Special vectors

The zero vector

The zero vector is the vector with the components

(000)\left(\begin{array}{r} 0\\0\\0 \end{array}\right)

Moving from head to tail along the zero-vector means moving not at all, as the tail and head occupy the same location. Clearly, the magnitude of the zero vector is 00.

The unit vector

Generally speaking, a unit vector is a vector of magnitude 11. The unit vector of another vector u⃗\vec{u} is the vector that points into the same direction as u⃗\vec{u} and has magnitude 11.

The opposite vector

Consider a vector

u⃗=(xyz)\vec{u}=\left(\begin{array}{r} x\\y\\z \end{array}\right)

The opposite vector of u⃗\vec{u}, written −u⃗-\vec{u}, has the components

−u⃗=(−x−y−z)\boxed{-\vec{u}=\left(\begin{array}{r} -x\\-y\\-z \end{array}\right)}

Clearly, the components of the opposite vector of u⃗\vec u describe how to get from the head to the tail of the arrow represented by u⃗\vec u.

The position vector of a point

The vector from the coordinate origin 00 to a point AA is called the position vector of AA, and clearly has the components

OA→=(AxAyAz)\boxed{\overrightarrow{OA}=\left(\begin{array}{r} A_x\\ A_y\\ A_z\end{array}\right)}

Instead of OA→\overrightarrow{OA} we also use the notation A⃗\vec{A}.

Exercise 1
Q1

Determine the opposite vector of v⃗=(−22−3)\vec{v}=\left(\begin{array}{r} -2\\ 2\\ -3 \end{array}\right).

Q2
  1. Determine the magnitude of the zero vector.
  2. How many other vectors of magnitude 00 do exist?
Q3
  1. Determine a unit vector with at least 22 non-zero components.
  2. How many unit vectors exist? If you attach the arrow of all these vectors to the origin, what geometrical object is formed by the arrow heads?
Q4

Consider the arrows in the yzyz-plane shown below.

  1. Which of these arrows are formed by the opposite vector of a⃗\vec{a}?
  2. Which of these arrows have the same components?
  3. Also, determine the components of all the arrows, and also of −a⃗-\vec{a}.
Q5

Determine the position vector of C(4∣5∣10)C(4\vert 5\vert 10) and indicate it in a 3d-Coordinate system.

Solution
A1

−v⃗=(−(−2)−2−(−3))=(2−23)-\vec{v}=\left(\begin{array}{r} -(-2)\\ -2\\ -(-3) \end{array}\right)=\left(\begin{array}{r} 2\\ -2\\ 3 \end{array}\right)

A2
  1. u⃗=(000)\vec{u}=\left(\begin{array}{r} 0\\ 0\\ 0 \end{array}\right) and thus ∣u⃗∣=02+02+02=0\vert \vec{u}\vert = \sqrt{0^2+0^2+0^2}=0.
  2. It is the only one. Because if any component of a vector u⃗\vec{u} is different from zero, then ux2+uy2+uz2>0u_x^2+u_y^2+u_z^2>0 and thus ∣u⃗∣>0\vert \vec{u}\vert >0.
A3

  1. e.g. for u⃗=(uxuy0)\vec{u}=\left(\begin{array}{r} u_x\\ u_y\\ 0 \end{array}\right) we have the condition

    ∣u⃗∣=ux2+uy2+02=1∣()2\vert \vec{u}\vert = \sqrt{u_x^2+u_y^2+0^2}=1 \quad \vert ()^2 ux2+uy2=1u_x^2+u_y^2=1

    Choose, for example, ux=0.5u_x=0.5 (or any other value between −1-1 and 11), then it follows

    0.52+uy2=10.5^2+u_y^2=1

    And thus uy=±0.75u_y=\pm \sqrt{0.75}

  2. There are infinitely many unit vectors (as there are infinitely many arrows of length 11, all pointing in different directions). Attached to the origin, the arrow heads for a sphere of radius 11.

A4
  1. c⃗\vec{c}
  2. a⃗,d⃗\vec{a}, \vec{d}
  3. a⃗=(0−14)\vec{a}=\left(\begin{array}{r} 0\\ -1\\ 4 \end{array}\right), b⃗=(0−3−3)\vec{b}=\left(\begin{array}{r} 0\\ -3\\ -3 \end{array}\right), c⃗=(01−4)=−a⃗\vec{c}=\left(\begin{array}{r} 0\\ 1\\ -4 \end{array}\right)=-\vec{a}, d⃗=(0−14)=a⃗\vec{d}=\left(\begin{array}{r} 0\\ -1\\ 4 \end{array}\right)=\vec{a}
A5

OC→=(4510)\overrightarrow{OC}=\left(\begin{array}{r} 4\\ 5\\ 10\end{array}\right)