63 replies to this topic

[sparky]

Hmmm...it appears to be one of those horrible vector ones...

I know you go a.(b+c)
= a.b + a.c
but after that I am lost.

Edited by sparky, 30 April 2004 - 08:10 PM.

[Steve]

Here goes,

a is the line from P to Q and b+c is the line joining Q and S.

The angle at R is 120° (since the shape is a regular hexagon - look here for more details).

Looking at triangle QRS, the angles at Q and S are both 30° (angles in a triangle add up to 180).

So the angle between PQ and QS is 90° (120-30).

You can then use the geometric form of the scalar product formula (given). You don't need to know the magnitude of b+c since the angle between the vectors is 90° and cos90°=0.

Therefore a.(b+c)=0

[Ally]

Thanks for that!

I'll print off your working & get my past paper book, because right now I'm so confused.

Are we meant to know about all those shapes in the link?
It's got me worried now

Thanks again for the solution.

[sparky]

Three vectors, p, q and r are such that p.(q+r) = q.(p+r).

Show that r must be perpendicular to p-q
3 marks

Any help would be appreciated.

[George]

OK, expanding p.(q+r) = q.(p+r) gives

p.q + p.r = q.p + q.r

And since p.q = q.p they can be cancelled out giving

p.r = q.r
p.r - q.r = 0
r.(p - q) = 0

So r is perpendicular to p - q as required.

[sparky]

Thanks, I did expand, but never thought about cancelling!!

[hssc11045]

Hey......am new to these forums and was jus wonderin if any1 can provide some help with questions:

2c) ii)
6)
11b)

of the 2000 calculator paper.

Thanks

P.S a knw the working has already been posted 4 question 6 but a cnt open tha link

[Steve]

I've fixed the link to the working for Question 6 from 2000 - here is a link to it.

[Ally]

[Dave]

i tried doin it the wrong way and it didn't work out i got (-13,9) which is wel lout i must have made a hell of a mistake somewhere.

Question 11)

You should ave worked the equation to be P = 0.6Q + 1.8

if you sub the logs into this you get:

log base e(p) = 0.6 log base e (q) + 1.8 log base e (e)

one of the log rules states log xy = log x + log y so:

ln (p) = ln (e^1.8) ln (q^0.6)

if you calculate this out you get

ln(p) = 6.05 ln q^0.6)

remove the ln and you get

P = 6.05 q ^ 0.6

a = 6.05
b = 0.6

[hssc11045]

thanks guys.....all makes sense nw

[Ally]

[Dave]

Not likily i can't find where i did the working but it doesn't look familer

[sparky]

Find algebraically the exact value of
sinø + sin(ø +120) +cos(ø+150)

I know you multiply these out using the double angle rules, and then substitute in the values of the related angles, but I still dont get this is cancel down to O which is the correct answer!

Any help appreciated

[$impl¥_®i¢h]

Is this question from one of the pastpapers?

[AndyW]

Using the shorthand c = cos theta and s = sin theta, it expands to:

s + s cos 120 + c sin 120 + c cos 150 - s sin 150

= s + s ( - 0.5 ) + c ( root3over2) + c ( - root3over2) - s ( 0.5)

= s + s ( - 0.5 ) - s ( 0.5)

= s - 0.5 s - 0.5 s

= 0

Quite tricky that one.

[hssc11045]

alrite everyone its me again with a couple more questions:

5a from tha 2001 non cal..........i narrowed it down 2 tha expanision or tha wave function but i canot seem to get any to work

7b) ii) get all tha previous questions in 7 correct but cannot seem to do this one

8) the answer I get is 80 whereas tha answer at tha bk says 81


FINALLY

question 11c)

Much appreciated

Thanks

[Allan]

[Allan]

[Allan]