# Thread: Vector's (proving 1 line is parallel to another)

1. ## Vector's (proving 1 line is parallel to another)

I have this question that I am really stuck with, I have done the first part and believe I am close to solving the second part but really cannot see what to do next.

Q) ABCD is a trapezium with BC parallel to AD.
M is the midpoint of BC and N is the midpoint of AD
a) Find the vector MN in terms of a and b - which i found to be 3a-3b

b) Prove that XY is parallel to AD

if anybody could help me with this part it would be much appreciated

Thanks

2. What is XY here?

3. sorry missed out 1 line of the question here it is again:

I have this question that I am really stuck with, I have done the first part and believe I am close to solving the second part but really cannot see what to do next.

Q) ABCD is a trapezium with BC parallel to AD.
M is the midpoint of BC and N is the midpoint of AD
a) Find the vector MN in terms of a and b - which i found to be 3a-3b

X is the midpoint of MN and Y is the midpoint of CD.
b) Prove that XY is parallel to AD

if anybody could help me with this part it would be much appreciated

Thanks

4. $\\* \vec{MN} = 3a - 3b \\* \vec{CD} = (-3a - 3b + 9a) = 6a - 3b$

X and Y are the mid-points of MN and CD, respectively.
$\\* X = \frac{1}{2}\vec{MN} \\* Y = \frac{1}{2}\vec{CD}$

Subtract the two midpoint vectors to get the resulting vector.
$\\* \vec{XY} = Y - X \\* = \frac{1}{2}(6a - 3b) - \frac{1}{2}(3a - 3b) \\* = \frac{1}{2}(3a)$

$\\* \vec{AD} = 9a$

Both vectors consist of only 'a' vectors, so they are parallel.

5. Hello, wahhdoe!

I'll solve it in my usual baby-talk way . . .

$ABCD\text{ is a trapezium with }BC \parallel AD.$
$\overrightarrow{AB} = 3\vec b,\;\; \overrightarrow{BC} = 3\vec a,\;\; \overrightarrow{AD} = 9\vec a$
$M\text{ is the midpoint of }BC\text{ and }N\text{ is the midpoint of }AD.$

$\text{a) Find the vector }\overrightarrow{MN}\text{ in terms of }\vec a\text{ and }\vec b$
. . $\text{-- which I found to be }3\vec a-3\vec b$ . Yes!

$X\text{ is the midpoint of }MN\text{ and }Y\text{ is the midpoint of }CD.$

$\text{b) Prove that }XY \parallel AD$
Code:
            : - -  3a - - :
M
B o * * o * * o C
*       *       *
*         *         *   Y
3b *         X o * * * * * o
*             *             *
*               *               *
A o  *  *  *  *  *  o  *  *  *  *  *  o D
N
: - - - - - - - -9a - - - - - - - - :

$(a)\;\overrightarrow{MN} \:=\:\overrightarrow{MB} + \overrightarrow{BA} + \overrightarrow{AN}$

. . . . . . . $=\;\text{-}\tfrac{1}{2}\overrightarrow{BC} - \overrightarrow{AB} + \tfrac{1}{2}\overrightarrow{AD}$

. . . . . . . $=\;\text{-}\tfrac{1}{2}(3\vec a) - (3\vec b) + \tfrac{1}{2}(9\vec a)$

. . . . . . . $=\;\text{-}\tfrac{3}{2}\vec a - 3\vec b + \tfrac{9}{2}\vec a$

. . . . . . . $=\;3\vec a - 3\vec b$

$(b)\;\overrightarrow{CD} \;=\;\overrightarrow{CB} + \overrightarrow{BA} + \overrightarrow{AD}$

. . . . . . $=\;-3\vec a - 3\vec b + 9\vec a$

. . . . . . $=\;6\vec a - 3\vec b$

$\overrightarrow{XY} \;=\;\overrightarrow{XM} + \overrightarrow{MC} + \overrightarrow{CY}$

. . . . $=\;\text{-}\tfrac{1}{2}\overrightarrow{MN} + \tfrac{1}{2}\overrightarrow{BC} + \tfrac{1}{2}\overrightarrow{CD}$

. . . . $=\;\text{-}\tfrac{1}{2}(3\vec a - 3\vec b) + \tfrac{1}{2}(3\vec a) + \tfrac{1}{2}(6\vec a - 3\vec b)$

. . . . $=\;\text{-}\tfrac{3}{2}\vec a + \tfrac{3}{2}\vec b + \tfrac{3}{2}\vec a + 3\vec a - \tfrac{3}{2}\vec b$

. . . . $=\; 3\vec a$

$\text{We have: }\:\overrightarrow{XY} \:=\:3\vec a \:=\:\tfrac{1}{3}\overrightarrow{AD}$

$\text{Therefore: }\:\overrightarrow{XY} \parallel \overrightarrow{AD}$