1. ## Vectors

I got AY = -3a + 1/2 b for the first one. The other two I can't work out.

2. Oops I forgot part (iii)!

3. Originally Posted by GAdams
I got AY = -3a + 1/2 b for the first one. The other two I can't work out.
why did you get 1/2 b? shouldn't it be 3/2 b?

4. Originally Posted by Jhevon
why did you get 1/2 b? shouldn't it be 3/2 b?
Sorry, I meant to type 1 1/2b , i.e. 1.5b

5. i) $\displaystyle \overrightarrow{AY}=\overrightarrow{AO}+\overright arrow{OY}=-\overrightarrow{OA}+\frac{1}{2}\overrightarrow{OB} =-3\mathbf{a}+\frac{3}{2}\mathbf{b}$.
ii) $\displaystyle \overrightarrow{OX}=\frac{1}{2}(\overrightarrow{OA }+\overrightarrow{OB})=\frac{3}{2}\mathbf{a}+\frac {3}{2}\mathbf{b}$
iii) $\displaystyle \overrightarrow{AZ}=\overrightarrow{AO}+\overright arrow{OZ}=-\overrightarrow{OA}+\frac{2}{3}\overrightarrow{OX} =-2\mathbf{a}+\mathbf{b}$.
If $\displaystyle A,Z,Y$ are colinear, then $\displaystyle Z$ is baricentre, so $\displaystyle \frac{AZ}{ZY}=2$

6. Originally Posted by GAdams
I got AY = -3a + 1/2 b for the first one. The other two I can't work out.
Hello,

as Jhevon pointed out:

$\displaystyle \overrightarrow{AY} = -3 \vec{a} + \frac{3}{2} \vec{b}$

(ii):

1. $\displaystyle \overrightarrow{AB} = -3 \vec{a} + 3 \vec{b}$

2. $\displaystyle \overrightarrow{OX} = 3 \vec{a} + \frac{1}{2} \overrightarrow{AB}$

3. $\displaystyle \overrightarrow{OX} = 3 \vec{a} + \frac{1}{2} (-3 \vec{a} + 3 \vec{b}) = 3 \vec{a} - \frac{3}{2} \vec{a} + \frac{3}{2} \vec{b}) = \frac{3 \vec{a} + 3 \vec{b}}{2}$

As you can see $\displaystyle \overrightarrow{OX}$ is the mean (or average) of A and B.

(iii)

$\displaystyle \overrightarrow{AZ} = -3\vec{a} + \frac{2}{3} \cdot \overrightarrow{OX}$

$\displaystyle \overrightarrow{AZ} = -3\vec{a} + \frac{2}{3} \cdot \frac{3 \vec{a} + 3 \vec{b}}{2} = -2\vec{a} + \vec{b}$

7. Originally Posted by earboth
Hello,

as Jhevon pointed out:

$\displaystyle \overrightarrow{AY} = -3 \vec{a} + \frac{3}{2} \vec{b}$

(ii):

1. $\displaystyle \overrightarrow{AB} = -3 \vec{a} + 3 \vec{b}$

2. $\displaystyle \overrightarrow{OX} = 3 \vec{a} + \frac{1}{2} \overrightarrow{AB}$

3. $\displaystyle \overrightarrow{OX} = 3 \vec{a} + \frac{1}{2} (-3 \vec{a} + 3 \vec{b}) = 3 \vec{a} - \frac{3}{2} \vec{a} + \frac{3}{2} \vec{b}) = \frac{3 \vec{a} + 3 \vec{b}}{2}$

As you can see $\displaystyle \overrightarrow{OX}$ is the mean (or average) of A and B.

(iii)

$\displaystyle \overrightarrow{AZ} = -3\vec{a} + \frac{2}{3} \cdot \overrightarrow{OX}$

$\displaystyle \overrightarrow{AZ} = -3\vec{a} + \frac{2}{3} \cdot \frac{3 \vec{a} + 3 \vec{b}}{2} = -2\vec{a} + \vec{b}$
Thanks, I got it except the last bit, why is it 2/3 * OX?

8. Originally Posted by red_dog
i) $\displaystyle \overrightarrow{AY}=\overrightarrow{AO}+\overright arrow{OY}=-\overrightarrow{OA}+\frac{1}{2}\overrightarrow{OB} =-3\mathbf{a}+\frac{3}{2}\mathbf{b}$.
ii) $\displaystyle \overrightarrow{OX}=\frac{1}{2}(\overrightarrow{OA }+\overrightarrow{OB})=\frac{3}{2}\mathbf{a}+\frac {3}{2}\mathbf{b}$
iii) $\displaystyle \overrightarrow{AZ}=\overrightarrow{AO}+\overright arrow{OZ}=-\overrightarrow{OA}+\frac{2}{3}\overrightarrow{OX} =-2\mathbf{a}+\mathbf{b}$.
If $\displaystyle A,Z,Y$ are colinear, then $\displaystyle Z$ is baricentre, so $\displaystyle \frac{AZ}{ZY}=2$

I don't quite get the last part about colinear and baricentre

9. If $\displaystyle X$ is the midpoint of $\displaystyle AB$ and $\displaystyle Y$ is the midpoint of $\displaystyle OB$, then $\displaystyle OX$ and $\displaystyle AY$ are medians in the triangle $\displaystyle OAB$ and $\displaystyle Z=OX\cap AY$. The intersection point of medians is the baricenter which splits each median in two segments in ratio 2:1.