A regular tetrahedron with edge a lies in the first octant…

A regular tetrahedron with edge $a$ lies in the first octant with one vertex at the origin $O$ and the edge $OA$ along $Ox$. $1^\circ$ Find the unit vectors along the edges $OA$, $OB$, $OC$. $2^\circ$ Forces of $1$, $2$ and $3\ \mathrm{kg}$ act along these edges. Find the magnitude of the resultant and the cosines of the angles it makes with each force.

Step 1: Vertices

Put $B$ in the plane $xOy$ ($y\gt0$) and $C$ above it ($z\gt0$). All edges have length $a$: $A=(a,0,0),\qquad B=\left(\tfrac a2,\tfrac{a\sqrt3}{2},0\right)$.

For $C=(x,y,z)$, use the three distance conditions one at a time.

• $|OC|=|AC|$ gives $x=\tfrac a2$.
• $|OC|=|BC|$ gives $\tfrac a2x+\tfrac{a\sqrt3}{2}y=\tfrac{a^2}{2}$, so $y=\tfrac{a}{2\sqrt3}=\tfrac{a\sqrt3}{6}$.
• $|OC|^2=a^2$ gives $z^2=a^2-\tfrac{a^2}{4}-\tfrac{a^2}{12}=\tfrac{2a^2}{3}$.

\[ C=\left(\tfrac a2,\ \tfrac{a\sqrt3}{6},\ \tfrac{a\sqrt6}{3}\right). \]

This is the centroid of $OAB$ raised by the height $a\sqrt{2/3}$ of the tetrahedron.

Step 2: $1^\circ$ Unit vectors

\[\vec e_{OA}=(1,0,0),\qquad \vec e_{OB}=\left(\tfrac12,\tfrac{\sqrt3}{2},0\right),\qquad \vec e_{OC}=\left(\tfrac12,\tfrac{\sqrt3}{6},\tfrac{\sqrt6}{3}\right)\]

Check for $\vec e_{OC}$: $\tfrac14+\tfrac1{12}+\tfrac23=1$ $\checkmark$. The unit vectors do not depend on $a$. Each pair has scalar product $\tfrac12$, i.e. makes $60^\circ$.

Step 3: $2^\circ$ Resultant

\[ \vec F_1=(1,0,0),\qquad \vec F_2=2\vec e_{OB}=(1,\sqrt3,0),\qquad \vec F_3=3\vec e_{OC}=\left(\tfrac32,\tfrac{\sqrt3}{2},\sqrt6\right), \]

\[ \vec R=\left(\tfrac72,\ \tfrac{3\sqrt3}{2},\ \sqrt6\right),\qquad |\vec R|^2=\tfrac{49}{4}+\tfrac{27}{4}+6=25. \]

\[|\vec R|=5\ \mathrm{kg}\]

Check without coordinates:

\[ |\vec R|^2=1+4+9+2\cdot\tfrac12(1\cdot2+1\cdot3+2\cdot3)=14+11=25.\ \checkmark \]

Step 4: Angles with the forces

\[ \vec R\cdot\vec F_1=\tfrac72,\qquad \vec R\cdot\vec F_2=\tfrac72+\tfrac92=8,\qquad \vec R\cdot\vec F_3=\tfrac{21}4+\tfrac94+6=\tfrac{27}{2}. \]

\[ \cos\angle(\vec R,\vec F_1)=\frac{7/2}{5\cdot1},\qquad \cos\angle(\vec R,\vec F_2)=\frac{8}{5\cdot2},\qquad \cos\angle(\vec R,\vec F_3)=\frac{27/2}{5\cdot3}. \]

\[\cos\angle(\vec R,\vec F_1)=\tfrac{7}{10},\quad \cos\angle(\vec R,\vec F_2)=\tfrac45,\quad \cos\angle(\vec R,\vec F_3)=\tfrac{9}{10}\]

The angles are $\approx45.6^\circ$, $36.9^\circ$ and $25.8^\circ$. Here "kg" means kilogram-force (kgf), so $|\vec R|=5\ \mathrm{kgf}\approx49\ \mathrm N$.