Define vector and scalar .

A vector has both a magnitude and a direction , while a scalar is an ordinary number with size only. Temperature is a scalar, but a change in temperature is a vector, since it has a direction as well as a size.

What is a vector path, and why is more than one route allowed?

A vector path is any chain of vectors leading from a start point to an end point on a diagram. Different routes between the same two points always simplify to the same answer, so any correct path will do.

Points are collinear when they all lie on the same straight line. Proving three points collinear is a standard vector argument, and it needs two things established rather than one.

How do you find the point where two lines on a vector diagram cross?

Write two different paths to that point, each carrying its own unknown scalar, then set the two expressions equal. Equating the coefficients of the two base vectors gives simultaneous equations in those scalars, which you solve and substitute back.

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Vectors (Edexcel IGCSE Further Pure Maths): Flashcards

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Define vector and scalar.

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Cards in this collection (33)

Define vector and scalar.
A vector has both a magnitude and a direction, while a scalar is an ordinary number with size only.
Temperature is a scalar, but a change in temperature is a vector, since it has a direction as well as a size.
What are the two ways of writing a vector from its components?
As a column vector $(\begin{matrix} 3 \\ 2 \end{matrix})$ , or in i and j notation as $3 i + 2 j$ .
Both say exactly the same thing: $3$ across and $2$ up.
How are vectors printed, and how should you write them by hand?
Printed vectors are set in bold, as in $a$ , but bold cannot be produced by hand.
By hand you underline the letter instead, and the two mean exactly the same thing.
What does $\vec{A B}$ mean, and does the order matter?
It is the vector from $A$ to $B$ , and the order matters a great deal.
Reversing the letters reverses the direction, so $\vec{B A} = - \vec{A B}$ .
Complete the results of adding and subtracting these column vectors:
$(\begin{matrix} 2 \\ 1 \end{matrix}) + (\begin{matrix} 1 \\ 4 \end{matrix}) = (\begin{matrix} ______ \\ 5 \end{matrix}) and (\begin{matrix} 2 \\ 1 \end{matrix}) - (\begin{matrix} 1 \\ 4 \end{matrix}) = (\begin{matrix} 1 \\ ______ \end{matrix})$
The completed results are:
$(\begin{matrix} 2 \\ 1 \end{matrix}) + (\begin{matrix} 1 \\ 4 \end{matrix}) = (\begin{matrix} 3 \\ 5 \end{matrix}) and (\begin{matrix} 2 \\ 1 \end{matrix}) - (\begin{matrix} 1 \\ 4 \end{matrix}) = (\begin{matrix} 1 \\ - 3 \end{matrix})$
Tops combine with tops and bottoms with bottoms; the two components never mix.
What does multiplying a vector by a scalar $k$ do?
Every component is multiplied by $k$ , so the whole vector is scaled by that factor.
A negative $k$ also reverses the direction, so $- 2 a$ is twice as long as $a$ and points the opposite way.
How do you travel the wrong way along a vector on a diagram?
Add its negative instead, so going backwards along $u$ contributes $- u$ to the journey.
A route that follows $t$ and then reverses $u$ therefore simplifies to $t - u$ .
True or False?
$a + b$ and $b + a$ take you to the same place.
True.
Following $a$ and then $b$ ends at the same point as following $b$ and then $a$ .
The two routes are the two pairs of sides of a parallelogram, and both reach the opposite corner.
Define position vector.
A position vector gives the location of a point relative to a fixed origin $O$ .
The point $A$ has position vector $a = \vec{O A}$ , and its components are simply its coordinates.
What is the position vector of the point $(3, - 2)$ ?
It is $3 i - 2 j$ , because a position vector's components are just the point's coordinates.
That correspondence is what lets you move between coordinate geometry and vector work freely.
Define displacement vector.
A displacement vector gives the direction and distance between two points, rather than a position relative to the origin.
The displacement from $A$ to $B$ is written $\vec{A B}$ .
Complete the rule linking a displacement vector to two position vectors:
$\vec{A B} =______-______$
The completed rule is:
$\vec{A B} = b - a$
The order looks backwards but is not: going from $A$ to $B$ means going $A$ to $O$ first, which is $- a$ , and then $O$ to $B$ .
$P$ and $Q$ have position vectors $3 i + 2 j$ and $6 i - 10 j$ . Find $\vec{P Q}$ .
Use $\vec{P Q} = q - p$ , which gives $(6 i - 10 j) - (3 i + 2 j)$ .
Subtracting component by component leaves $3 i - 12 j$ .
True or False?
$\vec{A B}$ depends on where the origin has been placed.
False.
Moving the origin changes both $a$ and $b$ by the same amount, so the difference $b - a$ is left untouched.
A displacement is about the two points alone, which is why the same arrow can be drawn anywhere on a diagram.
Define the magnitude of a vector.
The magnitude of a vector is its length, also called its modulus, and it is written $|a|$ .
It is a scalar: a magnitude has size but carries no direction.
How do you find the magnitude of $a = x i + y j$ ?
Use $|a| = \sqrt{x^{2} + y^{2}}$ .
The two components form a right-angled triangle with the vector itself as the hypotenuse, so squaring, adding and rooting gives its length.
True or False?
$|a + b|$ is the same as $|a| + |b|$ .
False.
Add the vectors first, and only then take the magnitude of the result.
Two vectors pointing in different directions partly cancel, so the combined length is usually less than the two lengths added together.
Given $a = 4 i + x j$ and $|a| = 5$ , how do you find $x$ ?
Write $\sqrt{4^{2} + x^{2}} = 5$ and square both sides to clear the root.
That gives $16 + x^{2} = 25$ , so $x = \pm 3$ , and both signs are valid unless the question rules one out.
Define unit vector.
A unit vector is a vector of length exactly $1$ .
It carries direction only, which is why it is the natural way to state the direction of something.
Complete the rule for a unit vector in the direction of $a$ :
$\hat{a} = \frac{______}{______}$
The completed rule is:
$\hat{a} = \frac{a}{|a|}$
Dividing any vector by its own length always leaves something of length $1$ , still pointing the same way.
How do you turn $3 i - 4 j$ into a unit vector?
Divide by its magnitude, which is $\sqrt{3^{2} + {(- 4)}^{2}} = 5$ .
Each component is divided separately, giving $\frac{3}{5} i - \frac{4}{5} j$ .
What is a vector path, and why is more than one route allowed?
A vector path is any chain of vectors leading from a start point to an end point on a diagram.
Different routes between the same two points always simplify to the same answer, so any correct path will do.
If $A X : X B = 3 : 5$ , what fraction of $\vec{A B}$ is $\vec{A X}$ ?
It is $\frac{3}{8}$ , because the ratio has $3 + 5 = 8$ parts altogether.
Read the question carefully though: $\vec{A X}$ is $\frac{3}{5}$ of $\vec{X B}$ , which is a different comparison.
How do you write a long vector path in terms of the vectors you are given?
Break the journey into steps that each match a given vector, reversing the sign wherever you travel backwards.
Adding the steps and collecting like terms gives the answer, so $b + b - a - a$ becomes $2 b - 2 a$ .
Complete the condition for $b$ to be parallel to $a$ :
$b =______a, where k is any non-zero______$
The completed condition is:
$b = k a, where k is any non-zero scalar$
A negative $k$ still means parallel, just pointing in the opposite direction.
How does factorising show that $9 a + 6 b$ and $12 a + 8 b$ are parallel?
They factorise to $3 (3 a + 2 b)$ and $4 (3 a + 2 b)$ , sharing a common bracket.
Two vectors built on the same bracket are scalar multiples of each other, so here the second is $\frac{4}{3}$ times the first.
Show that $2 p - 4 q$ and $6 q - 3 p$ are parallel.
Factorise both, giving $2 (p - 2 q)$ and $- 3 (p - 2 q)$ .
The second is therefore $- \frac{3}{2}$ times the first, so the two are parallel and point in opposite directions.
Define collinear.
Points are collinear when they all lie on the same straight line.
Proving three points collinear is a standard vector argument, and it needs two things established rather than one.
What two things must you show to prove $A$ , $B$ and $C$ are collinear?
That two of the line segments are parallel, and that they share a common point.
Parallel alone is not enough: two parallel segments sitting side by side never meet, so the shared point is what forces them onto one line.
If $\vec{O B} = 4 \vec{O Y}$ , are $O$ , $Y$ and $B$ collinear?
Yes. The scalar multiple shows $O B$ and $O Y$ are parallel, and both segments contain the point $O$ .
Parallel together with a common point is exactly the test, so the three points lie on one line.
For non-parallel $a$ and $b$ , complete the principle of equating coefficients:
$if α_{1} a + β_{1} b = α_{2} a + β_{2} b then α_{1} =______and β_{1} =______$
The completed principle is:
$if α_{1} a + β_{1} b = α_{2} a + β_{2} b then α_{1} = α_{2} and β_{1} = β_{2}$
It works only because $a$ and $b$ are not parallel; if they were, one could be absorbed into the other.
How do you find the point where two lines on a vector diagram cross?
Write two different paths to that point, each carrying its own unknown scalar, then set the two expressions equal.
Equating the coefficients of the two base vectors gives simultaneous equations in those scalars, which you solve and substitute back.
True or False?
Two different vector paths to the same point should use the same scalar letter.
False.
Each path needs its own unknown, usually $k$ and $λ$ , because the fractions travelled along the two lines are different.
Using one letter for both would force those fractions to be equal and give the wrong point.

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