Geometrical proof using vectors

[nobahar]

Hello again, another vector question.

The diagram (not included) shows a parallelogram OABC with OA=a and OC=c; AC and OB intersect at point D such that OD=hOB and AD=kAC.
How do I show what h and k equal?
I managed to achieve:
a+c=2DB+AD-DC=2OD+DC-AD=OB
and
c-a=2DC+OD-DB=2AD+DB-OD=AC
Furthermore, I know that AD+DB=OD+DC and OD-AD=DB-DC
I can 'see' that h and k = 1/2; but how do I show it?
As always, the help is appreciated. I love you people :).

 

[tiny-tim]

Hi nobahar!

Hint: what is the vector equation for a general point on AC?

 

[nobahar]

The equation of the line AC could be: r=zAD+c; r=zAD+a; r=zAD+xOB? (xOB being OD.)
I'm guessing any of these are suitable since it satisfies the need for a parallel vector and a position vector?
Then, any point that meets the line AC should =r, with different values for z. Where do I go from here? (If anywhere!)

 

[tiny-tim]

Hi nobahar!

The equation of the line AC could be: r=zAD+c; r=zAD+a

hmm … sort-of the right idea …

but AD is unknown …

look for something similar that only involves a and c (which are given) and numbers.

 

[nobahar]

Hello again!
Right, my maths tutor offered me this solution by e-mail:
"OD = hOB = h(a+c) = ha +hc
OD = OA + AD = a + kAC = a + k(c - a)

Hence ha + hc= a + kc -ka
Equating coefficients of a: h = 1 - k
Equating coefficients of c: h = k
Hence k=1-k
2k = 1
k=0.5 = h"

From this (and other comments from her e-mail) I established the 'relationships' I first offered served no purpose; and (thanks to Tiny Tims input) neither did the vector equations of a line I proposed.
However, I still have some further questions; since I like to understand as much as possible and as clearly as possible before moving on.
I'm guessing the above isn't a vector equation of a line at any point, since it doesn't contain a parallel vector?
On that basis, presumably there is going to be an alternative method?
As always, thanks in advance.

 

[tiny-tim]

OD = hOB = h(a+c) = ha +hc
OD = OA + AD = a + kAC = a + k(c - a)
…
I'm guessing the above isn't a vector equation of a line at any point, since it doesn't contain a parallel vector?
On that basis, presumably there is going to be an alternative method?

Hello nobahar!

The vector equation for the line AC (ie for a general point P on AC) is OP = OA + kAC for any number k,

(of course, it's the same as OP = OC + k'AC, with k' = k + 1)

because you go to A first, and then as far as you like along the direction of AC.

(This is the equation your teacher used, with D = P. )

But I don't understand what you mean by "it doesn't contain a parallel vector".

 

[nobahar]

For the parallel vector I was just going by the textbook, which proposes that: "...a line is defined if we know:
i) a vector that is parallel to the line,
and ii) the position vector of a point on the line.
(Sadler, A.J., Thorning, D.W.S (2007, pg.65). Understanding Pure Mathematics, Glasgow: Oxford Universty Press).
I now realize that a parallel vector is unnecessary if the direction of the vector in question is already identified (?).
One final question
why k'=k+1 in OP=OC+k'AC?
OP=OC-AC+kAC or OP=OC-(AC-kAC)
OP=c-(c-a)+k(c-a)
OP=c-c+a+kc-ka
OP=a+kc-ka= OA+kAC

 

[tiny-tim]

"...a line is defined if we know:
i) a vector that is parallel to the line,
and ii) the position vector of a point on the line.
…
I now realize that a parallel vector is unecessary if the direction of the vector in question is already identified (?).

don't forget … c - a isn't the vector AC, it's a vector from O parallel to AC.

so c - a was "a vector that is parallel to the line"

One final question
why k'=k+1 in OP=OC+k'AC?
OP=OC-AC+kAC or OP=OC-(AC-kAC)
OP=c-(c-a)+k(c-a)
OP=c-c+a+kc-ka
OP=a+kc-ka= OA+kAC

hmm … perhaps that should have been k' = k - 1 :

OP = c - (c-a) + k(c-a)
OP = c + (k - 1)(c-a)

 

[nobahar]

That is excellent!
Thankyou for your help; as always Mr. Tiny Tim, it's much appreciated.