About of a Regular Tetrahedron

Question

Among the Properties of a regular tetrahedron one is "The perpendicular from vertices to the opposite faces meet the faces at their centroid". How can I prove this by vector medhod?

Answer 1

Let $ABCD$ be our tetrahedron and $M$ be the centroid of $\Delta ABC$.

Thus, $$\vec{DM}=\frac{1}{3}\left(\vec{DA}+\vec{DB}+\vec{DC}\right)$$ and $$\vec{DM}\cdot\vec{AC}=\frac{1}{3}\left(\vec{DA}+\vec{DB}+\vec{DC}\right)\left(\vec{DC}-\vec{DA}\right)=$$ $$=\frac{1}{3}\left(DC^2-DA^2+\vec{DB}\cdot\vec{DC}-\vec{DB}\cdot\vec{DA}\right)=$$ $$=\frac{1}{3}\left(DB\cdot DC\cos60^{\circ}-DB\cdot DA\cos60^{\circ}\right)=0,$$ which says $$DM\perp AC.$$ By the same way we can prove that $$DM\perp AB$$ and since $\vec{AB}$ and $\vec{AC}$ they are the base of the plane $ABC$, we obtain $$DM\perp(ABC).$$ But there is an unique perpendicular from $D$ to the plane $ABC$ and we are done.

Answer 2

Here is a method which chooses a co-ordinate frame to put the tetrahedron in a standard form, and allow some simple calculations. This may not be quite permitted under your "vector method" requirement, but it does make things a deal easier.

Let the vertices of the tetrahedron be $\vec a=(1,1,1); \vec b=(1,-1,-1); \vec c=(-1,1,-1); \vec d=(-1,-1,1)$

The centroid of the last three points is $\frac 13(\vec b+\vec c+\vec d)=\frac 13(-1,-1,-1)$. The line joining this centre to the other vertex consists of the points $t \vec a$.

Now compute the scalar product of $t\vec a$ with two linearly independent vectors in the plane of the relevant face (eg $\vec b - \vec c, \vec b-\vec d$).

Alternatively, a vector perpendicular to the plane of the final three points is $(\vec b-\vec c)\times (\vec b-\vec d)$. Show that this is parallel to $\vec a$ and that the vector in direction $\vec a$ through the point with position vector $\vec a$ passes through the centroid.

Answer 3

For a pure vector method, let the vertices be at positions $0, \vec a, \vec b, \vec c$ so that the centroid of the final three is at $\frac 13 (\vec a+\vec b +\vec c)$.

If the side length is $d$ we have $$d^2=\vec a\cdot \vec a=\vec b\cdot \vec b=\vec c\cdot \vec c=(\vec a-\vec b)\cdot (\vec a-\vec b)=(\vec a-\vec c)\cdot (\vec a-\vec c)=(\vec c-\vec b)\cdot (\vec c-\vec b)$$From the last three we obtain $$\vec a\cdot \vec b =\vec a\cdot \vec c =\vec c\cdot \vec b =\frac 12d^2$$

Then we have $$\frac 13 (\vec a+\vec b +\vec c)\cdot(\vec a-\vec b)=\frac 13d^2(1-\frac 12+\frac 12-1+\frac 12-\frac 12)=0=\frac 13 (\vec a+\vec b +\vec c)\cdot(\vec a-\vec c)$$

Then we invoke symmetry to say this is true for each of the centroids of the sides.

Answer 4

Set the basis of the tetrahedron (an equilateral triangle) with the vertices at coordinates

$$v_k=\left(\cos\frac{2k\pi}3,\sin\frac{2k\pi}3,0\right)$$ for $k=0,1,2$. It is an easy matter to check that the origin is the centroid of the face.

Now it suffices to show that the remaining vertex lies on the $z$ axis, $v_3=(0,0,z)$.

Indeed,

$$\|\vec{v_kv_3}\|^2=\cos^2\frac{2k\pi}3+\sin^2\frac{2k\pi}3+z^2=1+z^2,$$

is independent of $k$. (And you can choose $z$ such that $\|\vec{v_kv_3}\|^2=\|\vec{v_0v_1}\|^2$). With $v_3$ off-axis, the $\|\vec{v_kv_3}\|$ would differ.

Answer 5

Let the unit vectors $a$,$b$, $c$ be the three edges emanating from the vertex $O$. Dropping the perpendicular from $C$ to the plane spanned by $a$ and $b$ requires solving $$(H=)\qquad c+\lambda(a\times b)=\mu a+\nu b$$ for $\lambda$, $\mu$, $\nu$. Taking the scalar product with $a$ and $b$ gives the equations $$c\cdot a=\mu+\nu\>b\cdot a,\qquad c\cdot b=\mu\>a\cdot b+\nu\ .$$ Since $a\cdot b=b\cdot c=c\cdot a={1\over2}$ this amounts to $\mu+{1\over2}\nu={1\over2}$,$\>{1\over2}\mu+\nu={1\over2}$ with the solution $\mu=\nu={1\over3}$. It follows that $$H={a+b+0\over3}\ .$$