# Thread: Vectors in 2D space

1. ## Vectors in 2D space

question: ABCD is a quadrilateral. P,Q,R and S divide AB, AD, CD and CB respectively in the ration 2:1.
(a) Find the position vectors p, q, r, s in therms of a, b, c, d.
(b) Prove that PQRS is a parallelogram.

(a) is easy:
p = (a+2b)/3
q = (a+2d)/3
r = (c+2d)/3
s = (c+2b)/3
But how to do (b) ?

2. Originally Posted by afeasfaerw23231233
question: ABCD is a quadrilateral. P,Q,R and S divide AB, AD, CD and CB respectively in the ration 2:1.
(a) Find the position vectors p, q, r, s in therms of a, b, c, d.
(b) Prove that PQRS is a parallelogram.

(a) is easy:
p = (a+2b)/3
q = (a+2d)/3
r = (c+2d)/3
s = (c+2b)/3
But how to do (b) ?
Hello,

you have to prove that the sides $\displaystyle PQ \parallel RS~\wedge~|PQ| = |RS|$

3. $\displaystyle PQRS$ paralelogram $\displaystyle \Leftrightarrow p+r=q+s$
$\displaystyle \displaystyle p+r=\frac{a+2b+c+2d}{3}$
$\displaystyle \displaystyle q+s=\frac{a+2b+c+2d}{3}$

4. so far i have done:
$\displaystyle \overrightarrow{P Q} \ = (2a+2b+2d)/3$
$\displaystyle \overrightarrow{R S} \ = (2c+2d+2b)/3$

it seems that i have to prove a = c in order to get

$\displaystyle \overrightarrow{P Q} \ = \overrightarrow{S R} \$

but i have no idea how to prove a=c

5. You are wrong.
$\displaystyle \overrightarrow{PQ}=(q-p)=\left(\frac{2(d-b)}{3}\right)$
$\displaystyle \overrightarrow{RS}=(s-r)=\left(\frac{2(b-d)}{3}\right)$.
Then $\displaystyle \overrightarrow{PQ}\parallel \overrightarrow{RS}$ and $\displaystyle |\overrightarrow{PQ}|=|\overrightarrow{RS}|$

### vectors abcd parallelogram. show that 2DC -DB=AC

Click on a term to search for related topics.