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    (Original post by notnek)
    You say you are iffy but what you wrote is correct. Can you explain what you're unsure about?

    To summarise, the important thing is that you remember these:

    When \mathbf{a} and \mathbf{b} are perpendicular, \mathbf{a}\cdot \mathbf{b} = 0

    When \mathbf{a} and \mathbf{b} are parallel, \mathbf{a}\cdot \mathbf{b} = |\mathbf{a}| |\mathbf{b}|

    These results come from the scalar/dot product formula where \theta = 90 and \theta = 0

    For C4 you don't need to know how the scalar product formula is derived. You just need to be able to use the formula and use the results above that come from the formula.
    Thank you!!

    I think I'm just struggling to absorb/remember everything from these chapter on vectors in general but I've been making notes as I'm going along so hopefully they will make sense when I need to read them back
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    (Original post by Philip-flop)
    Thank you!!

    I think I'm just struggling to absorb/remember everything from these chapter on vectors in general but I've been making notes as I'm going along so hopefully they will make sense when I need to read them back
    A issue with self-teaching is that it isn't always clear from the textbook what you actually need to know for the exam.

    That's why the Examsolutions are really good because he will focus on the important parts. I really recommend them when you start learning about the vector equation of a line. Students often get confused by the topic but he explains it very well in my opinion.
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    (Original post by notnek)
    A issue with self-teaching is that it isn't always clear from the textbook what you actually need to know for the exam.

    That's why the Examsolutions are really good because he will focus on the important parts. I really recommend them when you start learning about the vector equation of a line. Students often get confused by the topic but he explains it very well in my opinion.
    Yeah I do need to visit the ExamSolutions website more tbh. I sometimes find that watching a video on ExamSolutions is much clearer/faster to learn a topic than from the textbook!

    Thanks notnek
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    So I don't think I quite understand how to start any of these parts from this question...
    Name:  C4 EXE5G Q9.png
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    I assume that for part (a) it looks a little something like this...
    Attachment 610950610952
    But besides that, I'm not really sure how to go about simplifying
    Attached Images
     
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    (Original post by Philip-flop)
    So I don't think I quite understand how to start any of these parts from this question...
    Name:  C4 EXE5G Q9.png
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    I assume that for part (a) it looks a little something like this...

    But besides that, I'm not really sure how to go about simplifying
    Use the distributive property of the dot product which states that \mathbf{a} \cdot (\mathbf{b}+\mathbf{c}) = \mathbf{a} \cdot \mathbf{b} + \mathbf{a} \cdot \mathbf{c} and if \mathbf{b} and \mathbf{c} are perpendicular then \mathbf{b} \cdot \mathbf{c} = 0.
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    (Original post by RDKGames)
    Use the distributive property of the dot product which states that \mathbf{a} \cdot (\mathbf{b}+\mathbf{c}) = \mathbf{a} \cdot \mathbf{b} + \mathbf{a} \cdot \mathbf{c} and if \mathbf{b} and \mathbf{c} are perpendicular then \mathbf{b} \cdot \mathbf{c} = 0.
    Oh yeah of course!!

    So I would do...
    Name:  Photo 12-01-2017, 13 19 02.jpg
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    right?
    Sorry I wrote my workings out in the longest way possible so I could understand it better in future.

    Thanks for the help RDKGames
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    So tonight I managed to complete the main content of the chapter on vectors. It's a topic that I know I will have to keep coming back to but its time for me to start working through the last chapter of the Edexcel C4 textbook, Integration!! Wish me luck waaaaaa.

    Thanks everyone for your help
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    So last night I attempted the Edexcel C4 June 2014 paper, overalll it wasn't too bad apart from the last part of Question 8 (part f).

    Exam paper here: http://qualifications.pearson.com/co...-June-2014.pdf

    I was able to answer the question completely but not in surd form!! So then I worked through the whole part again but with ExamSolutions. Here are my answers (sorry about the layout!!)...
    Name:  C4 June 2014 Q8.png
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    The part I'm particularly having trouble is the right side when we are trying to find h in surd form. I don't understand the bit where we have to compare the two Pythag Triangles. Why do the two triangles give a different length for the hypotenuse? I thought since the angle  \theta is the same then the hypotenuse would be the same for both triangles?

    What I'm trying to say is, why doesn't  h = 2 \sqrt2 since that's what the opposite sideis for the second triangle?? Why do we have to compare the two triangles when  \theta is the same angle?
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    (Original post by Philip-flop)
    So last night I attempted the Edexcel C4 June 2014 paper, overalll it wasn't too bad apart from the last part of Question 8 (part f).

    I was able to answer the question completely but not in surd form!! So then I worked through the whole part again but with ExamSolutions. Here are my answers (sorry about the layout!!)...
    Name:  C4 June 2014 Q8.png
Views: 23
Size:  240.7 KB
    The part I'm particularly having trouble is the right side when we are trying to find h in surd form. I don't understand the bit where we have to compare the two Pythag Triangles. Why do the two triangles give a different length for the hypotenuse? I thought since the angle  \theta is the same then the hypotenuse would be the same for both triangles?

    What I'm trying to say is, why doesn't  h = 2 \sqrt2 since that's what the opposite sideis for the second triangle?? Why do we have to compare the two triangles when  \theta is the same angle?
    I don't have time today to go through the question (someone else might) but I can possibly help if you post the key information here. You don't have to show the triangles but just tell us what lines of working you understand and the part that you don't.

    Also, please always post a link to a paper if your post is about an exam question
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    (Original post by notnek)
    I don't have time today to go through the question (someone else might) but I can possibly help if you post the key information here. You don't have to show the triangles but just tell us what lines of working you understand and the part that you don't.

    Also, please always post a link to a paper if your post is about an exam question
    Oops sorry, Ive now posted a link of the paper for others to see.

    Ah no worries. The part I don't understand is how the right angle triangle gives two different hypotenuse lengths  \sqrt27 for the first triangle and  3 for the other one. I thought they are meant to be the same right-angle triangle (as seen on the vector diagram) so will give the same length for the hypotenuse since they have the same angle  \theta
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    (Original post by Philip-flop)
    Oops sorry, Ive now posted a link of the paper for others to see.

    Ah no worries. The part I don't understand is how the right angle triangle gives two different hypotenuse lengths  \sqrt27 for the first triangle and  3 for the other one. I thought they are meant to be the same right-angle triangle (as seen on the vector diagram) so will give the same length for the hypotenuse since they have the same angle  \theta
    Looks like I'd have to try the question and then go through your working to understand this, which I can do tomorrow. Tag me tomorrow if you haven't received a reply by then.

    Or is there a section of an Examsolutions video that I can watch to see the part you don't understand?
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    (Original post by notnek)
    Looks like I'd have to try the question and then go through your working to understand this, which I can do tomorrow. Tag me tomorrow if you haven't received a reply by then.

    Or is there a section of an Examsolutions video that I can watch to see the part you don't understand?
    https://youtu.be/KaH1DcgBRCg
    Start from 03:19

    Some background info from the previous parts..

    Length of PB = \sqrt{27}

    Angle  \theta = 78.8... which was given by  cos \theta = \frac{1}{3}
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    (Original post by Philip-flop)
    So last night I attempted the Edexcel C4 June 2014 paper, overalll it wasn't too bad apart from the last part of Question 8 (part f).

    Exam paper here: http://qualifications.pearson.com/co...-June-2014.pdf

    I was able to answer the question completely but not in surd form!! So then I worked through the whole part again but with ExamSolutions. Here are my answers (sorry about the layout!!)...
    Name:  C4 June 2014 Q8.png
Views: 23
Size:  240.7 KB
    The part I'm particularly having trouble is the right side when we are trying to find h in surd form. I don't understand the bit where we have to compare the two Pythag Triangles. Why do the two triangles give a different length for the hypotenuse? I thought since the angle  \theta is the same then the hypotenuse would be the same for both triangles?

    What I'm trying to say is, why doesn't  h = 2 \sqrt2 since that's what the opposite sideis for the second triangle?? Why do we have to compare the two triangles when  \theta is the same angle?
    The second triangle that he drew is used to find \sin \theta using the fact that \cos \theta = \frac{1}{3}. This triangle is not part of the diagram.

    If you know that \cos \theta = \frac{1}{3} then a triangle that has adjacent 1 and hypotenuse 3 must have an angle \theta that satisfies \cos \theta = \frac{1}{3} so you can use this arbitrary triangle to work out what \sin \theta must be. You first work out the hypotenuse using Pythagoras (no relation to the hypotenuse in the vector diagram) and then \sin \theta must be equal to the opposite over the hypotenuse in this triangle.

    (I may still be misunderstanding your question since I still haven't actually tried it).
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    (Original post by notnek)
    The second triangle that he drew is used to find \sin \theta using the fact that \cos \theta = \frac{1}{3}. This triangle is not part of the diagram.

    If you know that \cos \theta = \frac{1}{3} then a triangle that has adjacent 1 and hypotenuse 3 must have an angle \theta that satisfies \cos \theta = \frac{1}{3} so you can use this arbitrary triangle to work out what \sin \theta must be. You first work out the hypotenuse using Pythagoras (no relation to the hypotenuse in the vector diagram) and then \sin \theta must be equal to the opposite over the hypotenuse in this triangle.

    (I may still be misunderstanding your question since I still haven't actually tried it).
    Yeah I understand the whole "SOHCAHTOA" thing and how we find the opposite side (of the second triangle) by using the fact that H^2 = O^2 + A^2 \Rightarrow O^2 = H^2 - A^2

    But I'm not sure why we use the second triangle and then compare sin \theta with the  sin \theta of the first triangle. Why can't we just use the first triangle to work all this out? Originally I did this question just using the first triangle but obviously my answer wasn't in exact (surd) form

    EDIT (27/04/2017 14:29): notnek so I may still need help understanding this
 
 
 
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