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.**

Posted 01 May 2004 - 04:04 PM

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 120Â° (since the shape is a regular hexagon - look here for more details).

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

So the angle between PQ and QS is 90Â° (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 90Â° and cos90Â°=0.

Therefore a.(b+c)=0

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

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

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

So the angle between PQ and QS is 90Â° (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 90Â° and cos90Â°=0.

Therefore a.(b+c)=0

Posted 04 May 2004 - 08:42 PM

QUOTE (hssc11045 @ May 4 2004, 05:09 PM) |

Hey......am new to these forums and was jus wonderin if any1 can provide some help with questions: 2c) ii) |

Well for:

2 © (iii) - I was stuck on this question aswell but i thought it through and I think this is the answer :

I think that you put y=9 (answer to previous question) into the equation

Also for this question, I think that you can do it the long way by finding the tangent at P and equating this to y=9 which also gives (-9,9), but this is probably not the right way to do it as its only worth 2 marks. Maybe someone could clear this up?

Posted 04 May 2004 - 08:58 PM

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

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

Posted 04 May 2004 - 09:06 PM

QUOTE (Dave @ May 4 2004, 08:58 PM) |

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. |

Did you get the tangent at P to be

Posted 05 May 2004 - 07:58 PM

Find algebraically the **exact** value of

sinÃ¸ + sin(Ã¸ +120) +cos(Ã¸+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

sinÃ¸ + sin(Ã¸ +120) +cos(Ã¸+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

Posted 06 May 2004 - 05:37 PM

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

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

Posted 06 May 2004 - 06:02 PM

QUOTE (hssc11045 @ May 6 2004, 06:37 PM) |

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 |

You can't use the wave function here as you have an "x" and a "2x" value.

So you need to use the double angle forumla as given in the formula sheet.

Subsituting you get:

2sinxcosx - cosx = 0

Taking out the common factor cos x:

cos x(2sinx - 1)=0

Set both parts to 0:

cos x = 0 2sinx - 1 = 0

Then you can solve it from there like a normal trig equation to get x = 30,90,150

Posted 06 May 2004 - 06:07 PM

QUOTE (hssc11045 @ May 6 2004, 06:37 PM) |

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

g(h(x)) = cosxcos(pi/4) - sinxsin(pi/4)

= 1/√2 cosx - 1/√2 sinx

Substitue these into the formula given:

(1/√2sinx + 1/√2 cosx) - (1/√2 cosx - 1/√2sinx)

Simplifying this gives:

2/√2 sin x = 1

sin x = √2/2

inverse sin (√2/2) = (pi/4)

so x = (pi/4) and (3pi/4) [1st and 2nd quadrants]

Posted 06 May 2004 - 06:12 PM

QUOTE (hssc11045 @ May 6 2004, 06:37 PM) |

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

4 log [x] 6 - 2log [x]4 = 1

Using log rules:

log [x] 6[power 4] - log [x] 4Â²

log [x] 1296 - log [x] 16 = 1

Using more log rules:

log [x] 1296/16 = 1

log [x] 81 = 1

More log rules

81 = x[power 1]

x = 81

[NB The log rules should be in your textbook, you've just got to apply them here]