Vectors

Vector notation Paper 3

A vector has a size and a direction. In two dimensions it is written as a column, $\left(\begin{array}{c}3\\ -2\end{array}\right)$, or as $3\mathbf{i}-2\mathbf{j}$, where $\mathbf{i}$ and $\mathbf{j}$ are vectors of length 1 along the $x$- and $y$-axes. In three dimensions a third unit vector $\mathbf{k}$ points along the $z$-axis, so $\left(\begin{array}{c}2\\ -1\\ 5\end{array}\right)=2\mathbf{i}-\mathbf{j}+5\mathbf{k}$. Questions use both notations, often in the same question, so move between them freely.

Printed vectors are bold, $\mathbf{a}$; by hand, underline them. $\overrightarrow{AB}$ is the displacement from $A$ to $B$. The position vector of a point $A$ is $\overrightarrow{OA}$, the displacement from the origin $O$, and its components are the coordinates of $A$: the point $(2,-1,5)$ has position vector $2\mathbf{i}-\mathbf{j}+5\mathbf{k}$. Writing $\mathbf{a}$ and $\mathbf{b}$ for the position vectors of $A$ and $B$,
$$\overrightarrow{AB}=\overrightarrow{AO}+\overrightarrow{OB}=\mathbf{b}-\mathbf{a}$$
which is always "end minus start". Everything in this topic works in two or three dimensions; three is the usual setting at A Level.

Adding, subtracting and scaling vectors Paper 3

Vectors add component by component. Geometrically, $\mathbf{p}+\mathbf{q}$ is the single displacement with the same effect as $\mathbf{p}$ followed by $\mathbf{q}$ (the triangle law). $\mathbf{p}-\mathbf{q}$ means $\mathbf{p}+(-\mathbf{q})$, where $-\mathbf{q}$ has the same length as $\mathbf{q}$ and the opposite direction. Multiplying by a scalar $\lambda$ multiplies every component by $\lambda$: the result is parallel to the original, $\left|\lambda \right|$ times as long, and reversed if $\lambda <0$.

So two vectors are parallel exactly when one is a scalar multiple of the other. $\left(\begin{array}{c}2\\ -4\\ 6\end{array}\right)$ and $\left(\begin{array}{c}-1\\ 2\\ -3\end{array}\right)$ are parallel because the first is $-2$ times the second.

These facts turn geometry into algebra:

• "$OABC$ is a parallelogram" means $\overrightarrow{CB}=\overrightarrow{OA}$, which is the same as $\overrightarrow{OB}=\overrightarrow{OA}+\overrightarrow{OC}$;
• the midpoint $M$ of $AB$ has position vector $\frac{1}{2}(\mathbf{a}+\mathbf{b})$, because $\overrightarrow{OM}=\mathbf{a}+\frac{1}{2}(\mathbf{b}-\mathbf{a})$;
• points $A$, $B$ and $C$ are collinear when $\overrightarrow{AB}$ is a multiple of $\overrightarrow{AC}$.

The general ratio theorem is not needed, but the midpoint result is.

Magnitude and unit vectors Paper 3

The magnitude (length) of $\mathbf{v}=x\mathbf{i}+y\mathbf{j}+z\mathbf{k}$ is
$$\left|\mathbf{v}\right|=\sqrt{x^2+y^2+z^2}$$
which is Pythagoras' theorem used twice: $\sqrt{x^2+y^2}$ across the base, then combined with the height $z$. The distance between points $A$ and $B$ is $\left|\overrightarrow{AB}\right|$.

A unit vector has magnitude 1. The unit vector in the direction of $\mathbf{v}$ is $\frac{1}{\left|\mathbf{v}\right|}\mathbf{v}$: dividing by the length keeps the direction and makes the length 1. For a vector of length 10 in that direction, multiply the unit vector by 10.

The vector equation of a line Paper 3

A line is fixed by one point on it and its direction. If $A$, with position vector $\mathbf{a}$, lies on the line and $\mathbf{b}$ is parallel to the line, then every point $R$ of the line is reached by going to $A$ and then moving some multiple of $\mathbf{b}$:
$$\mathbf{r}=\mathbf{a}+t\mathbf{b}$$
Every symbol has a job:

• $\mathbf{r}$ is the position vector of a general point $R$ on the line;
• $\mathbf{a}$ is the position vector of one particular point on the line;
• $\mathbf{b}$ is a direction vector, parallel to the line;
• $t$ is a scalar parameter: $t=0$ gives $A$, $t=1$ gives the point one step of $\mathbf{b}$ along, and negative values go the other way. Each value of $t$ gives exactly one point.

For the line through two points $A$ and $B$, use $\mathbf{a}$ as the point and $\overrightarrow{AB}=\mathbf{b}-\mathbf{a}$ as the direction.

The equation is not unique. Any point of the line can replace $\mathbf{a}$, and any non-zero multiple of the direction can replace $\mathbf{b}$: all of them describe the same line. Where an answer here has to be checked by its form, the question says which point and direction to use.

To test whether a point lies on a line, set it equal to $\mathbf{a}+t\mathbf{b}$, find $t$ from one component, and check that the same $t$ fits the other two.

Parallel, intersecting or skew Paper 3

In three dimensions two distinct lines are related in one of three ways.

• Parallel: their direction vectors are multiples of each other.
• Intersecting: not parallel, and they meet at exactly one point.
• Skew: not parallel and never meeting. Skew lines cannot happen in a plane: think of a road crossing a railway on a bridge.

To decide, first compare the directions. If they are not parallel, write the general point of each line with different parameters, $s$ and $t$, and equate components. That gives three equations in two unknowns. Solve two of them for $s$ and $t$, then test the third. If it is satisfied, the lines intersect, and substituting $s$ into the first line gives the point of intersection. If it is not, the lines are skew. Say clearly which equation fails: that is where the final mark is.

You will not be asked for the shortest distance between skew lines, or for their common perpendicular.

The scalar product Paper 3

The scalar product of $\mathbf{a}=a_1\mathbf{i}+a_2\mathbf{j}+a_3\mathbf{k}$ and $\mathbf{b}=b_1\mathbf{i}+b_2\mathbf{j}+b_3\mathbf{k}$ is a number:
$$\mathbf{a}·\mathbf{b}=a_1{b}_1+a_2{b}_2+a_3{b}_3=\left|\mathbf{a}\right|\left|\mathbf{b}\right|\cos \theta$$
where $\theta$ is the angle between the two vectors drawn from the same point. The two forms agree because of the cosine rule: in the triangle with sides $\mathbf{a}$, $\mathbf{b}$ and $\mathbf{b}-\mathbf{a}$, expanding $|\mathbf{b}-\mathbf{a}{|}^2$ in components leaves exactly $-2(a_1{b}_1+a_2{b}_2+a_3{b}_3)$ where the cosine rule has $-2\left|\mathbf{a}\right|\left|\mathbf{b}\right|\cos \theta$. So
$$\cos \theta =\frac{\mathbf{a}·\mathbf{b}}{\left|\mathbf{a}\right|\left|\mathbf{b}\right|}$$

• Non-zero vectors are perpendicular exactly when $\mathbf{a}·\mathbf{b}=0$, since $\cos {90}^{°}=0$.
• A negative scalar product means an obtuse angle between the vectors.
• The angle between two lines is the angle between their direction vectors, normally given as the acute angle: if $\cos \theta <0$, use ${180}^{°}-\theta$.
• For the angle $ABC$ in a triangle use $\overrightarrow{BA}$ and $\overrightarrow{BC}$, both pointing away from $B$. Using $\overrightarrow{AB}$ with $\overrightarrow{BC}$ gives the supplement.

Knowledge of the vector product is not required.

The foot of a perpendicular Paper 3

The foot of the perpendicular from a point $P$ to a line is the point $N$ of the line nearest to $P$, and $PN$ is the perpendicular distance from $P$ to the line. The method uses the scalar product:

1. write the general point of the line, $\overrightarrow{ON}=\mathbf{a}+t\mathbf{b}$;
2. form $\overrightarrow{PN}=\overrightarrow{ON}-\overrightarrow{OP}$, which depends on $t$;
3. solve $\overrightarrow{PN}·\mathbf{b}=0$, because $PN$ is perpendicular to the line;
4. substitute that $t$ to find $N$, and work out $\left|\overrightarrow{PN}\right|$ for the distance.

The reflection of $P$ in the line is the point $P\prime$ for which $N$ is the midpoint of $PP\prime$, so $\overrightarrow{OP\prime }=2\overrightarrow{ON}-\overrightarrow{OP}$.

Questions may set all this inside a solid, such as a cuboid or a pyramid: put the origin at a corner, read off the position vectors of the vertices, and the methods are unchanged.