Part 1
According to the Rectangular solid in the following image, express the vectors $\overrightarrow {AF}$, $\overrightarrow {GD}$, $\overrightarrow {FC}$, $\overrightarrow {EC}$ in terms of $\vec a$, $\vec b$ and $\vec c$. M1 and M2 are the midpoints of their respective faces.
$\overrightarrow {AF}=$$\vec a$ + $\vec c$
$\overrightarrow {GD}=$-$\vec a$ - $\vec c$
$\overrightarrow {FC}=$$\vec b$ - $\vec c$
$\overrightarrow {EC}=$$\vec a$ + $\vec b$ - $\vec c$
Did I express it alright? This is my first exercise and I did not really understand the question when they said EXPRESS. Or should it be expressed in a different form?
Find $\overrightarrow {AM1}$ and $\overrightarrow {AM2}$.
$\overrightarrow {AM1}$ = $\overrightarrow {AB}$ + $\overrightarrow {BM1}$ = $\vec a$ + $\frac{1}{2}$$\vec b$ + $\frac{1}{2}$$\vec c$
$\overrightarrow {AM2}$= $\overrightarrow {AE}$ + $\overrightarrow {EM2}$ = $\vec c$ + $\frac{1}{2}$$\vec b$ + $\frac{1}{2}$$\vec a$
Again I am not sure about my expression. And is it even correct or did I completely misunderstand the methodology?
Form the scalar product $\overrightarrow {AM1}$ ·$\overrightarrow {AM2}$
$\overrightarrow {AM1}$ ·$\overrightarrow {AM2}$=$\frac{1}{2}$($\vec {|a|^2}$ + $\frac{1}{2}$$\vec {|b|^2}$ + $\vec {|c|^2}$)
Part 2
Using the same image, but let |$\vec a$| = |$\vec b$| = |$\vec c$| = 1. Now the rectangular solid is a cube. Find |$\overrightarrow {AM1}$| and |$\overrightarrow {AM2}$| and then find the angle of it.
|$\overrightarrow {AM1}$| = 1+0.5+0.5= 2
|$\overrightarrow {AM2}$| = 1+0.5+0.5= 2
$\overrightarrow {AM1}$ ·$\overrightarrow {AM2}$= |$\overrightarrow {AM1}$||$\overrightarrow {AM2}$| $cos \varphi$
$\varphi$=71.8
Find the area of the triangle $\overrightarrow {AM1M2}$.
Area = $\frac{1}{2}$|$\overrightarrow {AM2}$||$\overrightarrow {M1M2}$| $sin \varphi$
Now here is where I am very confused because I get $0$ for |$\overrightarrow {M1M2}$|. Is that right? But the Area can't be $0$ though...
If my answers are wrong I would appreciate it if you'd guide me through my misunderstanding. I missed almost all the classes because I got very sick but I will catch up with some OT
$AM_1 = a + \frac 12 b + \frac 12 c\\ AM_2 = \frac 12a + \frac 12 b + c\\ AM_1\cdot AM_2 = \frac 12 |a|^2 + \frac 14 |b|^2 + \frac 12 |c|$
If $|a| = |b| = |c| = 1$
$|AM_1|^2 = AM_1\cdot AM_1 = 1 + \frac 14 + \frac 14\\ |AM_1| = \sqrt {\frac {3}{2}}\\ |AM_2| = \sqrt {\frac {3}{2}}\\ \cos \phi = \frac{AM_1\cdot AM_2}{|AM_1||AM_2|} = \frac {\frac {5}{4}}{\frac 32} = \frac 56$
For the area you can say $\frac 12 |AM_1\times AM_2|$
or you can say $\frac 12 |AM_1||AM_2| \sin \phi$
or you can use the Pythagorean theorem to say $\sin^2\phi + \cos^2 \phi = 1$ to find $\sin\phi$
One last note:
${M_1M_2} = \frac 12 a + \frac 12 b + c - (a +\frac 12 b + \frac 12 c) = -\frac 12 a + \frac 12 c$