1. ## veectors

I have a question on vectors which probably doesn't belong here but I can't see an appropriate place to post it. If anyone has a better idea please tell me.

In the diagram, OABCDEFG is a cuboid in which the length of OA is 4 units, AB is 3 units and BF is 2 units. Unit vectors i, j, and k are parallel to OA, OC and OD respectively. The midpoints of AB and FG are M and N respectively.

a) Express each of the vectors ON and MG in terms of i, j, and k.
I would start this problem by finding ON→=OD→+DG→+1/2GN→
I have place the arrow to the right of the signs as I can't put them above it.
ON→=1/2i+j+k
MG=1/2AB+BF+FG
-i+1/2j+k
Am I doing the right thing anyone?

b) Find the acute angle between the direction of ON and MG, correct to the nearest 0.1°.
Will tackle this next.

2. Hello, stealthmaths!

In the diagram, $\displaystyle OABCD{E}FG$ is a cuboid
. . in which: .$\displaystyle |\overrightarrow{OA}| = 4,\; |\overrightarrow{AB}| = 3.\;|\overrightarrow{BF}| = 2$ units.

Unit vectors $\displaystyle \vec i, \vec j, \vec k$ are parallel to $\displaystyle \overrightarrow{OA}, \,\overrightarrow{OC}, \,\overrightarrow{OD}$, respectively.

The midpoints of $\displaystyle AB$ and $\displaystyle FG$ are $\displaystyle M$ and $\displaystyle N$, respectively.

$\displaystyle \text{a) Express each of the vectors }\overrightarrow{ON}$ and $\displaystyle \overrightarrow{MG}\text{ in terms of }\vec i,\, \vec j,\, \vec k.$

We see that: . $\displaystyle \begin{array}{ccccc} \overrightarrow{OA} \:=\:\overrightarrow{CB} \:=\:\overrightarrow{DE} \:=\:\overrightarrow{GF} &=& 4i \\ \overrightarrow{OC} \:=\:\overrightarrow{AB} \:=\:\overrightarrow{DG} \:=\:\overrightarrow{EF} &=& 3j \\ \overrightarrow{OD} \:=\:\overrightarrow{AE} \:=\:\overrightarrow{BF} \:=\:\overrightarrow{CG} &=& 2k \end{array}$

$\displaystyle \overrightarrow{ON} \;=\;\overrightarrow{OC} + \overrightarrow{CG} + \tfrac{1}{2}\overrightarrow{GF} \;=\;3\vec j + 4\vec k + \tfrac{1}{2}(4\vec i) \;=\;2\vec i + 3\vec j + 4\vec j$

$\displaystyle \overrightarrow{MG} \;=\;\overrightarrow{MB} + \overrightarrow{BC} + \overrightarrow{CG} \;=\;\tfrac{1}{2}(3\vec j) - 4\vec i + 2\vec k \;=\;-4\vec i + \tfrac{3}{2}\vec j + 2\vec k$

$\displaystyle \text{b) Find the acute angle between }\overrightarrow{ON} \text{ and }\overrightarrow{MG}\text{, to the nearest 0.1}^o.$
Angle $\displaystyle \theta$ between $\displaystyle \vec u$ and $\displaystyle \vec v$ is given by: . $\displaystyle \cos\theta \;=\;\frac{\vec u\cdot\vec v}{|\vec u||\vec v|}$

We have: .$\displaystyle \overrightarrow{ON} \:=\:\langle 2,3,4\rangle,\;\;\overrightarrow{MG} \:=\:\langle -4,\tfrac{3}{2},2\rangle$

. . $\displaystyle \overrightarrow{ON}\cdot \overrightarrow{MG} \:=\:\langle2,3,4\rangle \cdot\langle -4,\tfrac{3}{2},2\rangle \:=\:-8 + \tfrac{9}{2} + 8 \:=\:\tfrac{9}{2}$

. . $\displaystyle |\overrightarrow{ON}| \:=\:\sqrt{4+9+16} \:=\:\sqrt{29}$

. . $\displaystyle |\overrightarrow{MG}| \;=\;\sqrt{16 + \tfrac{9}{4} + 4} \;=\;\sqrt{\tfrac{89}{4}} \;=\;\frac{\sqrt{89}}{2}$

Hence: .$\displaystyle \cos\theta \;=\;\frac{\frac{9}{2}}{\sqrt{29}\,\frac{\sqrt{89} }{2}} \;=\;0.177152998$

Therefore: .$\displaystyle \theta \;=\;79.79602628 \;\approx\;79.8^o$

3. Hi Soroban:

Soroban:
On the first line of question a)
ON=OC+CG+1/2CF=3j+4k+1/2(4i)
Should it not be: 3j+2k+1/2(4i)?

4. I have calculated with the correction to get the following:

ON=(2, 3, 2)
MG=(-4, 3/2, 2)

ON.MG=(2, 3, 2)(-4, 3/2, 2)
=-8+9/2+4=1/2

|ON|=√4+9+4=√17
|MG|=√16+9/4+4=√89/2

hence: cosθ=1/2 / (√17.√89/2)=0.02570872484
=88.5° (1d.p.)

Solution found anyone?