Describe vector inside a triangle as other vectors

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Homework Statement

In triangle ABC, the point P bisects the side BC, the point Q is the point on side AB for which |AQ| = 2|QB|, and the point R is the point of intersection between AP and CQ. Designate vectors a=AB and b=AC. Describe vector AR in terms of vectors a and b.

Homework Equations

The Attempt at a Solution

I drew a triangle representation for the problem:

http://img338.imageshack.us/img338/2571/triangulus.jpg

I can't find a way to express either of vectors AR, PR, QR or CR in terms of AB, AC, BC, CQ or AP no matter how much I've tried vector addition and subtraction alone. I realized that triangles ABP and APC are equal in area, and triangle QBC has half the area of triangle AQC. Consequently, I tried to use the fact that the cross product of vectors CQ and CB is equal to the area of triangle AQC, but I've consistently failed to find a good relationship between AR and other vectors. Attempts to describe the areas of the smaller triangles, or the quadrilateral QBPR in terms of other parts haven't gotten me anywhere either.

Any hints would be appreciated.

 

[lanedance]

so once you find AQ & QR or just RC in terms of a & b you're pretty much there.

Note that as a & b are not parallel, any 2D vector can be written in ternms of them. So using brute force, you could just go teough and start finding all the segments in terms of a & b,

eg. strat by writing AQ, QB in terms of a (think fractions), CB in terms of a & b, then CP & PB, should fall rather easily & so on...

 

[Entertainment]

so once you find AQ & QR or just RC in terms of a & b you're pretty much there.

That's the thing. I'm unable to find any of the interior vectors QR, AR, PR or CR in terms of any of the other vectors. If I could do that, the problem would be trivial.

Note that as a & b are not parallel, any 2D vector can be written in ternms of them. So using brute force, you could just go teough and start finding all the segments in terms of a & b, eg. strat by writing AQ, QB in terms of a (think fractions), CB in terms of a & b, then CP & PB, should fall rather easily & so on...

I don't think that brute force is applicable, since no relation is given between sides AB and BC (not that I haven't tried). Obviously, |QB| = |AQ|/2 = |AB|/3, and |BP| = |PC| = |BC|/2, but I don't see how this helps me per se.

If you try some addition of vectors, say, AR = AQ + QR = AC + CR = AP + RP, you're just adding another unknown for every relation, so using an equation system to solve this is of no avail. The same seems to hold true if you attempt to directly express CR, PR and QR in terms of vector addition. I couldn't find a way with vector subtraction either. This makes me think that there is some other geometric relation that should be utilized.

 

[lanedance]

ok so i get what you mean... you can find all the full vectros, but the split to get R itself, try this:

write RP as an unknown fraction of AP
you may have to do similar for AR, QR, QC

so now you have 4 unknown constants, but the 2 parts must sum to one along AP & QC so there are actually only really 2 unknowns...

if you write down a the vector sum of a few different triangles, you should hopefully be able to solve for the unkowns

 

[lanedance]

i haven't tried it, but i would start with 2 triangles out of ARQ, ARC, ARP - in each triangle the only unknowns will be the fraction

 

[Entertainment]

i haven't tried it, but i would start with 2 triangles out of ARQ, ARC, ARP - in each triangle the only unknowns will be the fraction

I'm not sure if I understood you entirely, but I assigned

\overrightarrow{AR} = n \cdot \overrightarrow{AP}
\overrightarrow{RP} = (1-n) \cdot \overrightarrow{AP}
\overrightarrow{RC} = m \cdot \overrightarrow{QC}
\overrightarrow{QR} = (1-m) \cdot \overrightarrow{AP}.

So, we've got \overrightarrow{AR} + \overrightarrow{RC} = n \cdot \overrightarrow{AP} + m \cdot \overrightarrow{QC} = \overrightarrow{AC}. (1)

...crap. I created another four vector equations based on the aforementioned assignments to describe the smaller triangles and the quadrilateral QBPR. When compared to (1), they have all turned out to be dead-ends since the variables n and m just cancel each other out.

 

[lanedance]

ok is see what you mean, you always end up with a similar m, n combination...

but have you tried subtituting in for the vectors in terms of a&b? i mean that's the whole idea at the end of the day... i think you might only need one triangle to do it...

(i've tried to preserve your notation & hopefully I didn't stuff up anything up)
triangle ARC
\vec{AR} = n\vec{AP} = \vec{AC} + \vec{CR}= \vec{AC} - \vec{RC}= \vec{AC} - m\vec{QC}
now try substituting in for a & b

the thing to notice when you do it is that a & b are linearly independent, so you need the co-efficients of each to cancel in the expression (this is similar to decomposing the equation into x & y components, which will give 2 equations to solve for n & m),

if they happen to cancel, add another triangle, (notice you could also look at QRP)

 

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No, that looks good. Thanks for your help, by the way. :)

Okay, so, since \mathbf{a}=\vec{AB} and \mathbf{b}=\vec{AC}, we get...

\vec{QC} = \vec{AC} - \vec{AQ} = \mathbf{b} - (2/3) \mathbf{a}

...and...

\vec{AR} = ... = \vec{AC} - m \vec{QC} = \mathbf{b} - m(\mathbf{b} - (2/3) \mathbf{a}) = (1-m) \mathbf{b} + (2/3)m \mathbf{a}Hmm, I tried to use QRP, but by using it, I either end up with two unknowns n and m, or m still just cancels out. Bah. It's the triangle combinations that seem to mess up my attempts to solve this. Then again, I guess \vec{AR} is indeed known in terms of \mathbf{a} and \mathbf{b} now, which was kind of what the problem asked for, only that m is an unknown variable. If only it was a parameter. Blah.

 

[lanedance]

no worries, i think we're getting close now ;)

No, that looks good. Thanks for your help, by the way. :)

Okay, so, since \mathbf{a}=\vec{AB} and \mathbf{b}=\vec{AC}, we get...

\vec{QC} = \vec{AC} - \vec{AQ} = \mathbf{b} - (2/3) \mathbf{a}

...and...

\vec{AR} = ... = \vec{AC} - m \vec{QC} = \mathbf{b} - m(\mathbf{b} - (2/3) \mathbf{a}) = (1-m) \mathbf{b} + (2/3)m \mathbf{a}
.

ok so the only thing we need to add is AR in terms of nAP
\vec{AR} = n\vec{AP} = \vec{AC} - m\vec{QC}

re-arranging a little
0 = \vec{AC} - m\vec{QC} - n\vec{AP}

subtituting in for a & b looks pretty similar to what you got
0 = (\vec{b}) - m(\vec{b}-(2/3)\vec{a}) -n(\vec{a}+\vec{b})/2

re-arrange as co-efficients of b & a
0 = \vec{b}(1-m -n/2) + \vec{a}(-(2/3)m -n/2)
now as a & b are linearly independent you, both theier co-efficients must be zero...

Hmm, I tried to use QRP, but by using it, I either end up with two unknowns n and m, or m still just cancels out. Bah. It's the triangle combinations that seem to mess up my attempts to solve this. Then again, I guess \vec{AR} is indeed known in terms of \mathbf{a} and \mathbf{b} now, which was kind of what the problem asked for, only that m is an unknown variable. If only it was a parameter. Blah.

doubt we'll need the extra traingle, as basically we know one side of a triangle & the other 2 directions, so it should be sufficient to triangulate for the distances, (m&n) and find R...

 

[Entertainment]

re-arrange as co-efficients of b & a
0 = \vec{b}(1-m -n/2) + \vec{a}(-(2/3)m -n/2)
now as a & b are linearly independent you, both theier co-efficients must be zero...

Awesome! That works out! Thanks again for your assistance! :D

By the way, I think it should be a plus sign rather than a minus sign in front of (2/3)m, but that's trivial.

Okay, so finally:

\left\{ \begin{array}{ccc} 1-m- \frac{n}{2} & = & 0 \\ \frac{2}{3} m - \frac{n}{2} & = & 0 \end{array} \right. \Rightarrow \left\{ \begin{array}{ccc} m & = & \frac{3}{5} \\ n & = & \frac{4}{5} \end{array} \right.

\vec{AR} = n \cdot \vec{AP} = n \cdot (\frac{ \vec{a} + \vec{b} }{2}) = \frac{2}{5} \vec{a} + \frac{2}{5} \vec{b}