Vectors

Geometry: Parametric curves

Vectors

The concept of directed arrows of a certain length in the plane can be made precise using vectors. Vectors are objects with length (or magnitude) and direction. We start with the notion of a "displacement vector".

Let #\blue A# and #\green B# be points in the plane. Define the displacement vector #\vec{\blue A \green B}# as the (straight) arrow starting at #\blue A# and ending at #\green B#.

We say that #\blue A# is the starting (or initial) point of #\vec{ \blue A \green B}# and #\green B# is the ending (or terminal) point of #\vec{ \blue A \green B}#.

The origin is denoted with the letter #O#

Example

The following picture denotes #\vec{O \green B}# where #\green B=\ivcc{1}{2}#.

Two displacement vectors can be added together as long as the ending point of one coincides with the starting point of the other.

Let #\blue A, \green B# and #\orange C# be three points in the plane. Define the sum of #\vec{\blue A \green B}# and #\vec{ \green B \orange C}# as

\[ \vec{\blue A \green B} + \vec{ \green B \orange C} = \vec{ \blue A \orange C }\]

A displacement vector can always be "moved". This gives the notion of a vector. Intuitively, a vector is a displacement vector without any specified starting point.

A vector #\blue{\vec{x}}# is an arrow in the plane which can be repositioned, without rotating or stretching/shrinking, to the displacement vector #\vec{OA}# for some point #A# in the plane.

If #A=\ivcc{x_1}{x_2}# we denote the vector #\blue{\vec{x}}# by #\cv { x_1 \\x_2 }#.

Note that there are multiple representations of the same vector. The picture shows various representations of the vector #\cv{1\\2}#.

Example

As said, it is common to reposition our vectors to a vector of the form #\vec{OA}#. In this case we can also decide to write #A=[x_1,x_2]# for the vector.

The length of a vector can be computed using the Pythagorean theorem.

Given a vector #\blue{\vec{x}}=\cv{ \green{x_1}\\ \orange{x_2} }#, the length of the vector, denoted by #\| \blue{\vec{x}} \|#, is given by

\[\| \blue{\vec{x}} \| = \sqrt{\green{x_1}^2+\orange{x_2}^2}\]

Example

The length of #\blue{\vec{x}}=\cv{\green{1}\\ \orange{2}}# is given by \[\|\blue{\vec{x}}\|=\sqrt{\green 1^2+ \orange 2^2}=\sqrt{5}\]

Pythagorean Theorem

As mentioned, this length is calculated using the Pythagorean theorem. Consider the picture on the right.

We see a right angled triangle with sides #\green{x_1}# and #\orange{x_2}#. The length of the hypotenuse is given by #\sqrt{\green{x_1}^2 + \orange{x_2}^2 }#.

This corresponds with the length of the vector #\blue{\vec{x}}#

Picture

Notice that the length of a vector is always positive, except for the zero vector, which has length zero.

Adding vectors is considerably easier than adding displacement vectors. For example, any two vectors can be added together regardless whilst two displacement vectors need to align in order to be added. We can also define "scalar multiplication".

Let #\blue{\vec{x}} =\cv{\blue{x_1}\\\blue{x_2}}# and #\green{\vec{y}}=\cv{\green{y_1}\\\green{y_2}}# be two vectors and #c# a real number. We define the sum of #\blue{\vec{x}}# and #\green{\vec{y}} # to be \[\blue{\vec{x}}+\green{\vec{y}}=\cv{\blue{x_1}+\green{y_1}\\\blue{x_2}+\green{y_2}}\]

and the scalar multiple #\orange{c} \cdot \blue {\vec{x}}#

\[\orange{c} \cdot \blue{\vec{x}}=\cv{\orange{c}\cdot \blue{x_1}\\ \orange{c}\cdot \blue{ x_2}}\]

Example

The geometry of adding two vectors is best visualised by the so called head-and-tail rule. This rule states that given two vectors #\blue{\vec{x}}# and #\green{\vec{y}}#, the sum of these two is obtained by positioning #\green{\vec{y}}# at the endpoint of #\blue{\vec{x}}#. This can be seen in the picture.

Picture of scalar multiple

Scalar multiple

Here is a picture of the scalar multiple of a vector. You can adjust the value of #\orange c# using the slider in the top right corner.

Note that the direction of #\blue {\vec{x}} # changes only if #\orange c# is negative.

Picture

Displacement vector

Let #\blue A = \rv{\blue{x_1},\blue{x_2}}# and #\green B = \rv{\green{y_1},\green{y_2}}# be points in the plane. Then we know that \[\vec{O\blue{A}}+ \vec{\blue A \green B} = \vec{O\green{B}}\]

Thus, the displacement vector #\vec{\blue A \green B}# can be calculated as the difference of #\vec{O\green{B}}# and #\vec{O\blue{A}}#, \[ \begin{array}{rcl}
\vec{\blue A \green B} &=& \vec{O\green{B}} - \vec{O\blue{A}} \\ &=& \cv{\green{y_1 \\ y_2}} - \cv{\blue{x_1 \\ x_2}} \\
&=& \cv{\green{y_1} - \blue{x_1} \\ \green{y_2} - \blue{x_2} }
\end{array}\]

Example

#\blue{A} = \rv{\blue 1,\blue 2}# and #\green{B} = \rv{\green 4,\green3 }# then \[\vec{\blue A \green B} = \cv{\green 4-\blue 1 \\ \green 3-\blue 2} = \cv{3 \\ 1} \]

Based on the vectors #\vec{u}=\rv{10,14}# and #\vec{v}=\rv{-1,13}#, calculate the vector

\[ 3\cdot \vec{u}-4\cdot \vec{v}\tiny.\]

\(3\cdot\vec{u}-4 \cdot \vec{v}=\) # \left[ 34 , -10 \right]#

This follows from the following calculation:
\[\begin{array}{rcl} 3\cdot\vec{u}-4\cdot\vec{v}&=&3\cdot\rv{10,14}-4\cdot\rv{-1,13}\\
&&\qquad \blue{\text{definitions }\vec{u}, \vec{v}}\\
&=&\rv{3\cdot 10, 3\cdot 14}+\rv{-4\cdot-1, -4\cdot 13}\\
&&\qquad \blue{\text{multiplication per coordinate}}\\
&=&\rv{30, 42}+\rv{4, -52}\\
&&\qquad \blue{\text{simplified}}\\
&=&\rv{30+4, 42-52}\\
&&\qquad \blue{\text{addition per coordinate}}\\
&=&\rv{34,-10}\\
&&\qquad \blue{\text{simplified}}\end{array}\]

The figure below shows the vectors #\vec{u}# and #\vec{v}# drawn in black, the vectors #3\cdot \vec{u}# and #-4\cdot\vec{v}# in blue dashed lines, and the vector #3\cdot\vec{u}-4\cdot\vec{v}# in a red dotted line.

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