1
$\begingroup$

Point $A, B, C, D$ have position vectors $a, b, c, d$ respectively relative to an origin O.

If $P$ divides $AB$ in the ratio $1:2$ and $Q$ divides $CD$ in the ratio $1:2$, obtain an expression for the position vector of $X$, where $X$ is the midpoint of $PQ$.

If $ABCD$ is a parallelogram show that $X$ is the point in which the diagonals $AC$ and $BD$ intersect.

I did the following diagram to work out this problem.

Diagram

I figured that to get $\overrightarrow{OX}$, I need $\overrightarrow{PQ}$, $\overrightarrow{PX}$ and $\overrightarrow{AP}$, which I got as below,

$$ \begin{align} \overrightarrow{PQ} &= \overrightarrow{PA} + \overrightarrow{AD} + \overrightarrow{DQ} \\ &= -\frac{2}{3} a - \frac{1}{3} b + \frac{2}{3} c + \frac{1}{3} d \\ \\ \overrightarrow{PX} &= \frac{1}{2}\overrightarrow{PQ} \\ &= -\frac{1}{3} a - \frac{1}{6} b + \frac{1}{3} c + \frac{1}{6} d \\ \\ \overrightarrow{AP} + \overrightarrow{PX} &= \overrightarrow{AX} \\ \overrightarrow{AX} &= \frac{1}{3} \overrightarrow{AB} + \overrightarrow{PX} \\ &= -\frac{2}{3}a + \frac{1}{6}b + \frac{1}{3}c + \frac{1}{6}d \\ \\ \overrightarrow{OX} &= \overrightarrow{OA} + \overrightarrow{AX} \\ &= \frac{1}{3}a + \frac{1}{6}b + \frac{1}{3}c + \frac{1}{6}d \\ &= \frac16(2a+b+2c+d) \end{align} $$

For the second part I figure I need to show that $X$ lies on $AC$ and $BD$. ie:- $\overrightarrow{AX} = k\overrightarrow{AC} |k\overrightarrow{XC}$ and $\overrightarrow{BX} = k\overrightarrow{BC} |k\overrightarrow{XD}$.

I tried doing this but end up with vectors without all the components. So I am unable to represent the vectors as multiples of each other. What am I missing? How do you solve the last part of this question?

Thanks for all your help!

$\endgroup$

1 Answer 1

Reset to default
1
$\begingroup$

In a parallelogram, you have $b-a=c-d$, so $b-a+d-c=0$. Adding one twelfth of that to your last displayed equation yields $\overrightarrow{OX}=\frac14(a+b+c+d)$, which is the midpoint of the parallelogram and the point where the diagonals intersect.

$\endgroup$
4
  • $\begingroup$ Thanks. Sorry if my question sounds stupid! Can you please clarify how you arrived at one twelfth? The equation I got is $\overrightarrow{OX} = \frac14(2a+b+2c+d)$, As per your suggestion, if $\overrightarrow{OX}=\frac14(a+b+c+d)$, Shouldn't $\overrightarrow{OX}$ be unique? $\endgroup$
    – mathguy80
    Jul 22, 2011 at 12:46
  • $\begingroup$ I don't understand. You added the line $=\frac16(2a+b+2c+d)$ in an edit after I wrote my answer; this is correct. I don't know how you arrived at $\overrightarrow{OX} = \frac14(2a+b+2c+d)$. This is $\frac32\overrightarrow{OX}$, as you can see from $\overrightarrow{OX}=\frac16(2a+b+2c+d)$. If you add one twelfth of $0=b-a+d-c$ to $\overrightarrow{OX}=\frac16(2a+b+2c+d)$, the result is $\overrightarrow{OX}=\frac14(a+b+c+d)$, since $\frac26-\frac1{12}=\frac16+\frac1{12}=\frac14$. $\endgroup$
    – joriki
    Jul 22, 2011 at 13:05
  • $\begingroup$ Sorry about the edit. I saw your answer, and realized it is easier to compare if the common factor is taken out. $\endgroup$
    – mathguy80
    Jul 22, 2011 at 13:12
  • $\begingroup$ I see what you mean about the one twelfths now. I had interpreted it as add one twelfth + $\overrightarrow{OX}$ instead of add both the equations together which makes more sense now. $\endgroup$
    – mathguy80
    Jul 22, 2011 at 13:15

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .