Vector operations

Adding a vector to a point

Consider a point $A$ and a vector $\vec v$ . What do we mean with Adding a vector to a point? If we start at point $A$ and follow the instructions given by vector $\vec v$

\begin{array}{lcl} A & \rightarrow & B \\ \hline A_x & \overset{+v_x}{\rightarrow} & A_x+v_x \\ A_y & \overset{+v_y}{\rightarrow} & A_y+v_y \\ A_z & \overset{+v_z}{\rightarrow} & A_z+v_z \\ \end{array}

we end up at new point with the coordinates

(A_x+v_x \vert A_y+v_y \vert A_z + v_z)

We denote this new point by

\boxed{A+\vec v}

or, if give it a new name, e.g. $B$ , we write

B = A+\vec v

Adding two vectors

Consider two vectors $\vec u$ and $\vec v$ . The sum of the two vectors, denoted by

\boxed{\vec u+\vec v}

is a new vector with the components

\vec u+\vec v = \left(\begin{array}{r} u_x+v_x\\ u_y+v_y\\ u_z+v_z \end{array}\right)

What is the relationship between the arrows represented by $\vec u$ , $\vec v$ , and $\vec u+\vec v$ ? We can actually construct an arrow of $\vec u+\vec v$ with the help of the arrows of $\vec u$ and $\vec v$ as follows:

arrange the arrows of $\vec u$ and $\vec v$ such that the tail of $\vec v$ is attached to the head of $\vec u$
the arrow of $\vec u+\vec v$ is obtained by drawing an arrow from the tail of $\vec u$ to the head of $\vec v$

This construction is called completing the triangle.

Note: To find the arrow that is represented by the sum of three vectors $\vec u+\vec v+\vec w$ , we form a chain of the arrows (attach the tail of $\vec v$ to the head of $\vec u$ , and the tail of $\vec w$ to the head of $\vec v$ ) and form the arrow from the tail of $\vec u$ to the head of

Subtracting two vectors

The subtraction of \vec v from \vec u is defined as adding to $\vec u$ the opposite vector of $\vec v$ :

\boxed{\vec u-\vec v = \vec u+ (-\vec v)}

The components are therefore

\vec u - \vec v = \left(\begin{array}{r} u_x-v_x\\ u_y-v_y\\ u_z-v_z \end{array}\right)

It also follows that we can construct the arrow representing the vector $\vec u-\vec v$ by simply completing the triangle of the vectors $\vec u$ and $-\vec v$ . Recall that the arrow of $-\vec v$ is obtained by reflecting the arrow of $\vec v$ about its tail.

Multiplication of a vector with a scalar

Consider a vector $\vec u$ and a constant $c$ , where $c$ can by any real number. In the context of vectors, a number is called a scalar. The multiplication of c with \vec u, denoted by

\boxed{c\cdot \vec{u}}

is defined as the vector with the components

c\cdot \vec u = \left(\begin{array}{r} c\cdot u_x\\ c \cdot u_y\\ c \cdot u_z \end{array}\right)

Thus each component of $\vec u$ is multiplied by the constant $c$ . Make sure you understand that

The arrow that is represented by $c \cdot \vec{u}$ is $c$ times longer than $\vec u$ .
If $c$ is positive, the arrows point in the same direction, if $c$ is negative, the arrow points in the opposite direction (see figure below).
For $c=2$ the arrow length is doubled, for $c=0.5$ the arrow length is halved.
More generally, for a $c>1$ the arrow length increases, for $0<c<1$ the arrow length decreases.

Exercise 1

Q1

Consider the vector $\vec{u}=\left(\begin{array}{r} -2\\ 2\\ -1 \end{array}\right)$ Determine the vector $\vec{v} = 3\cdot \vec{u}$ and verify that the magnitude of $\vec v$ is $3$ times bigger than the magnitude of $\vec{u}$ .

Q2

Show that $0\cdot \vec{a}$ is the zero vector.

Q3

Show that $4\cdot (5\cdot \vec{a})=20\cdot \vec{a}$

Q4

Show the following:

$\vec{a}+\vec{b}=\vec{b}+\vec{a}$
$\vec{a}+(\vec{b}+\vec{c})=(\vec{a}+\vec{b})+\vec{c}$
$\vec{b}-\vec{a}$ is the opposite vector of $\vec{a}-\vec{b}$
$\vec{a}+\vec{a}=2 \vec{a}$
$4 \vec{a} + 6 \vec{a} =10 \vec{a}$
$4 \vec{a} - 6 \vec{a} =-2 \vec{a}$
$3 (\vec{a} +\vec{b}) =3 \vec{a}+3 \vec{b}$

Q5

Construct arrows that represent the following vectors: $\vec{a}+\vec{b}$ , $\vec{a}-\vec{b}$ , where all the arrows are in the yz-plane.

Q6

Construct arrows that represent the following vectors: $1.5\vec{a}+3\vec{b}$ , $0.5\vec{a}-2\vec{b}$ , where all the arrows are in the yz-plane.

Q7

Find the point in the middle between the two points $A(1\vert 4\vert -2)$ and $B(10 \vert 2 \vert 4)$ .

Q8

Consider the vectors $\vec{a}=\left(\begin{array}{r} 1\\ -2\\ 1 \end{array}\right)$ and $\vec{b}=\left(\begin{array}{r} 2\\ 3\\ -1 \end{array}\right)$ .

Determine $\vert \vec a + \vec b \vert$ and show that this is not $\vert \vec a \vert + \vert \vec b \vert$ .
Indeed, use a figure to show that for any two vectors $\vec a$ and $\vec b$ it is always true that $\vert \vec a + \vec b \vert \leq \vert \vec a \vert + \vert \vec b \vert$ . For what vectors is $\vert \vec a + \vec b \vert = \vert \vec a \vert + \vert \vec b \vert$ ?
Determine $\vert \vec a - \vec b \vert$ . Again, show that $\vert \vec a - \vec b \vert \neq \vert \vec a \vert - \vert \vec b \vert$ .
Determine the unit vector of $\vec a$ (that is, a vector of length one pointing in the same direction as $\vec a$ ).

Solution

A1

$\vert\vec{u}\vert=\sqrt{(-2)^2+2^2+(-1)^2}=3$ , $\vec v = 3\cdot \vec u = \left(\begin{array}{r} -6\\ 6\\ -3 \end{array}\right) \rightarrow \vert \vec{v}\vert =\sqrt{(-6)^2+6^2+(-3)^2}=9$ . So indeed, three times bigger.

A2

$0\cdot \vec{a}=\left(\begin{array}{r} 0\cdot a_x\\ 0\cdot a_y\\ 0\cdot a_z \end{array}\right) = \left(\begin{array}{r} 0\\ 0\\ 0 \end{array}\right)$

A3

$4\cdot (5\cdot \vec{a})= 4\cdot \left(\begin{array}{r} 5\cdot a_x\\ 5\cdot a_y\\ 5\cdot a_z \end{array}\right) = \left(\begin{array}{r} 20\cdot a_x\\ 20\cdot a_y\\ 20\cdot a_z \end{array}\right)= 20\cdot \vec{a}$

A4

$\vec{a}+\vec{b}=\left(\begin{array}{r} a_x+b_x\\ a_y+b_y\\ a_z+b_z \end{array}\right) =\left(\begin{array}{r} b_x+a_x\\ b_y+a_y\\ b_z+a_z \end{array}\right) =\vec{b}+\vec{a}$
$\vec{a}+(\vec{b}+\vec{c})=\vec{a}+\left(\begin{array}{r} b_x+c_x\\ b_y+c_y\\ b_z+c_z \end{array}\right) = \left(\begin{array}{r} a_x+b_x+c_x\\ a_y+b_y+c_y\\ a_z+b_z+c_z \end{array}\right)$ and similar, $(\vec{a}+\vec{b})+\vec{c}=\left(\begin{array}{r} a_x+b_x\\ a_y+b_y\\ a_z+b_z \end{array}\right)+\vec{c}=\left(\begin{array}{r} a_x+b_x+c_x\\ a_y+b_y+c_y\\ a_z+b_z+c_z \end{array}\right)$ , so the same vectors.
$\vec{b}-\vec{a}=\left(\begin{array}{r} b_x-a_x\\ b_y-a_y\\ b_z-a_z \end{array}\right)=\left(\begin{array}{r} -(a_x-b_x)\\ -(a_y-b_y)\\ -(a_z-b_z) \end{array}\right)=-(\vec{a}-\vec{b})$ , and this is the opposite vector of $\vec{a}-\vec{b}$
$\vec{a}+\vec{a}=\left(\begin{array}{r} a_x+a_x\\ a_y+a_y\\ a_z+a_z \end{array}\right)=\left(\begin{array}{r} 2 a_x \\ 2 a_y\\ 2 a_z \end{array}\right)=2 \vec{a}$
$4 \vec{a} + 6 \vec{a} =\left(\begin{array}{r} 4a_x+6a_x\\ 4a_y+6a_y\\ 4a_z+6a_z \end{array}\right) = 10 \vec{a}$
$4 \vec{a} - 6 \vec{a} =\left(\begin{array}{r} 4a_x-6a_x\\ 4a_y-6a_y\\ 4a_z-6a_z \end{array}\right) =-2 \vec{a}$
$3 (\vec{a} +\vec{b}) = 3\left(\begin{array}{r} a_x+b_x\\ a_y+b_y\\ a_z+b_z \end{array}\right) = \left(\begin{array}{r} 3(a_x+b_x)\\ 3(a_y+b_y)\\ 3(a_z+b_z) \end{array}\right) = \left(\begin{array}{r} 3a_x+3b_x\\ 3a_y+3b_y\\ 3a_z+3b_z \end{array}\right) =3\vec{a}+3 \vec{b}$

A5

A6

A7

A8

$\vec a + \vec b = \left(\begin{array}{r} 3\\ 1\\ 0 \end{array}\right) \rightarrow \vert \vec a + \vec b \vert = \sqrt{3^2+1^2+0^2}=\sqrt{10}$ . But $\vert \vec a \vert + \vert \vec b \vert = \sqrt{6} +\sqrt{14} \neq \sqrt{10}$ .
Follows immediately from the completion of the triangle figure. The equation holds if the two vectors are parallel and pointing in the same direction.
$\vec a - \vec b = \left(\begin{array}{r} -1\\ -5\\ 2 \end{array}\right) \rightarrow \vert \vec a - \vec b \vert = \sqrt{(-1)^2+(-5)^2+2^2}=\sqrt{30}$ . But $\vert \vec a \vert - \vert \vec b \vert = \sqrt{6} -\sqrt{14} \neq \sqrt{30}$
As $\vert \vec a \vert =\sqrt{6}$ , the length of arrow $\vec a$ is $\sqrt{6}$ . As multiplying $\vec a$ the scalar $c=\frac{1}{\sqrt{6}}$ makes the arrow $c$ times "longer", the resulting length will be $1$ . Thus, the unit vector is $\frac{1}{\sqrt{6}}\cdot \vec a = \left(\begin{array}{r} 1/\sqrt{6}\\ -2/\sqrt{6}\\ 1/\sqrt{6} \end{array}\right)$