Vectors

Syllabus coverage

Use the standard notations for vectors, including the column vector $(\begin{matrix} x \\ y \\ z \end{matrix})$ , the component form $x \vec{i} + y \vec{j} + z \vec{k}$ , the arrow form $\vec{A B}$ , and a single letter such as $\vec{a}$
Carry out addition and subtraction of vectors and multiplication of a vector by a scalar, and interpret these operations in geometrical terms
Calculate the magnitude of a vector, and use unit vectors, displacement vectors and position vectors, in two or three dimensions
Understand the significance of all the symbols used when the equation of a straight line is expressed in the form $\vec{r} = \vec{a} + t \vec{b}$ , and find the equation of a line given sufficient information
Determine whether two lines are parallel, intersect or are skew, and find the point of intersection of two lines when it exists
Use formulae to calculate the scalar product of two vectors, and use scalar products in problems involving lines and points, including the angle between two lines and the foot of the perpendicular from a point to a line

Part of the AS & A Level Maths syllabus 9709.

The parallelogram OABC with

\vec{O A} = \vec{a}

and

\vec{O C} = \vec{b}

. Its diagonals are

\vec{a} + \vec{b}

and

\vec{b} - \vec{a}

, and they cross at the midpoint of each.

The four ways Cambridge writes a vector

A vector has both a magnitude and a direction; a scalar has magnitude only. The 9709 syllabus lists the notations you have to read and write, and they all say the same thing: a column vector $(\begin{matrix} 3 \\ - 1 \\ 4 \end{matrix})$ , the component form $3 \vec{i} - \vec{j} + 4 \vec{k}$ , the arrow form $\vec{A B}$ , and a single letter in bold or underlined, $\vec{a}$ .

$\vec{A B}$ is the displacement vector from A to B. It records how far, and in which direction, you travel to get from A to B. The order of the letters matters, because $\vec{B A} = - \vec{A B}$ .

A position vector is a displacement measured from the origin O. The point A(3, -1, 4) has position vector $\vec{O A} = 3 \vec{i} - \vec{j} + 4 \vec{k}$ . Every point has exactly one position vector, which is why position vectors describe lines so neatly.

In two dimensions the third component is absent, so $(\begin{matrix} x \\ y \end{matrix}) = x \vec{i} + y \vec{j}$ . Pure 3 questions are usually set in three dimensions, but the algebra is identical in two.

Notation	What it means	Example
Column vector	the components listed vertically, in order	$(\begin{matrix} 3 \\ - 1 \\ 4 \end{matrix})$
Component form	$x \vec{i} + y \vec{j} + z \vec{k}$	$3 \vec{i} - \vec{j} + 4 \vec{k}$
Arrow form	the displacement from the first letter to the second	$\vec{A B} = \vec{O B} - \vec{O A}$
Single letter	a vector named for convenience, printed in bold or underlined	$\vec{a} = (\begin{matrix} 3 \\ - 1 \\ 4 \end{matrix})$
Magnitude	the length of the vector, which is a scalar	$\| \vec{a} \| = \sqrt{3^{2} + 1^{2} + 4^{2}} = \sqrt{26}$

\vec{A B}

and

\vec{B A}

are not the same vector. A question that asks for

\vec{A B}

will not accept

\vec{B A}

, and a sign slip in the first line of a vectors question usually costs every mark after it.

Adding, subtracting and scaling vectors

Vectors are added component by component: the x-components add, then the y-components, then the z-components. Subtraction is the same with the signs changed. A scalar multiple $k \vec{a}$ has every component multiplied by k, so $k \vec{a}$ is parallel to $\vec{a}$ and k times as long. If k is positive the direction is unchanged; if k is negative the direction is reversed. Two vectors are parallel exactly when one is a scalar multiple of the other.

Geometrically, addition is the triangle law: following $\vec{a}$ and then $\vec{b}$ lands you at the same point as following the single vector $\vec{a} + \vec{b}$ . The syllabus states the parallelogram case explicitly - if OABC is a parallelogram then $\vec{O B} = \vec{O A} + \vec{O C}$ - because travelling along two adjacent sides gets you to the opposite corner.

Subtraction has a geometric reading too. If $\vec{O A} = \vec{a}$ and $\vec{O B} = \vec{b}$ then $\vec{A B} = \vec{O B} - \vec{O A} = \vec{b} - \vec{a}$ . Always subtract the position vector of the starting point from the position vector of the finishing point. That one rule removes almost every sign error students make in this topic.

$\vec{a} = (\begin{matrix} 1 \\ 3 \end{matrix})$ and $\vec{b} = (\begin{matrix} 5 \\ 1 \end{matrix})$ give $\vec{a} + \vec{b} = (\begin{matrix} 6 \\ 4 \end{matrix})$
$\vec{b} - \vec{a} = (\begin{matrix} 4 \\ - 2 \end{matrix})$ , the displacement from the tip of $\vec{a}$ to the tip of $\vec{b}$
$3 \vec{a} - 2 \vec{b} = (\begin{matrix} 3 \\ 9 \end{matrix}) - (\begin{matrix} 10 \\ 2 \end{matrix}) = (\begin{matrix} - 7 \\ 7 \end{matrix})$
the vectors $(\begin{matrix} 3 \\ 3 \\ 9 \end{matrix})$ and $(\begin{matrix} 1 \\ 1 \\ 3 \end{matrix})$ are parallel, because the first is 3 times the second

Do not add magnitudes.

| \vec{a} + \vec{b} |

is not

| \vec{a} | + | \vec{b} |

unless the two vectors point in the same direction, and that is almost never the case in a question.

The triangle law on a coordinate grid: the path

\vec{a}

then

\vec{b}

ends at the same point as the direct route

\vec{a} + \vec{b}

, and the midpoint M of OB has position vector

\frac{1}{2} (\vec{a} + \vec{b})

Midpoints, and the diagonals of a parallelogram

The midpoint M of the segment AB has position vector $\vec{O M} = \frac{1}{2} (\vec{O A} + \vec{O B}) = \frac{1}{2} (\vec{a} + \vec{b})$ . It is the average of the two position vectors, component by component. This is the only point-on-a-segment result you need: the syllabus notes say the general form of the ratio theorem is not included, but this midpoint case is expected.

For example, A(2, 5, -1) and B(6, 1, 3) give $\vec{O M} = \frac{1}{2} (\begin{matrix} 2 + 6 \\ 5 + 1 \\ - 1 + 3 \end{matrix}) = (\begin{matrix} 4 \\ 3 \\ 1 \end{matrix})$ , so M is the point (4, 3, 1).

The same idea proves the standard parallelogram fact. In the parallelogram OABC, $\vec{O A} = \vec{a}$ and $\vec{O C} = \vec{b}$ , and because opposite sides are equal and parallel, $\vec{A B} = \vec{b}$ and $\vec{B C} = \vec{a}$ . So $\vec{O B} = \vec{a} + \vec{b}$ and $\vec{A C} = \vec{b} - \vec{a}$ . The midpoint of OB has position vector $\frac{1}{2} (\vec{a} + \vec{b})$ , and the midpoint of AC has position vector $\frac{1}{2} (\vec{a} + \vec{b})$ as well. Two points with the same position vector are the same point, so the diagonals bisect each other.

midpoint of the segment joining $\vec{a}$ and $\vec{b}$ : $\frac{1}{2} (\vec{a} + \vec{b})$
diagonal OB of the parallelogram OABC: $\vec{O A} + \vec{O C} = \vec{a} + \vec{b}$
diagonal AC of the parallelogram OABC: $\vec{O C} - \vec{O A} = \vec{b} - \vec{a}$
points A, B and C are collinear when $\vec{A B}$ is a scalar multiple of $\vec{B C}$

A "Show (that)" question about a parallelogram is almost always the same three lines of algebra: write every vector in terms of the two the question gives you, simplify, and compare the two results.

OABC is a parallelogram with

\vec{O A} = \vec{a}

and

\vec{O C} = \vec{b}

. The diagonals give

\vec{O B} = \vec{a} + \vec{b}

and

\vec{A C} = \vec{b} - \vec{a}

, and they cross at the midpoint of each.

Magnitude and unit vectors

The magnitude of $\vec{a} = (\begin{matrix} x \\ y \\ z \end{matrix})$ is $| \vec{a} | = \sqrt{x^{2} + y^{2} + z^{2}}$ . That is Pythagoras in three dimensions - the Pure 1 result applied twice, once inside the base and once up the side - and in two dimensions you simply drop the z term. A magnitude is never negative, and it is zero only for the zero vector.

The distance between two points is the magnitude of the displacement between them: the distance AB is $| \vec{A B} | = | \vec{b} - \vec{a} |$ . For A(2, -1, 5) and B(5, 3, 17), $\vec{A B} = (\begin{matrix} 3 \\ 4 \\ 12 \end{matrix})$ and $| \vec{A B} | = \sqrt{9 + 16 + 144} = 13$ .

A unit vector has magnitude 1. The unit vector in the direction of $\vec{a}$ is $\hat{a} = \frac{1}{| \vec{a} |} \vec{a}$ , found by dividing every component by the magnitude. For $\vec{a} = 2 \vec{i} + 3 \vec{j} + 6 \vec{k}$ , $| \vec{a} | = 7$ , so $\hat{a} = \frac{1}{7} (2 \vec{i} + 3 \vec{j} + 6 \vec{k}) = \frac{2}{7} \vec{i} + \frac{3}{7} \vec{j} + \frac{6}{7} \vec{k}$ . Dividing by the magnitude rescales the vector without turning it, which is what "in the direction of" means.

$| 3 \vec{i} - \vec{j} + 4 \vec{k} | = \sqrt{9 + 1 + 16} = \sqrt{26} = 5.10$ to 3 significant figures
the unit vector in the direction of $\vec{A B}$ is $\frac{\vec{b} - \vec{a}}{| \vec{b} - \vec{a} |}$ , so it points from A to B
a unit vector always satisfies $x^{2} + y^{2} + z^{2} = 1$ : check $(\frac{2}{7})^{2} + (\frac{3}{7})^{2} + (\frac{6}{7})^{2} = \frac{49}{49} = 1$
the unit vector in the direction of the negative of a vector is the negative of the unit vector

An exact unit vector is expected unless the question asks for a given accuracy.

(\begin{matrix} 0.286 \\ 0.429 \\ 0.857 \end{matrix})

is the same vector, but

\frac{1}{7} (\begin{matrix} 2 \\ 3 \\ 6 \end{matrix})

is the form the mark scheme is written for.

The equation of a line: r = a + t b

The line through the point A with position vector $\vec{a}$ , parallel to the vector $\vec{b}$ , has equation $\vec{r} = \vec{a} + t \vec{b}$ . The syllabus asks you to understand the significance of every symbol, so take them one at a time. $\vec{r}$ is the position vector of a general point on the line. $\vec{a}$ is the position vector of one known point on the line, the point where t = 0. $\vec{b}$ is a direction vector: any non-zero vector parallel to the line will do. $t$ is a scalar parameter, and as t runs through all real values, $\vec{a} + t \vec{b}$ runs through every point of the line and no other point.

Two points give you both ingredients. For the line through A and B, take $\vec{a} = \vec{O A}$ and $\vec{b} = \vec{A B} = \vec{O B} - \vec{O A}$ . Through A(1, 2, -1) and B(4, 5, 8): $\vec{A B} = (\begin{matrix} 3 \\ 3 \\ 9 \end{matrix})$ , so the line is $\vec{r} = (\begin{matrix} 1 \\ 2 \\ - 1 \end{matrix}) + t (\begin{matrix} 3 \\ 3 \\ 9 \end{matrix})$ . The line $\vec{r} = (\begin{matrix} 4 \\ 5 \\ 8 \end{matrix}) + t (\begin{matrix} 1 \\ 1 \\ 3 \end{matrix})$ is the same line written from the other end, with a direction vector scaled by 3, and both are accepted.

To test whether a point lies on a line, substitute its coordinates and read off a value of t from each component separately. If all three components give the same t, the point is on the line. For the line above, (7, 8, 17) gives t = 2 from the x-component, t = 2 from the y-component and t = 2 from the z-component, so it lies on the line.

t = 0 gives the point $\vec{a}$ ; t = 1 gives the point $\vec{a} + \vec{b}$
any non-zero scalar multiple of $\vec{b}$ is a valid direction vector for the same line
the line through the origin in the direction of $\vec{b}$ is simply $\vec{r} = t \vec{b}$
a two-dimensional line is written the same way: $\vec{r} = (\begin{matrix} 1 \\ 2 \end{matrix}) + t (\begin{matrix} 3 \\ - 1 \end{matrix})$

The parameter belongs to the line, not to space. When a question gives you two lines, use t for one and s for the other, and never assume that t = s means the same point.

Parallel, intersecting and skew lines

Two lines in three dimensions are related in exactly one of three ways. They are parallel when their direction vectors are scalar multiples of each other, and then they either coincide or never meet. They intersect when they pass through a single common point. They are skew when they are not parallel and never meet - the lines cross when you look at them from above, but they pass at different heights.

The test is mechanical, and showing it earns the method marks. First compare the direction vectors: if one is a multiple of the other the lines are parallel, and there is nothing more to find. If not, write the lines as $\vec{a} + t \vec{b}$ and $\vec{c} + s \vec{d}$ , equate the three components, and solve two of the resulting equations simultaneously for t and s. Then substitute those values into the third component. If the third component agrees, the lines intersect at the point $\vec{a} + t \vec{b}$ ; if it disagrees, they are skew.

Worked: $l_{1} : \vec{r} = (\begin{matrix} 1 \\ 2 \\ - 1 \end{matrix}) + t (\begin{matrix} 1 \\ 3 \\ 4 \end{matrix})$ and $l_{2} : \vec{r} = (\begin{matrix} 7 \\ 2 \\ - 1 \end{matrix}) + s (\begin{matrix} - 2 \\ 3 \\ 4 \end{matrix})$ . The directions are not multiples, so the lines are not parallel. Equating the first two components gives $1 + t = 7 - 2 s$ and $2 + 3 t = 2 + 3 s$ , so t = s = 2. The third component then gives $- 1 + 8 = 7$ on both lines, which agrees, so the lines intersect at (3, 8, 7). If it had disagreed, they would be skew.

Two lines in two dimensions can never be skew: if they are not parallel they must meet. Skew lines are a genuinely three-dimensional idea.

The syllabus explicitly excludes the shortest distance between two skew lines and the equation of the common perpendicular to them. Neither is required, so do not spend exam time on either.

The scalar product

The scalar product of $\vec{a} = (\begin{matrix} x_{1} \\ y_{1} \\ z_{1} \end{matrix})$ and $\vec{b} = (\begin{matrix} x_{2} \\ y_{2} \\ z_{2} \end{matrix})$ is $\vec{a} \cdot \vec{b} = x_{1} x_{2} + y_{1} y_{2} + z_{1} z_{2}$ . The result is a scalar, not a vector, and it can be negative. The second form of the same quantity is $\vec{a} \cdot \vec{b} = | \vec{a} | | \vec{b} | \cos θ$ , where $θ$ is the angle between the two directions; this is the form that answers every angle question.

Rearranged, $\cos θ = \frac{\vec{a} \cdot \vec{b}}{| \vec{a} | | \vec{b} |}$ . For the direction vectors $(\begin{matrix} 2 \\ - 1 \\ 2 \end{matrix})$ and $(\begin{matrix} 1 \\ 2 \\ 2 \end{matrix})$ : the scalar product is $2 - 2 + 4 = 4$ , both magnitudes are 3, so $\cos θ = \frac{4}{9} = 0.444 \dots$ and $θ = {63.6}^{\circ}$ .

The most useful consequence is that $\vec{a} \cdot \vec{b} = 0$ exactly when the two vectors are perpendicular, provided neither is the zero vector. Two more special cases are worth knowing: $\vec{a} \cdot \vec{a} = | \vec{a} |^{2}$ , which is a quick check on a magnitude, and $\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}$ . A negative scalar product means the angle between the directions is obtuse; the angle between two lines is normally quoted as the acute angle, so if your $\cos θ$ comes out negative, the angle you want is $180^{\circ} - θ$ .

$\vec{i} \cdot \vec{i} = \vec{j} \cdot \vec{j} = \vec{k} \cdot \vec{k} = 1$ and $\vec{i} \cdot \vec{j} = \vec{i} \cdot \vec{k} = \vec{j} \cdot \vec{k} = 0$
perpendicular vectors: $\vec{a} \cdot \vec{b} = 0$
parallel vectors: $\vec{a} \cdot \vec{b} = | \vec{a} | | \vec{b} |$ , because $\cos 0^{\circ} = 1$
the angle between two lines is found from their direction vectors, so $\cos θ = \frac{{\vec{d}}_{1} \cdot {\vec{d}}_{2}}{| {\vec{d}}_{1} | | {\vec{d}}_{2} |}$

Angle questions use the direction vectors of the lines, never the position vectors of points on them. Substituting a position vector into the cosine formula is the classic wrong answer in this topic.

The foot of the perpendicular, and vectors inside 3D shapes

A question that gives you a line and a point P outside it, and asks for the closest point of the line to P, is a scalar product question in disguise. Write a general point of the line as $\vec{a} + t \vec{b}$ , form the vector from P to that general point, and force it to be perpendicular to the direction $\vec{b}$ by setting the scalar product to zero. That gives one linear equation in t, and t then fixes the foot of the perpendicular. The distance from P to the line is the magnitude of that perpendicular vector.

Worked: $l : \vec{r} = (\begin{matrix} 2 \\ 0 \\ 1 \end{matrix}) + t (\begin{matrix} 1 \\ 2 \\ - 1 \end{matrix})$ and P(5, 4, 0). A general point of l is $(\begin{matrix} 2 + t \\ 2 t \\ 1 - t \end{matrix})$ , so the vector from P to it is $(\begin{matrix} t - 3 \\ 2 t - 4 \\ - 1 - t \end{matrix})$ . Perpendicular to the line means $(\begin{matrix} t - 3 \\ 2 t - 4 \\ - 1 - t \end{matrix}) \cdot (\begin{matrix} 1 \\ 2 \\ - 1 \end{matrix}) = 0$ , so $(t - 3) + 2 (2 t - 4) - (- 1 - t) = 6 t - 12 = 0$ and t = 2. The foot is F(4, 4, -1), and $| \vec{P F} | = \sqrt{(- 1)^{2} + 0^{2} + (- 1)^{2}} = \sqrt{2}$ , which is the shortest distance from P to the line.

Questions set inside a cuboid, a tetrahedron or a pyramid are ordinary vector questions with coordinates attached. Put the origin at one vertex and take the edges along the axes, read off the coordinates of the vertices you need, and then use the same tools as always: a displacement vector from subtracting position vectors, a magnitude from Pythagoras, and an angle from the scalar product.

The syllabus stops short of the angle between a line and a plane, and of the distance from a point to a plane. The angle between two lines and the foot of the perpendicular from a point to a line are the two scalar-product applications you are examined on.

A cuboid with the space diagonal AG. Choosing A as the origin and the edges as axes turns a 3D shape into coordinates, and the angle at A is then an ordinary scalar product calculation.

Key terms you must be able to define

Vector: A quantity with both magnitude and direction, written as a column, in component form, or with an arrow such as $\vec{A B}$ .

Scalar: A quantity with magnitude only - an ordinary number. The scalar product of two vectors is a scalar.

Column vector: The components of a vector written one above the other inside brackets, top to bottom in the order x, y, z.

Component form: A vector written as $x \vec{i} + y \vec{j} + z \vec{k}$ , using the unit vectors along the three coordinate axes.

Position vector: The displacement of a point from the origin O, written $\vec{O A}$ for the point A. It fixes the point uniquely.

Displacement vector: The vector from one point to another: $\vec{A B} = \vec{O B} - \vec{O A}$ . It does not depend on where the origin is.

Magnitude: The length of a vector: $| \vec{a} | = \sqrt{x^{2} + y^{2} + z^{2}}$ for $\vec{a} = x \vec{i} + y \vec{j} + z \vec{k}$ .

Unit vector: A vector of magnitude 1. The unit vector in the direction of $\vec{a}$ is $\frac{1}{| \vec{a} |} \vec{a}$ .

Zero vector: The vector with all components zero. It has no direction, and it is the only vector with magnitude 0.

Scalar multiple: $k \vec{a}$ , with every component of $\vec{a}$ multiplied by the scalar k. Positive k keeps the direction, negative k reverses it.

Parallel vectors: Two vectors are parallel when one is a non-zero scalar multiple of the other.

Midpoint of a segment: The point halfway between $\vec{a}$ and $\vec{b}$ , with position vector $\frac{1}{2} (\vec{a} + \vec{b})$ .

Collinear points: Points lying on one straight line. A, B and C are collinear when $\vec{A B}$ is a scalar multiple of $\vec{B C}$ .

Direction vector: Any non-zero vector parallel to a line. In $\vec{r} = \vec{a} + t \vec{b}$ it is $\vec{b}$ , and any scalar multiple of it gives the same line.

Parameter: The scalar t in $\vec{r} = \vec{a} + t \vec{b}$ . Each value of t gives one point of the line; different lines need different letters.

Skew lines: Two lines in three dimensions that are not parallel and never meet. They do not lie in a common plane.

Scalar product: $\vec{a} \cdot \vec{b} = x_{1} x_{2} + y_{1} y_{2} + z_{1} z_{2} = | \vec{a} | | \vec{b} | \cos θ$ , where $θ$ is the angle between the directions.

Perpendicular vectors: Vectors whose scalar product is zero. For non-zero vectors this is exactly equivalent to the angle between them being $90^{\circ}$ .

Foot of the perpendicular: The point F of a line l closest to a point P, where $\vec{P F}$ is perpendicular to the direction of l. $| \vec{P F} |$ is the distance from P to l.

Applied

At AS & A Level the marks sit in the setup: choosing the model, stating the variable and writing the equation before any working. These show the shape of a full-mark answer, then you do the algebra yourself.

Midpoint and parallelogram

OABC is a parallelogram with $\vec{O A} = \vec{a}$ and $\vec{O C} = \vec{b}$ . Then $\vec{O B} = \vec{a} + \vec{b}$ , so the midpoint of OB is $\frac{1}{2} (\vec{a} + \vec{b})$ . And $\vec{A C} = \vec{b} - \vec{a}$ , so the midpoint of AC is $\frac{1}{2} (\vec{a} + \vec{b})$ too. The two midpoints are the same point, which is why the diagonals of a parallelogram bisect each other.

The angle between two lines

Find the angle between the lines with direction vectors ${\vec{d}}_{1} = (\begin{matrix} 2 \\ - 1 \\ 2 \end{matrix})$ and ${\vec{d}}_{2} = (\begin{matrix} 1 \\ 2 \\ 2 \end{matrix})$ . The scalar product is $2 - 2 + 4 = 4$ and $| {\vec{d}}_{1} | = | {\vec{d}}_{2} | = 3$ , so $\cos θ = \frac{4}{9}$ and $θ = {63.6}^{\circ}$ . The answer is a scalar: no direction survives an angle calculation.

Two lines that meet

$l_{1} : \vec{r} = (\begin{matrix} 1 \\ 2 \\ - 1 \end{matrix}) + t (\begin{matrix} 1 \\ 3 \\ 4 \end{matrix})$ and $l_{2} : \vec{r} = (\begin{matrix} 7 \\ 2 \\ - 1 \end{matrix}) + s (\begin{matrix} - 2 \\ 3 \\ 4 \end{matrix})$ . Equating the first two components gives $1 + t = 7 - 2 s$ and $2 + 3 t = 2 + 3 s$ , so t = s = 2. The third component gives 7 = 7, so the lines meet, at the point (3, 8, 7). If the third component had disagreed the lines would be skew.

Foot of the perpendicular

For $l : \vec{r} = (\begin{matrix} 2 \\ 0 \\ 1 \end{matrix}) + t (\begin{matrix} 1 \\ 2 \\ - 1 \end{matrix})$ and P(5, 4, 0), a general point of l is $(2 + t, 2 t, 1 - t)$ . Perpendicular to l means $(t - 3) + 2 (2 t - 4) - (- 1 - t) = 0$ , so t = 2 and the foot is (4, 4, -1). The distance from P to the line is $\sqrt{2} = 1.41$ to 3 significant figures.

Flashcards

Click a card to flip it. Mark honestly — the tracker uses this.

Q1What does

\vec{A B}

mean, and how is it written using position vectors?

The displacement from A to B:

\vec{A B} = \vec{O B} - \vec{O A} = \vec{b} - \vec{a}

. Also

\vec{B A} = - \vec{A B}

Q2How do you find the magnitude of a three-dimensional vector?

Square the three components, add, and take the square root:

| x \vec{i} + y \vec{j} + z \vec{k} | = \sqrt{x^{2} + y^{2} + z^{2}}

Q3How do you find a unit vector in the direction of

\vec{a}

Divide

\vec{a}

by its magnitude:

\hat{a} = \frac{1}{| \vec{a} |} \vec{a}

. The result always has magnitude 1.

Q4What is the position vector of the midpoint of AB?

\frac{1}{2} (\vec{O A} + \vec{O B}) = \frac{1}{2} (\vec{a} + \vec{b})

- the average of the two position vectors.

Q5Write down the equation of a line, and say what each symbol means.

\vec{r} = \vec{a} + t \vec{b}

\vec{r}

is a general point,

\vec{a}

is one known point on the line,

\vec{b}

is a direction vector, and t is a scalar parameter.

Q6How do you find the equation of the line through A and B?

Take

\vec{a} = \vec{O A}

and

\vec{b} = \vec{A B} = \vec{O B} - \vec{O A}

, then write

\vec{r} = \vec{a} + t \vec{b}

Q7How do you test whether a point lies on a line?

Substitute its coordinates and find t from each component. The point is on the line only if all three components give the same t.

Q8How do you decide whether two lines are parallel?

Compare their direction vectors: the lines are parallel exactly when one direction vector is a scalar multiple of the other.

Q9How do you decide whether two lines intersect or are skew?

Equate the components, solve two of the three equations for t and s, then substitute into the third. Agreement means they intersect; disagreement means they are skew.

Q10What is the scalar product in component form?

\vec{a} \cdot \vec{b} = x_{1} x_{2} + y_{1} y_{2} + z_{1} z_{2}

. It is a scalar, and it can be negative.

Q11What is the scalar product in terms of the angle between the vectors?

\vec{a} \cdot \vec{b} = | \vec{a} | | \vec{b} | \cos θ

, so

\cos θ = \frac{\vec{a} \cdot \vec{b}}{| \vec{a} | | \vec{b} |}

Q12What does

\vec{a} \cdot \vec{b} = 0

tell you?

That the two non-zero vectors are perpendicular. It is the standard test for a right angle.

Q13In the parallelogram OABC, what is

\vec{O B}

in terms of

\vec{a}

and

\vec{b}

\vec{O B} = \vec{O A} + \vec{O C} = \vec{a} + \vec{b}

, while

\vec{A C} = \vec{b} - \vec{a}

Q14What condition does the foot of the perpendicular from P to a line satisfy?

If F is the foot, then

\vec{P F}

is perpendicular to the direction vector of the line, so

\vec{P F} \cdot \vec{b} = 0

. That equation gives t.

Q15What are

\vec{i} \cdot \vec{i}

and

\vec{i} \cdot \vec{j}

\vec{i} \cdot \vec{i} = 1

, because

\vec{i}

is a unit vector;

\vec{i} \cdot \vec{j} = 0

, because the axes are perpendicular.

0 of 15 marked as known

Exam technique for this topic

Write the direction vector down as a separate line before you use the cosine formula or the perpendicular condition. Most lost marks in this topic come from reaching for a position vector where a direction vector was needed.
When you have to decide whether two lines are parallel, intersect or are skew, show all three component equations. Even when the lines turn out to be skew, the equations and the solved t and s are worth the method marks.
Keep the notation the question used. If it gives column vectors, answer in column vectors; if it gives $x \vec{i} + y \vec{j} + z \vec{k}$ , answer in that form.
Check what the question actually asks for at the end: a direction vector, a position vector, a unit vector, an angle or a distance. A question that asks for the foot of the perpendicular wants a point, not the value of t.
In a "Show (that)" question, write every vector in terms of the letters the question supplies and never introduce a new one. If the two expressions you are comparing are identical, say so explicitly - the final comparison line is a mark.

Why students lose marks here

A student wrote

\vec{A B} = \vec{O A} - \vec{O B}

, so

\vec{A B}

points from B to A.

Why it loses marksThe displacement from A to B is the position vector of B minus the position vector of A. Subtracting the other way round gives the negative vector, which reverses the direction and changes every later answer.

The answer that scores

\vec{A B} = \vec{O B} - \vec{O A} = \vec{b} - \vec{a}

, and

\vec{B A} = - \vec{A B}

A student wroteThe angle between the lines is

\cos^{- 1} \frac{\vec{a} \cdot \vec{c}}{| \vec{a} | | \vec{c} |}

, where

\vec{a}

and

\vec{c}

are the position vectors of points on the two lines.

Why it loses marksThe angle between two lines depends only on their directions, and position vectors depend on where the origin happens to be.

The answer that scoresUse the direction vectors:

\cos θ = \frac{{\vec{d}}_{1} \cdot {\vec{d}}_{2}}{| {\vec{d}}_{1} | | {\vec{d}}_{2} |}

A student wrote

| \vec{a} + \vec{b} | = | \vec{a} | + | \vec{b} |

Why it loses marksAdding the lengths of two journeys is only the length of the direct route when the two vectors point the same way. In general the direct route is shorter, which is the triangle inequality.

The answer that scoresAdd the components first, then find the magnitude. For

\vec{a} = (\begin{matrix} 4 \\ - 1 \\ 3 \end{matrix})

and

\vec{b} = (\begin{matrix} - 2 \\ 5 \\ 1 \end{matrix})

\vec{a} + \vec{b} = (\begin{matrix} 2 \\ 4 \\ 4 \end{matrix})

and

| \vec{a} + \vec{b} | = 6

, while

| \vec{a} | + | \vec{b} | = 10.58

A student wroteThe lines are parallel because their direction vectors look similar, so they never meet.

Why it loses marks"Similar" is not "a scalar multiple". Parallelism is an exact test: one direction vector must equal a constant times the other, with the same constant in every component.

The answer that scores

(\begin{matrix} 3 \\ 3 \\ 9 \end{matrix}) = 3 (\begin{matrix} 1 \\ 1 \\ 3 \end{matrix})

, so those two are parallel;

(\begin{matrix} 1 \\ 3 \\ 4 \end{matrix})

and

(\begin{matrix} - 2 \\ 3 \\ 4 \end{matrix})

are not, because no single constant works for all three components.

A student wroteTwo lines that fail to meet are skew.

Why it loses marksSkew means not parallel and not meeting. Two parallel lines also never meet, but they are parallel, not skew. The decision is a two-part test and both parts must be shown.

The answer that scoresCheck the direction vectors first. If they are not multiples, solve two of the three component equations and substitute into the third: agreement means the lines intersect, disagreement means they are skew.

A student wroteA unit vector in the direction of

\vec{a}

\vec{a}

divided by the number of components, or

\vec{a}

left as it is because the fractions are awkward.

Why it loses marksA unit vector must have magnitude exactly 1. Dividing by 3 when the magnitude is 7 gives a vector of magnitude

\frac{7}{3}

, and leaving

\vec{a}

unchanged leaves its original magnitude.

The answer that scoresDivide by the magnitude: for

\vec{a} = 2 \vec{i} + 3 \vec{j} + 6 \vec{k}

| \vec{a} | = 7

, so the unit vector is

\frac{1}{7} (2 \vec{i} + 3 \vec{j} + 6 \vec{k})

Syllabus coverage

The four ways Cambridge writes a vector

Adding, subtracting and scaling vectors

Midpoints, and the diagonals of a parallelogram

Magnitude and unit vectors

The equation of a line: r = a + t b

Parallel, intersecting and skew lines

The scalar product

The foot of the perpendicular, and vectors inside 3D shapes

Key terms you must be able to define

Applied

Flashcards

Exam-style questions

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Exam technique for this topic

Why students lose marks here