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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: /0 on April 29, 2008, 07:56:16 pm

Title: Vector Geometry
Post by: /0 on April 29, 2008, 07:56:16 pm: The position vectors of P, Q with reference to an origin O are $\bold{p}$ and $\bold{q}$ and M is the point on PQ such that $\beta \ \overrightarrow{PM} = \alpha\ \overrightarrow{MQ}$ .

a) Prove that the position vector of M is $\bold{m}$ , where

$\bold{m} = \frac{\beta \bold{p}+\alpha \bold{q}}{\alpha+\beta}$ SOLVED!

The vector $\bold{p}=k\bold{a}$ and the vector $\bold{q}=l\bold{b}$ where k and l are positive real numbers and $\bold{a}$ and $\bold{b}$ are unit vectors.

b) Prove that the position vector of any point on the internal bisector of $\angle POQ$ has the form $\lambda (\bold{a}+\bold{b})$ .

c) If M is the point where the internal bisector of $\angle POQ$ meets PQ, show that:

$\frac{\alpha}{\beta}=\frac{k}{l}$ .

Please help on questions b) and c)! Thanks
Title: Re: Vector Geometry
Post by: Mao on April 29, 2008, 09:21:31 pm: b)
i dont know if this will be adequate enough, but here it is:

if M is a point on the internal bisector (position vector m) in the form $\lambda (a+b)$ , then

$\frac{m\cdot p}{|p|} = |m|\cdot \cos(\theta) = \frac{m\cdot q}{|q|}$

$\frac{k\cdot (a\cdot m)}{k} = \frac{l\cdot ( b\cdot m)}{l}$

$a\cdot \lambda (a+b) = b \cdot \lambda (a+b)$

$\lambda \cdot (a\cdot (a+b)) = \lambda \cdot (b\cdot (a+b))$

$|a|^2 + a\cdot b = a\cdot b + |b|^2$

$1 + a \cdot b = a\cdot b + 1$

QED?

well, i showed it, i didnt exactly "prove" it.... =S
Title: Re: Vector Geometry
Post by: /0 on April 30, 2008, 12:43:06 am: Thanks Mao, we're not supposed to have done dot products yet tho ;p
Title: Re: Vector Geometry
Post by: /0 on May 02, 2008, 05:50:01 pm: My teacher demonstrated the solution in class:

$\overrightarrow{OM} = \overrightarrow{OP}+\overrightarrow{PM}$

a)
And from $\beta \overrightarrow{PM} = \alpha \overrightarrow{MQ}$ ,

$\overrightarrow{PM} = \frac{\alpha}{\beta} \overrightarrow{MQ}=\frac{\alpha}{\alpha+\beta} \overrightarrow{PQ}=\frac{\alpha}{\alpha+\beta}(\bold{q}-\bold{p})$

$\Rightarrow \overrightarrow{OM} = \bold{p} + \frac{\alpha}{\alpha+\beta}(\bold{q}-\bold{p})$

$\Rightarrow \overrightarrow{OM} = \frac{\alpha+\beta}{\alpha+\beta}\bold{p} + \frac{\alpha}{\alpha+\beta}\bold{q}-\frac{\alpha}{\alpha+\beta}\bold{p}$

$\Rightarrow \overrightarrow{OM}=\frac{\beta \bold{p}+\alpha \bold{q}}{\alpha+\beta}$

b)

Draw the vectors $\bold{\hat{a}}$ and $\bold{\hat{b}}$ . The internal angle bisector does not depend on the magnitude of the rays which bound the angle, but only their directions. Draw another two vectors such that a rhombus is formed with side lengths $\bold{\hat{a}}$ and $\bold{\hat{b}}$ . A property that the rhombus has is that its diagonals bisect its angles. The direction of the diagonal of the drawn rhombus is $\bold{\hat{a}}+\bold{\hat{b}}$ , so the position vector of any point on it must be of the form $\lambda (\bold{\hat{a}}+\bold{\hat{b}})$ .

c)

We have two expressions for $\overrightarrow{OM}$

1. $\overrightarrow{OM} = \frac{\beta \bold{p}+\alpha \bold{q}}{\alpha+\beta}$

2. $\overrightarrow{OM} = \lambda (\bold{\hat{a}}+\bold{\hat{b}})$

Equating, we get

$\frac{\beta \bold{p}+\alpha \bold{q}}{\alpha+\beta} = \lambda(\bold{\hat{a}}+\bold{\hat{b}})$

$\frac{\beta (k\bold{\hat{a}})}{\alpha+\beta}+\frac{\alpha (l \bold{\hat{b}})}{\alpha+\beta}=\lambda\bold{\hat{a}}+\lambda\bold{\hat{b}}$

Equating unit vector coefficients:

$\frac{\beta k}{\alpha+\beta} = \lambda$

$\frac{\alpha l}{\alpha+\beta} = \lambda$

$\Rightarrow \frac{\alpha l}{\alpha+\beta} = \frac{\beta k}{\alpha+\beta}$

$\alpha l = \beta k$

$\therefore \frac{k}{l} = \frac{\alpha}{\beta}$
Title: Re: Vector Geometry
Post by: Mao on May 02, 2008, 09:11:36 pm: *in class*

what class is that =S

ur school must be supersaiyan....
Title: Re: Vector Geometry
Post by: /0 on May 02, 2008, 10:43:13 pm: Lol... it is a unusually hard problem from Essential Advanced General Maths, that's why our teacher showed it, and he had to think for a while too :p