FP3 (Not MEI) - Friday June 1 2012, AM

For the June 11 paper, question 5 part ii: why must k = -1?

EDIT: Nevermind, I was being an idiot

For this question part (ii)



How is that first step justified; surely they're assuming the result there?

For 3. ii) ?
They have removed the left x and the right y, to end up with $yxyxyxy...$ in the middle, or $(yx)^{n-1}$.

Ohhhhh, I get it! Thank you!

Don't think i could think of that in an exam tbh

EDIT: Unfortunately, I've repped you too recently

12a)

$\frac{\begin{pmatrix} 30 \\ -3 \\ - 5 \end{pmatrix} \times \begin{pmatrix} 4 \\ -5 \\ - 3 \end{pmatrix}}{|\begin{pmatrix} 4 \\ -5 \\ - 3 \end{pmatrix}|}$

Length = 22

12b)

Length of AB = 22
Length of OA = $\sqrt{934}$
length of OB = $\sqrt{934 - 22^2}$

Length of OB is a multiple of the direction vector
$\sqrt{450} = t\sqrt{50}[br]t=3[br]B = \begin{pmatrix} 12 \\ -15 \\ - 9 \end{pmatrix}$

That's alright.
Did you see my response to the question you asked about the roots on the tan equation?

...who on earth is confident going out of an exam?

Jan 12 Q6iii. I've got the answer but I'm not quite sure why it makes sense. Could someone explain this to me?

You know a plane can be expressed parametrically with two direction vectors that lie on it.
(These vectors cannot be parallel - cannot be the same.)

In part i) you were asked to indirectly find the normal to the plane.
Now recall that the vector product of two direction vectors will give a direction vector that is perpendicular to both of them.

So the vector product of the normal to the plane and the direction vector after lambda should give c.

For a vector equation of a plane the two direction vectors in the plane may or may not be perpendicular, but the normal is obviously perpendicular to both of them in all forms of the vector equation, since these vectors lie in the plane.

I think there was another method to find c in the MS but I can't remember - I used the one above when doing the paper since it seemed quicker.