How to do a C4 Vectors exam question

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Notnek
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Guided solution to an exam question is in the post below this one

C4 Vectors exam questions give a lot of A Level students nightmares, often because they don't fully understand the underlying methods and also because textbooks/teachers don't always prepare students for the kinds of questions that you see in exams. I thought I'd write a solution to an exam question that shows in detail how to draw diagrams and explains all of the methods.

I wrote this solution a few months back but after noticing a lot of vectors queries this year on TSR, I decided today to LaTeX it up and upload it here. Of course an actual solution won't need to be this detailed. If you are a C4 student who hasn't done this question before then I highly recommend trying it yourself before going through my solution.

The question is Edexcel C4 January 2012 Question 7 but it will be useful for all exam boards. Also not all C4 questions are as long as this one, I chose it because it involves a lot of different techniques.

Please let me know if you have any questions or notice any mistakes!
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Notnek
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First I'll make a diagram showing all of the information given so far. I recommend really big diagrams since you may end up including a lot of information in it so you need it to be clear (and of course use a pencil). Points A, B and D can be anywhere on my diagram since I don't care about their actual 3D positions. But I can see from a later part that angle BAD is obtuse so I will use that information when I make my diagram. If your initial diagram showed angle BAD as being acute then it's best to edit the diagram once you realise that it's actually obtuse. I've marked an origin on my diagram (in an arbitrary place) and showed all of the position vectors:

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To find  \overrightarrow{AB} I can see on the diagram that \overrightarrow{AB} = \overrightarrow{AO} + \overrightarrow{OB} = -\overrightarrow{OA} + \overrightarrow{OB}.

So \overrightarrow{AB}= -\begin{pmatrix}2 \\ -1 \\ 5\end{pmatrix}+\begin{pmatrix}5 \\ 2 \\ 10\end{pmatrix} =\begin{pmatrix}3 \\ 3 \\ 5\end{pmatrix}.

In general for any points X and Y and origin O, it is true that \overrightarrow{XY} = \overrightarrow{OY} - \overrightarrow{OX}


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I have added the vector \overrightarrow{AB} to my diagram. It's a good idea to add to your initial diagram as you get more information. But sometimes information in later parts will force you to make a new diagram e.g. because you don't have enough space or some new information changes the positions of the vectors.

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The vector equation of a line is \mathbf{r} = \mathbf{a} + t\mathbf{b}.

\mathbf{r} represents the general position vector of any point on the line. \mathbf{a} is the position vector of a fixed point on the line and \mathbf{b} is any vector parallel to the line.

For \mathbf{a} I have a choice of using \overrightarrow{OA} or \overrightarrow{OB} since both A and B lie on l. I will use \overrightarrow{OB}.

For \mathbf{b} I require a vector parallel to the line. I can use the vector \overrightarrow{AB} for this since it obviously has the same direction as the line that passes through A and B.

So my vector equation of l is \mathbf{r} = \overrightarrow{OB} + t(\overrightarrow{AB})


\mathbf{r} = \begin{pmatrix}5 \\ 2 \\ 10\end{pmatrix}+ t\begin{pmatrix}3 \\ 3 \\ 5\end{pmatrix}

I could add these two terms together to get

\mathbf{r} = \begin{pmatrix}5+3t \\ 2+3t \\ 10+5t\end{pmatrix}

This tells me that the position vector of any point on the line l must have the form \begin{pmatrix}5+3t \\ 2+3t \\ 10+5t\end{pmatrix} .

For example, when I set t=0 I get \mathbf{r}=\begin{pmatrix}5 \\ 2 \\ 10\end{pmatrix} which is \overrightarrow{OB} and with t=-1 I get \mathbf{r}=\begin{pmatrix}2 \\ -1 \\ 5\end{pmatrix} which is \overrightarrow{OA}.

It's important to realise that \mathbf{r} represents any position vector of a point on l for a certain value of t. Here's a diagram illustrating this:

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If in a future part of the question you were told that a point lies on the line l but didn’t have any more information about that point then you could represent its position vector like this

\begin{pmatrix}5+3t \\ 2+3t \\ 10+5t\end{pmatrix}

where t is unknown.


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For two vectors \mathbf{a} and \mathbf{b} either pointing towards each other or away from each other, the angle between the vectors \theta satisfies

\mathbf{a}\cdot \mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos \theta

The vectors \overrightarrow{AB} and \overrightarrow{AD} are pointing away from each other so I will use these vectors to work out the angle \theta.


First I need to work out \overrightarrow{AD}, which I can do in a similar way to part a):

\overrightarrow{AD} = \overrightarrow{OD}-\overrightarrow{OA}

=\begin{pmatrix}-1 \\ 1 \\ 4\end{pmatrix}- \begin{pmatrix}2 \\ -1 \\ 5\end{pmatrix}= \begin{pmatrix}-3 \\ 2 \\ -1\end{pmatrix}

So now using the dot product formula:

\overrightarrow{AB} \cdot \overrightarrow{AD} = |\overrightarrow{AB}| \times |\overrightarrow{AD}| \times \cos \theta

\displaystyle \Rightarrow \cos \theta = \frac{\overrightarrow{AB} \cdot \overrightarrow{AD}}{| \overrightarrow{AB} | \times |\overrightarrow{AD}|}

\overrightarrow{AB} \cdot \overrightarrow{AD}=(3\times -3) + (3\times 2) + (5\times -1) = -8

| \overrightarrow{AB} | \times |\overrightarrow{AD}| = \sqrt{3^2+3^2+5^2}\times \sqrt{(-3)^2+2^2+(-1)^2} = \sqrt{43}\sqrt{14}

So \displaystyle \ cos \theta= -\frac{8}{\sqrt{43}\sqrt{14}} = -0.326...

\Rightarrow \theta = 109.02... = 109 \ (to the nearest degree)

I will store the full number in my calculator in case I need it later.


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These are the kinds of questions that are often done very badly by students but if you draw a good diagram with all the information on it then it becomes a simple GCSE vectors problem.

I have had to edit my diagram a bit for space reasons but I've still included all the information including the angle DAB found in c) as well as the vector \overrightarrow{AD}. Notice that c) didn't ask me to find \overrightarrow{AD}, but it was found as part of my working in c) so it's a good idea to add it to my diagram. More information is always better.

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Using the information given in the question, I've marked on the point C in my diagram in the only place that it could be (remember that if a shape is given as ABCD then A has to be joined to B and B to C etc.) The question is asking for the position vector of C which is the vector \overrightarrow{OC}, so I've marked that on my diagram with an arrow.

I can use a vector path to find \overrightarrow{OC}. There are the two shortest paths:

\overrightarrow{OC} = \overrightarrow{OB} + \overrightarrow{BC}
\overrightarrow{OC} = \overrightarrow{OD} + \overrightarrow{DC}

There is no benefit to using either path so I'm going to use the first one (if you had time in an exam you should use both paths to check your answer).

I already have \overrightarrow{OB} but not \overrightarrow{BC}. But since the line BC is parallel to AD and has the same length, it must be true that \overrightarrow{BC} = \overrightarrow{AD}.

So \overrightarrow{OC} =\overrightarrow{OB} + \overrightarrow{AD} = \begin{pmatrix}5 \\ 2 \\ 10\end{pmatrix} + \begin{pmatrix}-3 \\ 2 \\ -1\end{pmatrix} = \begin{pmatrix}2 \\ 4 \\ 9\end{pmatrix}


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I have added the vector \overrightarrow{OC} to my diagram:

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I am asked to find the area of a parallelogram so I will consider the two main ways of finding this : either I could use the formula base x perpendicular height or I could split the parallelogram into two congruent triangles then double the area of one of the triangles.

The first method would be okay but would require a bit of extra work to find the perpendicular height (it turns out that this is required for part e) anyway so this method doesn't work out too bad for this question!)

But loooking at triangle DAB, I can see that I can work out lengths of vectors to give me the lengths AD and AB plus I have the angle DAB so the \frac{1}{2}ab\sin{C} formula works well here. Even better, I've already worked out the lengths of AD and AB in part c) so I won't need to work them out again:

\displaystyle Area \ of \ {\bigtriangleup DAB} = \frac{1}{2} \times \sqrt{43} \times \sqrt{14} \times \sin 109.02... = 11.597...

\Rightarrow Area \ of \ ABCD = 2 \times 11.597... = 23.194... = 23.2 (3sf)

I will store the full number in my calculator in case I need it later.


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The shortest distance from a point P to a line l is the length of a line that starts at P and meets the line l at a right-angle. I have shown this distance in my diagram below and marked it with a d. This change to the diagram has created a new point on the line which I have labelled E:

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As for all vectors questions (and any other A Level maths questions), always look for the simplest method first. And the simplest method here is to use lengths, angles and SOHCAHTOA to find d. Using angle DAB I can find angle DAE and I already have the length of AD so that's enough to use SOHCAHTOA to find d.

D\hat{A}E = 180 - 109.029... = 70.971...

So \displaystyle \sin(70.971...) = \frac{d}{\sqrt{14}}

\Rightarrow d = \sqrt{14} \times \sin(70.971...) = 3.537... = 3.54 (3sf)

There are alternative approaches to this question which make use of the vector equation of the line. These approaches are definitely not recommended for this question because they are much longer. But below I will show one of these methods and I highly recommend understanding it because you may need to use a similar method in harder questions.


Alternative method for f) (I highly recommend understanding this):

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As soon as I see the word “perpendicular” or notice a right-angle in a C4 vector question, I know that the use of the dot product is probably going to be a possible method. The vectors \overrightarrow{AB} and \overrightarrow{DE} are perpendicular so I know that

\overrightarrow{AB} \cdot \overrightarrow{DE} = 0

I have \overrightarrow{AB} but the problem is \overrightarrow{DE} since I have no information about E, other than that it lies on l. But looking back at part b), I found that the position vector of any point on the line l has this form

\begin{pmatrix}5+3t \\ 2+3t \\ 10+5t\end{pmatrix}

So I can say \overrightarrow{OE} = \begin{pmatrix}5+3t \\ 2+3t \\ 10+5t\end{pmatrix}

for some unknown t and I have redrawn the diagram below showing this vector (I’ve removed some of the information so it is clearer):

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Now using vector paths I can work out the vector \overrightarrow{DE} in terms of t:

\overrightarrow{DE} = \overrightarrow{OE} – \overrightarrow{OD}

=\begin{pmatrix}5+3t \\ 2+3t \\ 10+5t\end{pmatrix}-\begin{pmatrix}-1 \\ 1 \\ 4\end{pmatrix}=\begin{pmatrix}6+3t \\ 1+3t \\ 6+5t\end{pmatrix}

So \overrightarrow{DE} = \begin{pmatrix}6+3t \\ 1+3t \\ 6+5t\end{pmatrix}


Then going back to the dot product result mentioned earlier I get

\begin{pmatrix}3 \\ 3 \\ 5\end{pmatrix} \cdot \begin{pmatrix}6+3t \\ 1+3t \\ 6+5t\end{pmatrix} = 0

\Rightarrow 3(6+3t) + 3(1+3t) + 5(6+5t) = 0

\Rightarrow 18 + 9t + 3 + 9t + 30 + 25t = 0

\displaystyle \Rightarrow 51 + 43t = 0 \Rightarrow t = -\frac{51}{43}


I know that \overrightarrow{DE} = \begin{pmatrix}6+3t \\ 1+3t \\ 6+5t\end{pmatrix}

So I can substitute t=-\frac{51}{43} into the above to get

(skipping a line of working)

\overrightarrow{DE}=\frac{1}{43}\begin{pmatrix}105 \\ -110 \\ 3 \end{pmatrix}


d is the length of \overrightarrow{DE} so I'm nearly done...

d = |\overrightarrow{DE}| = \frac{1}{43}\sqrt{105^2+(-110)^2+3^2}= \frac{\sqrt{23134}}{43} = 3.54 (3sf)

Phew!

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Philip-flop
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Thank you so much for this notnek you have no idea how much I appreciate this!
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username2982648
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Thanks for the tag! I'll definitely have a look at this further once I've finished learning the whole chapter!
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kiiten
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Thanks for this, very helpful


This is hard for Edexcel!? The AQA ones are way worse :cry:
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Notnek
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(Original post by kiiten)
Thanks for this, very helpful

This is hard for Edexcel!? The AQA ones are way worse :cry:
Have you looked at the whole question? I'll post the rest of the question now and the solutions over the next few days. The first 3 parts are standard but the 3 parts that are still to come are harder.
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kiiten
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(Original post by notnek)
Have you looked at the whole question? I'll post the rest of the question now and the solutions over the next few days. The first 3 parts are standard but the 3 parts that are still to come are harder.
Oh, no i thought that was one question, my bad . I didnt know that the rest will make up just one question! :O
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Notnek
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(Original post by kiiten)
Oh, no i thought that was one question, my bad . I didnt know that the rest will make up just one question! :O
I just posted the rest of the parts above. It's quite a long one!
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(Original post by notnek)
...
PRSOM. Looks helpful!
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kiiten
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(Original post by notnek)
I just posted the rest of the parts above. It's quite a long one!
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is AB = -DC ? (the arrow on my diagram goes in the opposite direction). Its A to B to C to D right?


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DFranklin
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(Original post by kiiten)
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is AB = -DC ? (the arrow on my diagram goes in the opposite direction). Its A to B to C to D right?

If the vertices are given as ABCD, then that would normally imply A to B to C to D. But

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if you draw a parallelogram ABCD, you'll find that AB and CD point in opposite directions (and so AB = -CD, and therefore AB = DC).
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Notnek
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(Original post by kiiten)
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is AB = -DC ? (the arrow on my diagram goes in the opposite direction). Its A to B to C to D right?

Is this right? I feel like ive done something wrong.


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No vector AB = vector DC. If you've drawn the points in the correct order around the parallelogram then I'm not sure why you think AB = -DC.
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(Original post by DFranklin)
If the vertices are given as ABCD, then that would normally imply A to B to C to D. But

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if you draw a parallelogram ABCD, you'll find that AB and CD point in opposite directions (and so AB = -CD, and therefore AB = DC).


Is this wrong?

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Notnek
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(Original post by kiiten)
Is this wrong?

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Your arrows show the vectors \overrightarrow{AB} and \overrightarrow{CD}. From your diagram you can see that \overrightarrow{AB}= -\overrightarrow{CD} or alternatively \overrightarrow{AB} = \overrightarrow{DC}

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kiiten
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(Original post by notnek)
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Your arrows show the vectors \overrightarrow{AB} and \overrightarrow{CD}. From your diagram you can see that \overrightarrow{AB}= -\overrightarrow{CD} or alternatively \overrightarrow{AB} = \overrightarrow{DC}


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? Im confused. You said my arrows show AB = CD but my diagram shows AB = -CD??
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Notnek
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(Original post by kiiten)
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? Im confused. You said my arrows show AB = CD but my diagram shows AB = -CD??





Too many spoilers - I'm getting fed up with them

The vector \overrightarrow{AB} is the vector that goes from A to B. In your diagram this vector starts at A and moves up to B.

The vector \overrightarrow{CD} is the vector that goes from C to D. In your diagram this vector starts at C and moves down to D.

Since these vectors are parallel but in opposite directions then \overrightarrow{AB} =-\overrightarrow{CD}.

EDITED
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Philip-flop
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Parts a to c were beautifully explained!! Thanks again for this! I will definitely use this as a reference for when I get stuck on vector questions! I'm excited to read parts d to f now
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kiiten
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(Original post by notnek)
Too many spoilers - I'm getting fed up with them

The vector \overrightarrow{AB} is the vector that goes from A to B. In your diagram this vector starts at A and moves up to B.

The vector \overrightarrow{CD} is the vector that goes from C to D. In your diagram this vector starts at C and moves down to B.

Since these vectors are parallel but in opposite directions then \overrightarrow{AB} =-\overrightarrow{CD}.
Oh haha i just added them so people who havent seen the answer wont see it

Do you mean moves down to D?

Oh i thought the arrow determined if the vector is +ve or -ve e.g. if my arrow was pointing to C from D it would be -CD. But youre saying that because CD is going 'downwards' its -CD?
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Notnek
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I've updated the second post in this thread with the rest of the solutions. I recommend understanding the alternative method in f) since it may help with harder questions.

If you take anything away from reading my solutions it should be that drawing a clear diagram showing all the information helps a lot with these kinds of questions. Don't be lazy!

Even though I had already written these solutions, formatting them on TSR took a lot longer than I expected. I wonder if I now have the record for most LaTeX in a single post?

Again, please let me know if you don't understand something or notice a typo/mistake.
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Notnek
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(Original post by kiiten)
Oh haha i just added them so people who havent seen the answer wont see it

Do you mean moves down to D?

Oh i thought the arrow determined if the vector is +ve or -ve e.g. if my arrow was pointing to C from D it would be -CD. But youre saying that because CD is going 'downwards' its -CD?
You're right about my mistake. I meant it moves down to D.

No an arrow on a vector diagram is the vector so if an arrow on your diagram goes from A to B then this arrow represents the vector \overrightarrow{AB}

My use of the words "up" and "down" in my last post was just so you can see on your diagram what I was talking about. But a vector going up or down on a diagram doesn't tell you anything about the vector.
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