This note covers vectors as quantities with both size and direction, in two and three dimensions. It includes position vectors, component form, magnitude, unit vectors, parallel vectors and dividing a line segment in a given ratio. Vectors are the language of motion, forces and geometry in space.
Key ideas
A vector has a magnitude (size) and a direction. A scalar has magnitude only. A vector can be drawn as an arrow. Two arrows represent the same vector if they have the same length and the same direction, wherever they start.
In component form a three-dimensional vector is written \(\mathbf{v}=\begin{pmatrix}v_1\\v_2\\v_3\end{pmatrix}=v_1\mathbf{i}+v_2\mathbf{j}+v_3\mathbf{k}\). Here \(\mathbf{i},\mathbf{j},\mathbf{k}\) are unit vectors along the \(x\), \(y\) and \(z\) axes. In two dimensions, drop the third component.
Position vectors. The position vector of a point \(A\) is the vector \(\overrightarrow{OA}=\mathbf{a}\) from the origin \(O\) to \(A\). Its components are the coordinates of \(A\). The vector from \(A\) to \(B\) is \[\overrightarrow{AB}=\mathbf{b}-\mathbf{a}.\] Remember "end minus start". Reversing direction reverses the sign: \(\overrightarrow{BA}=-\overrightarrow{AB}\).
Adding and scaling. Add or subtract vectors by adding or subtracting matching components. Multiplying by a scalar \(k\) multiplies each component by \(k\). If \(k\gt0\) the direction stays the same and the length is multiplied by \(k\). If \(k\lt0\) the direction reverses. In a diagram, adding two vectors means placing them head to tail.
Magnitude. The magnitude of \(\mathbf{v}\) is \(|\mathbf{v}|=\sqrt{v_1^2+v_2^2+v_3^2}\). The distance between points \(A\) and \(B\) is \(|\overrightarrow{AB}|\).
Unit vectors. A unit vector has magnitude 1. The unit vector in the direction of \(\mathbf{v}\) is \(\hat{\mathbf{v}}=\dfrac{\mathbf{v}}{|\mathbf{v}|}\). To make a vector of length \(m\) in the direction of \(\mathbf{v}\), use \(m\hat{\mathbf{v}}\).
Parallel vectors. Two non-zero vectors are parallel if one is a scalar multiple of the other: \(\mathbf{u}=k\mathbf{v}\) for some non-zero number \(k\). All components must give the same \(k\). Three points \(A\), \(B\), \(C\) are collinear if \(\overrightarrow{AB}\) is parallel to \(\overrightarrow{BC}\). Because the two vectors share the point \(B\), parallel implies the points lie on one line.
Dividing a segment. Suppose \(P\) lies on \(AB\) with \(AP:PB=m:n\). Then \(\overrightarrow{AP}=\dfrac{m}{m+n}\overrightarrow{AB}\), so \[\overrightarrow{OP}=\mathbf{a}+\dfrac{m}{m+n}(\mathbf{b}-\mathbf{a}).\] For the midpoint, \(m=n=1\), giving \(\overrightarrow{OM}=\tfrac12(\mathbf{a}+\mathbf{b})\).
Geometric proofs. Many geometry facts can be proved by writing each side of a shape as a vector in terms of two or three given vectors, then comparing. If two vectors are scalar multiples of each other, the lines are parallel. If they also share a point, the lines coincide. A parallelogram \(OABC\) with \(\overrightarrow{OA}=\mathbf{a}\) and \(\overrightarrow{OC}=\mathbf{c}\) has \(\overrightarrow{OB}=\mathbf{a}+\mathbf{c}\), and its diagonals bisect each other because both have midpoint \(\tfrac12(\mathbf{a}+\mathbf{c})\).
Vectors in context. Vectors describe displacement (change in position), velocity and force. A velocity vector holds both speed and direction. Its magnitude is the speed. For example, a ship with velocity \(\begin{pmatrix}3\\4\end{pmatrix}\) km/h travels at \(\sqrt{9+16}=5\) km/h. In two dimensions, a vector of magnitude \(r\) at angle \(\theta\) to the positive \(x\)-axis has components \(\begin{pmatrix}r\cos\theta\\r\sin\theta\end{pmatrix}\). Conversely, \(\tan\theta=v_2/v_1\), but check the quadrant of the vector, because a calculator gives only angles between \(-90^\circ\) and \(90^\circ\).
Equal vectors. Two vectors are equal exactly when all their matching components are equal. This is useful for finding unknowns: if \(\begin{pmatrix}x+1\\2y\end{pmatrix}=\begin{pmatrix}5\\-6\end{pmatrix}\), then \(x=4\) and \(y=-3\).
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| \(\overrightarrow{AB}=\mathbf{b}-\mathbf{a}\) | Vector from \(A\) to \(B\) using position vectors. |
| \(|\mathbf{v}|=\sqrt{v_1^2+v_2^2+v_3^2}\) | Magnitude; also the distance between two points. |
| \(\hat{\mathbf{v}}=\dfrac{\mathbf{v}}{|\mathbf{v}|}\) | Unit vector in the direction of \(\mathbf{v}\). |
| \(\mathbf{u}=k\mathbf{v}\) | Condition for \(\mathbf{u}\) and \(\mathbf{v}\) to be parallel. |
| \(\overrightarrow{OP}=\mathbf{a}+\dfrac{m}{m+n}(\mathbf{b}-\mathbf{a})\) | Position of \(P\) dividing \(AB\) in the ratio \(m:n\). |
| \(\overrightarrow{OM}=\tfrac12(\mathbf{a}+\mathbf{b})\) | Midpoint of \(AB\). |
Worked examples
Example 1 Two drones are at \(A(2,-1,3)\) and \(B(5,3,-9)\), with coordinates in metres. Find \(\overrightarrow{AB}\), the distance between the drones, and the unit vector in the direction of \(\overrightarrow{AB}\).
\(\overrightarrow{AB}=\mathbf{b}-\mathbf{a}=\begin{pmatrix}5-2\\3-(-1)\\-9-3\end{pmatrix}=\begin{pmatrix}3\\4\\-12\end{pmatrix}\).
Distance: \(|\overrightarrow{AB}|=\sqrt{3^2+4^2+(-12)^2}=\sqrt{9+16+144}=\sqrt{169}=13\) m.
Unit vector: \(\dfrac{1}{13}\begin{pmatrix}3\\4\\-12\end{pmatrix}\).
Answer: \(\overrightarrow{AB}=3\mathbf{i}+4\mathbf{j}-12\mathbf{k}\); distance 13 m; unit vector \(\tfrac{1}{13}(3\mathbf{i}+4\mathbf{j}-12\mathbf{k})\).
Example 2 The vectors \(\mathbf{p}=\begin{pmatrix}2\\-3\\k\end{pmatrix}\) and \(\mathbf{q}=\begin{pmatrix}-6\\9\\-12\end{pmatrix}\) are parallel. Find \(k\).
Parallel means \(\mathbf{q}=\lambda\mathbf{p}\) for some number \(\lambda\).
The first components give \(-6=2\lambda\), so \(\lambda=-3\).
Check the second components: \(-3\times(-3)=9\). This matches.
The third components give \(-12=\lambda k=-3k\), so \(k=4\).
Answer: \(k=4\).
Example 3 Points \(A\) and \(B\) have position vectors \(\mathbf{a}=\begin{pmatrix}1\\2\\3\end{pmatrix}\) and \(\mathbf{b}=\begin{pmatrix}6\\-3\\8\end{pmatrix}\). The point \(P\) lies on \(AB\) with \(AP:PB=2:3\). Find the coordinates of \(P\).
\(\overrightarrow{AB}=\mathbf{b}-\mathbf{a}=\begin{pmatrix}5\\-5\\5\end{pmatrix}\).
\(AP\) is \(\tfrac{2}{2+3}=\tfrac25\) of \(AB\), so \(\overrightarrow{AP}=\tfrac25\begin{pmatrix}5\\-5\\5\end{pmatrix}=\begin{pmatrix}2\\-2\\2\end{pmatrix}\).
\(\overrightarrow{OP}=\mathbf{a}+\overrightarrow{AP}=\begin{pmatrix}1+2\\2-2\\3+2\end{pmatrix}=\begin{pmatrix}3\\0\\5\end{pmatrix}\).
Answer: \(P=(3,0,5)\).
Common errors
- Computing \(\overrightarrow{AB}\) as \(\mathbf{a}-\mathbf{b}\). That gives \(\overrightarrow{BA}\), the opposite direction.
- Using \(m:n\) as the fraction \(\tfrac{m}{n}\) instead of \(\tfrac{m}{m+n}\) when finding a dividing point.
- Checking only one component when testing for parallel vectors. All components must give the same multiplier.
- Forgetting the square root in a magnitude, or squaring a negative component and keeping the sign.
- Treating a vector as a number: writing \(\mathbf{v}=5\) instead of giving components, or dividing by a vector.
Exam tip: Write the vector you are finding in the form "end minus start", for example \(\overrightarrow{AB}=\mathbf{b}-\mathbf{a}\), before substituting. It prevents sign slips.
Check yourself
- Find the magnitude of \(\mathbf{v}=2\mathbf{i}-6\mathbf{j}+3\mathbf{k}\).
Answer
\(\sqrt{4+36+9}=\sqrt{49}=7\). - Find the midpoint of the segment joining \(A(4,-2,1)\) and \(B(0,6,7)\).
Answer
\(\tfrac12\left(\begin{pmatrix}4\\-2\\1\end{pmatrix}+\begin{pmatrix}0\\6\\7\end{pmatrix}\right)=\begin{pmatrix}2\\2\\4\end{pmatrix}\), so the midpoint is \((2,2,4)\). - Find a vector of length 12 in the direction of \(\begin{pmatrix}2\\-1\\2\end{pmatrix}\).
Answer
The magnitude is \(\sqrt{4+1+4}=3\), so the unit vector is \(\tfrac13\begin{pmatrix}2\\-1\\2\end{pmatrix}\). Multiply by 12: \(\begin{pmatrix}8\\-4\\8\end{pmatrix}\). - Show that \(A(1,1,1)\), \(B(3,2,0)\) and \(C(7,4,2)\) are not collinear.
Answer
\(\overrightarrow{AB}=\begin{pmatrix}2\\1\\-1\end{pmatrix}\) and \(\overrightarrow{BC}=\begin{pmatrix}4\\2\\2\end{pmatrix}\). The first two components give a multiplier of 2, but \(2\times(-1)=-2\ne2\). The vectors are not parallel, so the points are not collinear.