Hey

OABCDEFG is a cuboid

-> -> ->

OA= 3j OC=2k OD=i

M is such that ->

OM= 1/3 ->

OE

N is the midpoint of BF. Find

a) ->

MN

not sure how to start, i drew a cuboid but dont know how to find the 1/3 of OE

thanks

Thanks

Printable View

- Sep 16th 2009, 02:16 AMsmmmcvector problem
Hey

OABCDEFG is a cuboid

-> -> ->

OA= 3j OC=2k OD=i

M is such that ->

OM= 1/3 ->

OE

N is the midpoint of BF. Find

a) ->

MN

not sure how to start, i drew a cuboid but dont know how to find the 1/3 of OE

thanks

Thanks - Sep 16th 2009, 04:52 AMGrandad
Hello smmmcI'm assuming that the vertices are labelled as in the attached diagram, so that, for instance, BF is parallel to CG, OD and AE.

If this is the case, then $\displaystyle \vec{BF}=\vec{OD} = \vec{i}$ and $\displaystyle \vec{CB}=\vec{OA}=3\vec{j}$

$\displaystyle \Rightarrow \vec{BN}=\tfrac12\vec{BF}=\tfrac12\vec{i}$

$\displaystyle \Rightarrow \vec{ON} = \vec{OC}+\vec{CB}+\vec{BN}$

$\displaystyle = 2\vec{k}+3\vec{j}+\tfrac12\vec{i}$

Also $\displaystyle \vec{OE} = \vec{OD}+\vec{DE} = \vec{i}+3\vec{j}$

$\displaystyle \Rightarrow \vec{OM} = \tfrac13\vec{OE}=\tfrac13\vec{i} + \vec{j}$

Now $\displaystyle \vec{ON}=\vec{OM}+\vec{MN}$

$\displaystyle \Rightarrow \vec{MN} = \vec{ON}-\vec{OM}$

$\displaystyle = 2\vec{k}+3\vec{j}+\tfrac12\vec{i}-\tfrac13\vec{i} - \vec{j}$

$\displaystyle =\tfrac16\vec{i}+2\vec{j}+2\vec{k}$

Grandad