Maths C4 - Vectors... Help???

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

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.

So I don't think I quite understand how to start any of these parts from this question...


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 $\mathbf{b} \cdot \mathbf{c} = 0$.

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!!)...

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 side is 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

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?

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/content/dam/pdf/A%20Level/Mathematics/2013/Exam%20materials/Question-paper-Unit-C4-(6666)-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!!)...

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 side is 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).