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How to do a C4 Vectors exam question

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Reply 80
Original post by Philip-flop
Ok so these types of vector questions seem to confuse me a little bit (see question below).

Let me get this straight...
So we call P(x, y, z)

We are told that "P lies on l" therefore we can say that OP=l \vec{OP} = l therefore (xyz)=(102λ2+λ3+λ)\begin{pmatrix}x \\ y \\ z\end{pmatrix} = \begin{pmatrix}10-2\lambda \\ 2+\lambda \\ 3+\lambda\end{pmatrix}

which is another way of saying that the point P lies somewhere on the line l.

But then how do we find the scalar parameter λ \lambda ?

It depends on the question.


Why do we use the Dot Product? I only know to use it because it mentions that CP \vec{CP} and AB \vec{AB} are perpendicular in the question.

We use it because the vectors are perpendicular as you say.

The solution has some unnecessary working in my opinion which may be confusing you. Here's the way I would go about it:

The vector CP\overrightarrow{CP} is perpendicular to ll which means that CP\overrightarrow{CP} is perpendicular to the direction vector of ll which is (211)\begin{pmatrix}-2\\1\\1\end{pmatrix}.

So we need to find CP\overrightarrow{CP} which is equal to OPOC\overrightarrow{OP}-\overrightarrow{OC}.

Since P lies on ll we can say that OP\overrightarrow{OP} is of the form

(102λ2+λ3+λ)\begin{pmatrix}10-2\lambda\\2+\lambda\\3+\lambda \end{pmatrix}.

Then if you do this minus OC\overrightarrow{OC} you get

CP=(72λλ10λ)\overrightarrow{CP} = \begin{pmatrix}7-2\lambda\\\lambda-10\\ \lambda \end{pmatrix}

Then you can dot this vector with (211)\begin{pmatrix}-2\\1\\1\end{pmatrix} and solve the find λ\lambda.
Original post by Philip-flop
...


I find their use of expressing P in terms of x,y,z pointless. All you need are direction vectors CP and L's direction. L's direction is straightforward. CP's direction is in terms of lambda. Dot the two and you have a linear eq. in lambda.

As you said, P has coordinates (102λ2+λ3λ)\begin{pmatrix} 10-2\lambda \\ 2+\lambda \\ 3\lambda \end{pmatrix} for some λR\lambda \in \mathbb{R}

So CP=OPOC=(72λ10+λ3λ)\overrightarrow{CP}= \overrightarrow{OP} -\overrightarrow{OC}= \begin{pmatrix} 7-2\lambda \\ -10+\lambda \\ 3\lambda \end{pmatrix}

So (72λ10+λ3λ)(211)=0\begin{pmatrix} 7-2\lambda \\ -10+\lambda \\ 3\lambda \end{pmatrix} \cdot \begin{pmatrix} -2 \\ 1 \\ 1 \end{pmatrix}=0 which gets you your λ=4\lambda = 4
(edited 6 years ago)
This thread has been useful.

Thanks guys.
Reply 83
Original post by Philip-flop
Ok so these types of vector questions seem to confuse me a little bit (see question below).


You're probably overcomplicating this. We have a point P for which we have some information about it (what line it lies on, some pernepdicular to it, blah blah we'll get back to this later) and we want to find the coordinates of the point. We automatically turn this into a concrete problem by letting P have coordinate (x,y,z) and then ask the equivalent question "what is x,y,z".

Right away, we're given that P lies on the given line, so we can immediately tell that (x,y,z)=(102λ,2+λ,3+λ)(x,y,z) = (10-2\lambda, 2+\lambda, 3+\lambda) and now we've used a bit of the given information in the problem to reduce the problem of "what is x,y,z" to the equivalent question "what is λ\lambda"? Now if we can find lambda, we immediately know what x,y,z is and we then immediately know what the position vector of P is.

Now we have one bit of information left, it is obvious that this piece of information is going to be instrumental in telling us what λ\lambda must be. In particular, the line CPCP is perpendicular to \ell. Right away, before even thinking about the problem, you should immediately write down that CPCP is in terms of λ\lambda (this is obvious, we want to convert all information in the problem to information about λ\lambda so we can determine it.

We then get

Unparseable latex formula:

\displaystyle [br]\begin{align*}CP = OP - OC &= (x,y,z) - (3,12,3) \\&= (10-2\lambda, 2+\lambda, 3+\lambda) - (3,12,3) \\&= (7-2\lambda, -10+ \lambda, \lambda) .\end{align*}



Now that that bit of automatic rewriting is out of the way, we can look at the given information is again; it says that CP is perpendicular to the line \ell. Right away, the word perpendicular should immediately scream dot product to you, given that in your course there is virtually no other way to determine perpendicularity apart from checking dot products. What does it mean to be perpendicular to \ell. It means that the dot product of CP and the direction vector of \ell is 0.

So we immediately write down (72λ,10+λ,λ)(2,1,1)=0(7-2\lambda, -10 + \lambda, \lambda) \cdot (-2,1,1) = 0 (I trust you know why this is the direction vector of \ell) and then expand that and solve for λ\lambda.

The solution you've linked is essentially the same thing, but done a bit more confusingly - they left CP in terms of (x,y,z), took the dot product and then used the information that (x,y,z) was intrinsically linked to λ\lambda, this means there's less writing but is not the true 'flow of thought' so to speak when going through the problem in the sense that we want to continually make our problem easier, so turning it from finding three different number (x,y,z) to being equivalent to finding a single number λ\lambda by using the fact that the point lies on the line, then turning the rest of the given information into information about λ\lambda because that's what we want to find, so making it natural to write down CP into something involving λ\lambda.

Any questions? Can you see why the above is a perfectly natural flow of thought? If not, please point out what's unclear.
(edited 6 years ago)
Thanks @notnek @RDKGames and @Zacken I read all of your comments 2 or 3 times to make sure I understand. It's amazing how you all chose to stay clear of appointing OP=(xyz) \vec{OP} = \begin{pmatrix}x \\ y \\ z\end{pmatrix} instead left it in the vector equation of any point on the line l.

You have all explained that it is much easier to see it as OP=(102λ2+λ3+λ) \vec{OP} = \begin{pmatrix} 10-2 \lambda\\ 2+\lambda \\ 3+\lambda \end{pmatrix} and then finding CP \vec{CP} by...

CP=OPOC=(102λ2+λ3+λ)(3123)=(72λ10+λλ) \vec{CP} = \vec{OP} - \vec{OC} = \begin{pmatrix} 10-2 \lambda\\ 2+\lambda \\ 3+\lambda \end{pmatrix} - \begin{pmatrix} 3\\ 12 \\3 \end{pmatrix} = \begin{pmatrix} 7 - 2 \lambda\\ -10 +\lambda \\ \lambda \end{pmatrix}

and then using the dot product on CPAB=0 \vec{CP} \cdot \vec{AB} = 0 to find that λ=4 \lambda = 4 then sub that into l l

Seriously can't thank you guys enough!! :smile: :smile: :smile:

Sorry I took so long to reply though.
What a beauty.
Original post by Notnek
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!

you bl oody Godsend :smile: tysm
Reply 87
Just bumping this since it's the last day of revision :smile:

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