Define the resultant vector.

The resultant is the single vector equivalent to two or more vectors added together. Drawn nose to tail, it is the direct route from the start of the first vector to the end of the last.

True or False? Finding the magnitude of a vector is the same calculation as finding the distance between two points.

True. Both are Pythagoras' theorem applied to the horizontal and vertical parts, with the vector or the line as the hypotenuse. So there is really only one formula here to learn, not two.

True or False? Multiplying a vector by a negative scalar changes only its magnitude.

False. It changes the magnitude and reverses the direction , so the vector ends up pointing the opposite way. Only a positive scalar leaves the direction alone.

True or False? If two vectors are scalar multiples of each other, the points involved must be collinear.

False. Scalar multiples are parallel , and parallel lines never meet, so the vectors could sit on two quite separate lines. Collinearity needs the two vectors to share a point as well, which is why both tests start them from the same place.

What is the difference between speed and velocity?

Velocity is a vector, carrying both size and direction; speed is a scalar, carrying size alone. Speed is the magnitude of the velocity, so it is found with Pythagoras and can never be negative.

IGCSEFurther MathsCambridge (CIE)Additional MathsFlashcardsVectorsVectors in Two Dimensions

Vectors in Two Dimensions (Cambridge (CIE) IGCSE Additional Maths): Flashcards

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FrontIntroduction to Vectors
Define a vector.

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Cards in this collection (33)

Define a vector.
A vector is a quantity with both a magnitude (size) and a direction.
Displacement, velocity and acceleration are all vector quantities, and each has components along the $x$ and $y$ axes.
How do you add or subtract two vectors?
Add or subtract the corresponding components separately.
In base vector form that means combining the $i$ terms with each other and the $j$ terms with each other, so $(2 i + j) - (i + 4 j) = i - 3 j$ .
What are the two ways of writing a vector?
As a column vector, such as $(\begin{matrix} 3 \\ - 2 \end{matrix})$ , with the $x$ component above the $y$ component.
Or in base vector form using the unit vectors, as $3 i - 2 j$ .
Complete the vector subtraction:
$(\begin{matrix} 2 \\ 1 \end{matrix}) - (\begin{matrix} 1 \\ 4 \end{matrix}) = (\begin{matrix} ______ \\ ______ \end{matrix})$
The completed subtraction is:
$(\begin{matrix} 2 \\ 1 \end{matrix}) - (\begin{matrix} 1 \\ 4 \end{matrix}) = (\begin{matrix} 1 \\ - 3 \end{matrix})$
The rows never mix: the top entries combine with each other and the bottom entries with each other.
How are vectors written in print, and how should you write them by hand?
In print a vector is set in bold, as $a$ .
You cannot write bold by hand, so underline the letter instead, which is what a written solution should show.
Define the resultant vector.
The resultant is the single vector equivalent to two or more vectors added together.
Drawn nose to tail, it is the direct route from the start of the first vector to the end of the last.
Complete the magnitude of the vector $v = v_{1} i + v_{2} j$ :
$|v| =______$
The completed formula is:
$|v| = \sqrt{v_{1}^{2} + v_{2}^{2}}$
The vertical bars mean magnitude, and the result is a scalar, since a length has no direction.
What is subtracting a vector the same as?
Adding the negative of that vector, so $a - b$ means $a + (- b)$ .
The negative of a vector has the same length but points the opposite way, so you travel back along it.
True or False?
Finding the magnitude of a vector is the same calculation as finding the distance between two points.
True.
Both are Pythagoras' theorem applied to the horizontal and vertical parts, with the vector or the line as the hypotenuse.
So there is really only one formula here to learn, not two.
When are two vectors parallel?
When one is a scalar multiple of the other, so $a = k b$ for some constant $k$ .
Every component must have been multiplied by that same $k$ , which is what you check when testing a pair.
Define a unit vector.
A unit vector is a vector whose magnitude is exactly 1.
The vectors $i$ and $j$ are the unit vectors along the positive $x$ and $y$ directions.
True or False?
Multiplying a vector by a negative scalar changes only its magnitude.
False.
It changes the magnitude and reverses the direction, so the vector ends up pointing the opposite way.
Only a positive scalar leaves the direction alone.
How do you find a unit vector in the direction of $v$ ?
Divide the vector by its own magnitude, giving $\frac{v}{|v|}$ .
So for $- 12 i + 5 j$ , whose magnitude is 13, the unit vector is $- \frac{12}{13} i + \frac{5}{13} j$ .
$O$ is the origin, with $\vec{O A} = a$ and $\vec{O B} = b$ . Find $\vec{A B}$ .
Travel from $A$ to $B$ by going back through the origin:
$\vec{A B} = \vec{A O} + \vec{O B} = - a + b$
So $\vec{A B} = b - a$ , which is the finishing position vector minus the starting one.
What is the difference between a position vector and a displacement vector?
A position vector is measured from the origin, so $\vec{O A} = a$ fixes where the point $A$ is.
A displacement vector runs between any two points and describes the journey from one to the other.
What is the zero vector?
It is the vector with no magnitude and no direction, written as a bold $0$ , or underlined by hand.
You get it by adding a vector to its own reverse, since $\vec{P Q} + \vec{Q P}$ returns you to where you started.
The point $A$ has coordinates $(3, - 2)$ . What is its position vector?
It is $3 i - 2 j$ .
The components of a position vector are the coordinates of the point, so the two carry the same information written differently.
Points $A$ and $B$ have position vectors $a$ and $b$ . How do you find the distance $A B$ ?
The distance is the magnitude of the vector between them, $|\vec{A B}|$ .
So find $\vec{A B}$ first and then take its magnitude, or work straight from the coordinates using Pythagoras.
Points $A$ and $B$ have position vectors $a$ and $b$ . Complete the position vector of the midpoint of $A B$ :
$\vec{O M} =______$
The completed result is:
$\vec{O M} = \frac{1}{2} (a + b)$
It is the average of the two position vectors, exactly as a midpoint of coordinates is.
Where does the midpoint result $\frac{1}{2} (a + b)$ come from?
Go from $O$ to $A$ , then half way along $\vec{A B}$ :
$\vec{O M} = a + \frac{1}{2} (b - a)$
Expanding gives $\frac{1}{2} a + \frac{1}{2} b$ , which is the stated result.
Define collinear.
Points are collinear if they all lie on the same straight line.
For instance $(- 2, - 2)$ , $(3, 3)$ and $(8, 8)$ are collinear, since all three sit on $y = x$ .
How do you show that three points $A$ , $B$ and $C$ are collinear?
Show that $\vec{A B} = k \vec{A C}$ for some constant $k$ .
Being scalar multiples makes them parallel, and because both start at the same point $A$ , the three points must lie on one line.
What is the difference between a specific and a general vector?
$\vec{A B}$ is specific: it is the one arrow starting at $A$ and ending at $B$ .
A labelled vector such as $a$ is general, so any arrow of the same length and direction anywhere on the diagram is also $a$ .
On a grid of identical parallelograms, how do you write the vector between two points in terms of $a$ and $b$ ?
Trace any path from one point to the other along the grid lines, and add the vectors as you go.
Each step is $a$ or $b$ , or its negative if you travel against the arrow, and different routes all give the same answer.
True or False?
If two vectors are scalar multiples of each other, the points involved must be collinear.
False.
Scalar multiples are parallel, and parallel lines never meet, so the vectors could sit on two quite separate lines.
Collinearity needs the two vectors to share a point as well, which is why both tests start them from the same place.
What does a velocity vector such as $(\begin{matrix} 3 \\ 7 \end{matrix}) {m s}^{- 1}$ tell you?
That the object moves at $3 {m s}^{- 1}$ in the positive $x$ direction and $7 {m s}^{- 1}$ in the positive $y$ direction.
The two components happen at once, so the object travels along the diagonal they produce.
What is the difference between speed and velocity?
Velocity is a vector, carrying both size and direction; speed is a scalar, carrying size alone.
Speed is the magnitude of the velocity, so it is found with Pythagoras and can never be negative.
A particle has velocity $(\begin{matrix} 6 \\ 8 \end{matrix}) {m s}^{- 1}$ . Find its speed.
Speed is the magnitude of the velocity:
$\sqrt{6^{2} + 8^{2}} = \sqrt{100} = 10$
So the speed is $10 {m s}^{- 1}$ , and the units stay the same because a magnitude is still a rate.
True or False?
Two objects travelling in opposite directions can have exactly the same speed.
True.
Velocities of $(\begin{matrix} 6 \\ 8 \end{matrix})$ and $(\begin{matrix} - 6 \\ - 8 \end{matrix})$ both have magnitude $10 {m s}^{- 1}$ .
Squaring the components destroys the signs, which is exactly why speed carries no direction.
Complete the position of a particle moving with constant velocity, at time $t$ :
$r =______+______$
The completed formula is:
$r = r_{0} + v t$
Here $r_{0}$ is the initial position, which is not necessarily the origin, and $v t$ is the displacement travelled in time $t$ .
A particle starts at $(2, - 3)$ and moves with constant velocity $(4 i + 7 j) {m s}^{- 1}$ . Where is it after 10 seconds?
Use $r = r_{0} + v t$ with $t = 10$ :
$(2 i - 3 j) + 10 (4 i + 7 j)$
This gives $r = (42 i + 67 j) m$ , so the particle is at the point $(42, 67)$ .
A boat's engine drives it at $(i + 2 j)$ while the current carries it at $(1.5 i + 2 j)$ . Find its resultant velocity.
Add the two velocity vectors, component by component:
$(i + 2 j) + (1.5 i + 2 j) = 2.5 i + 4 j$
So the resultant velocity is $(2.5 i + 4 j) {m s}^{- 1}$ , which is the velocity the boat actually travels at.
How do you find the time at which two particles collide?
Colliding means being in the same place at the same time, so set their position vectors equal to each other.
Two vectors are equal only when both components match, so equating either the $i$ or the $j$ components gives an equation to solve for $t$ .

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