+The absolute value of the scalar triple product reperesents the volume of the parallelepiped defined
+by those vectors:
+
+![](./images/vimscrot-2022-02-18T20:27:53,322534334+00:00.png)
+
+
+## The Unit Vector
+
+$$\hat a = \frac{a}{|a|}$$
+
+## Components of a Vector
+
+The component of a vector $\pmb a$ in the direction of the unit vector $\pmb n$ is
+
+$$\pmb a \cdot \pmb n$$
+
+![](./images/vimscrot-2022-02-18T20:34:50,128465689+00:00.png)
+
+Vectors are often written in terms of base vectors, such as the Cartesian system's $\pmb i$,
+$\pmb j$, and $\pmb k$ in three dimensions.
+
+![](./images/vimscrot-2022-02-18T19:31:50,686289623+00:00.png)
+
+These vectors have unit magnitude, are perpendicular to each other, and are right handed.
+
+If $\pmb a = a_1\pmb i + a_2\pmb j + a_3\pmb k$ then the scalars $a_1$, $a_2$, and $a_3$ are the
+*components* of the vector (relative to the base vectors).
+
+### Vector Projections
+
+The *vector projection* of $\pmb a$ onto $\pmb n$ is given by
+
+$$(\pmb a \cdot \pmb n)\pmb n$$
+
+![](./images/vimscrot-2022-02-18T21:40:15,724449945+00:00.png)
+
+They look like the same as [vector components](#components-of-a-vector) to me...
+no idea what the difference is but uh StackExchange says
+([permalink](https://physics.stackexchange.com/a/537690)):
+
+> As pointed out, the projection and component actually refers to the same thing.
+> To solve a problem like this it useful to introduce a coordinate system, as you mentioned yourself
+> you project onto the x-axis.
+> As soon as you introduce a coordinate system you can talk about the components of some vector.
+
+## Position Vectors
+
+If an origin $O$ is fixed, then any point $P$ in space may be represented by the vector $\pmb r$
+which has a magnitude and direction given by the line $\overrightarrow{OP}$.
+
+A point $(x, y, z)$ in Cartesian space has the position vector $r = x\pmb i + y\pmb j + z\pmb k$.
+
+# Applications of Vectors
+
+## Application of Vectors to Geometry
+
+### Equation of a Straight Line
+
+A straight line can be specified by
+
+- two points it passes
+- one point it passes and a direction
+
+If $\pmb a$ and $\pmb b$ are the position vectors of two distinct points, then the position vectors
+of an arbitrary point on the line joining these points is:
+
+$$\pmb r = \pmb a + \lambda(\pmb b - \pmb a)$$
+
+where $\lambda \in \Re$ is a parameter.
+
+![](./images/vimscrot-2022-02-18T21:55:30,367159917+00:00.png)
+
+Suppose $O$ is an origin and $\pmb a$, $\pmb b$, and $\pmb r$ are position vectors on the line such
+that
+
+\begin{align*}
+\pmb a &= (x_0, y_0, z_0) \\
+\pmb b &= (x_1, y_1, z_1) \\
+\pmb r &= (x, y, z)\\
+\\
+(x, y, z) &= (x_0, y_0, z_0) + \lambda((x_1, y_1, z_1) - (x_0, y_0, z_0)) \\
+\\
+x &= x_0 + \lambda(x_1-x_0) \\
+y &= y_0 + \lambda(y_1-y_0) \\
+z &= z_0 + \lambda(z_1-z_0) \\
+\\
+\lambda &= \frac{x-x_0}{x_1-x_0} = \frac{y-y_0}{y_1-y_0} = \frac{z-z_0}{z_1-z_0}
+\end{align*}
+
+In the above, the vector $\pmb b - \pmb a$ is in the direction of the line.
+Thus the equation of a line can be specified by giving a point it passes through ($\pmb a$, say) and
+the direction of the line ($\pmb d = (d_1, d_2, d_3)$, say).
+
+The vector equation is then
+
+$$\pmb r = \pmb a + \lambda\pmb d$$
+
+#### The Cartesian Equation
+
+$$\frac{x-x_0}{d_1} = \frac{y-y_0}{d_2} = \frac{z-z_0}{d_3}$$
+
+### Equation of a Plane
+
+A *plane* can be defined by specifying either:
+
+- three points (as long as they're not in a straight line)
+- a point on the plne and two directions (useful for a parametric form)
+- specifying a point on the plane and the normal vector to the plane
+
+#### Specifying a Point and a Normal Vector
+
+Let $\pmb a$ be the position vector of a point on the plane, and $\pmb n$ a normal vector to the
+plane.
+
+If $\pmb r$ is the position vector of an arbitrary point on the plane, then $\pmb r - \pmb a$ is a
+vector lying *in* the plane, so
+
+$$(\pmb r - \pmb a) \cdot \pmb n = 0$$
+
+So the *vector equation* of the plane is
+
+$$\pmb r \cdot \pmb n = \pmb a \cdot n = d$$
+
+where $\pmb r = (x, y, z)$ and the vectors $\pmb a$ and $\pmb n$ are known.
+
+Suppose $\pmb a$, $\pmb n$, and $\pmb r$ are given by
+
+\begin{align*}
+\pmb a &= (x_0, y_0, z_0) \\
+\pmb n &= (l, m, p) \\
+\pmb n &= (x, y, z)\\
+\text{then } 0 &= ((x, y, z) - (x_0, y_0, z_0))\cdot(l, m, p)
+\end{align*}
+
+#### Specifying Three Points on a Plane
+
+If we specify three points on a plane with position vectors $\pmb a$, $\pmb b$, and $\pmb c$ the
+vectors $\pmb c - \pmb a$ and $\pmb c - \pmb b$ lie *in* the plane.
+(The vectors $\pmb a$, $\pmb b$, and $\pmb c$ do not necessarily lie *in* the plane;
+rather they take you from $O$ **to** the plane.)
+
+The normal to the plane, $\pmb n$, is then parallel to
+
+$$(\pmb c - \pmb a)\times(\pmb c - \pmb b)$$
+
+and so the equation of the plane is
+
+$$(\pmb r - \pmb a)\cdot((\pmb c - \pmb a)\times(\pmb c - \pmb b)) = 0$$
+
+#### The Angle Between Two Planes
+
+... is the same as the angle between their normal vectors
+
+
+
+
+
+[^d_associative]: The grouping of elements in in an operation do not matter (e.g. scalar addition: $a+(b+c) = (a+b)+c$)
+[^d_commutative]: The order of elements in an operation do not matter (e.g. scalar addition: $a+b = b+a$)
+[^d_distributive]: Easiest to explain with examples. Scalar multiplication is said to be distributive because $(a+b)c = ac + bc$